A-Level Physics (9702) mencakup mulai dari pengukuran dan gerak hingga fisika kuantum. Lonjakan terbesar dari IGCSE adalah aljabar: Anda harus menyusun ulang dan menggabungkan persamaan, bukan hanya menghafalnya.
Tahun AS: mekanika, gelombang, listrik, dan awal fisika partikel. Tahun A2: gerak melingkar, gravitasi, listrik dan medan magnet, osilasi, termodinamika, dan fisika nuklir.
Baca satu topik, lalu kerjakan soal-soal tentangnya segera. Fisika tidak melekat hanya dengan membaca. Ketika jawaban salah, tanyakan pada diri sendiri: apakah itu fisika atau aljabar? Keduanya memerlukan perbaikan yang berbeda.
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A physical quantity 物理量 has two parts: a number (its magnitude 大小) and a unit 单位. The number on its own tells you nothing. You must also say what is measured and in which unit.
Example: "the length is 1.5" is not complete. "The length is 1.5 m" is a physical quantity.
You should be able to estimate 估算 the size of the physical quantities in this syllabus. Paper 1 usually opens with a question like this. Learn these rough values:
A good estimate has the right order of magnitude 数量级 (the right power of ten). For a human, 70 kg is a good guess; 7 kg is not.
To estimate a quantity that is not in the list, build it from ones that are. The kinetic energy of a moving car is $\tfrac{1}{2}mv^{2} \approx \tfrac{1}{2} \times 1000 \times 30^{2} \approx 5 \times 10^{5}\ \text{J}$. The pressure under a standing person is $700\ \text{N} / 0.02\ \text{m}^{2} \approx 4 \times 10^{4}\ \text{Pa}$. Always check that the power of ten looks sensible before you move on.
| English | Bahasa Indonesia |
|---|---|
| physical quantity/ˈfɪzɪkl ˈkwɒntɪti/ | besaran fisis |
| magnitude/ˈmæɡnɪtjuːd/ | besarnya |
| unit/ˈjuːnɪt/ | satuan |
| estimate/ˈestɪmət/ | estimasi |
| mass/mæs/ | massa |
| weight/weɪt/ | berat |
| vacuum/ˈvækjuːm/ | vakum |
| acceleration/əkˌseləˈreɪʃn/ | percepatan |
| free fall/friː fɔːl/ | jatuh bebas |
| density/ˈdensɪti/ | densitas |
| atmospheric pressure/ˌætməsˈferɪk ˈpreʃə/ | tekanan atmosfer |
| order of magnitude/ˈɔːdə ɒv ˈmæɡnɪtjuːd/ | orde besaran |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The SI system has seven base quantities 基本量; five of them are used in this topic. You must know these five and their units:
Every other unit in this syllabus is built from these five.
A derived unit 导出单位 is made by multiplying or dividing base units. You should be able to write any quantity in this syllabus in base units.
Build a derived unit from the equation that defines it:
When a question asks for the SI base units of a quantity, replace each named unit with its base units, then simplify. Example: the SI base units of the watt are $\text{kg m}^{2}\ \text{s}^{-3}$.
An equation is homogeneous 量纲一致 when both sides have the same base units. In plain words: the units on both sides match.
Write each side in base units and compare. Take the equation $v^{2} = u^{2} + 2as$:
Both sides give $\text{m}^{2}\ \text{s}^{-2}$, so the units match.
Be careful: matching units do not prove the whole equation is correct. It could still have a wrong number, or a missing factor of 2. But if the units do not match, the equation is wrong for sure.
Worked example. The drag force on a falling ball is given by $F = kv^{2}$, where $v$ is the speed. Find the SI base units of the constant $k$.
Rearrange: $k = F / v^{2}$. In base units, $F$ is $\text{kg m s}^{-2}$ and $v^{2}$ is $\text{m}^{2}\ \text{s}^{-2}$, so
A multiple-choice question often asks "which equation could be correct?". Check the base units of each option; only a homogeneous equation can be correct. A pure number (like 2, $\pi$ or $\tfrac{1}{2}$) has no unit, so it never changes the check.
A prefix 词头 is a letter put in front of a unit to make it bigger or smaller by powers of ten. You must know these:
| Prefix | Symbol | Factor |
|---|---|---|
| tera | T | $10^{12}$ |
| giga | G | $10^{9}$ |
| mega | M | $10^{6}$ |
| kilo | k | $10^{3}$ |
| deci | d | $10^{-1}$ |
| centi | c | $10^{-2}$ |
| milli | m | $10^{-3}$ |
| micro | $\mu$ | $10^{-6}$ |
| nano | n | $10^{-9}$ |
| pico | p | $10^{-12}$ |
To change a prefixed unit into base units, replace the prefix with its factor, then simplify. Example: change $0.25\ \text{kN mm}^{-2}$ into $\text{N m}^{-2}$:
Take special care with squared units like $\text{mm}^{2}$: you must square the factor too.
Only seven quantities are base quantities. Everything else is built from them, and its unit can be written in base units. · Hanya tujuh besaran yang merupakan besaran dasar. Segala sesuatu lainnya dibangun darinya, dan satuannya dapat ditulis dalam satuan dasar.
| English | Bahasa Indonesia |
|---|---|
| length/leŋθ/ | panjang |
| speed/spiːd/ | kecepatan (besaran skalar) |
| pressure/ˈpreʃə/ | tekanan |
| temperature/ˈtemprɪtʃə/ | suhu |
| power/ˈpaʊə/ | kuasa |
| energy/ˈenədʒi/ | tenaga |
| SI units/ˌes ˈaɪ ˈjuːnɪts/ | satuan SI |
| base quantity/beɪs ˈkwɒntɪti/ | kuantitas dasar |
| current/ˈkʌrənt/ | arus |
| ampere/ˈæmpeə/ | ampere |
| kelvin/ˈkelvɪn/ | kelvin |
| derived unit/dɪˈraɪvd ˈjuːnɪt/ | satuan turunan |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
| force/fɔːs/ | gaya |
| newton/ˈnjuːtn/ | newton |
| work/wɜːk/ | usaha |
| joule/dʒuːl/ | joule |
| watt/wɒt/ | watt |
| stress/stres/ | tegangan |
| pascal/ˈpæskl/ | pascal |
| homogeneous/ˌhɒməˈdʒiːniəs/ | homogen |
| prefix/ˈpriːfɪks/ | awalan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Every measurement 测量 has some uncertainty 不确定度 — we are never fully sure of the value. A good experimenter knows where the uncertainty comes from, makes a fair estimate of it, and carries it through to the final answer.
A systematic error 系统误差 changes every reading by the same amount, in the same direction. You cannot find it by repeating the measurement. Common causes:
A systematic error makes the accuracy 准确度 worse, but it does not change the precision 精密度.
A random error 随机误差 makes readings jump above and below the true value, with no pattern. Causes include how carefully you read the scale, changing conditions, and the smallest step the instrument 仪器 can show. If you repeat the measurement many times and take the mean 平均值 (the average), random errors partly cancel out.
A random error makes the precision worse. But with enough repeats, the mean can still be accurate.
On a graph the two errors look different. A systematic error moves every point by the same amount, so the line of best fit keeps its gradient but no longer passes through the origin: the intercept 截距 changes. Random errors scatter the points above and below the line; a best-fit line drawn through the middle of the scatter still gives a reliable gradient.
| Error | How to reduce it |
|---|---|
| systematic | check for a zero error and subtract it; calibrate the instrument against a known standard; read the scale from directly in front |
| random | repeat the reading and take the mean; use an instrument with smaller divisions; time many oscillations instead of one |
Precision is how close repeated readings are to each other. Precise readings are grouped very close together.
Accuracy is how close a reading (or the mean of several readings) is to the true value.
A set of readings can be:
When a question gives a table of repeated readings, look at the spread (precision) and the mean (accuracy) separately.
Before you combine uncertainties, you need a sensible uncertainty for each raw reading:
Be honest. If the edge of a shadow is hard to see, the uncertainty in its position is several millimetres, however fine the scale. Examiners do not accept "the smallest division" as the uncertainty of a difficult reading.
A measurement is often written as $x \pm \Delta x$. Here $\Delta x$ is the absolute uncertainty 绝对不确定度. The percentage uncertainty 百分比不确定度 is
A derived quantity 导出量 is one you calculate from measured values. Its uncertainty is found by simple rules:
Worked example. A ball's diameter 直径 is measured as $d = (5.26 \pm 0.02)\ \text{cm}$. The volume 体积 of a sphere is $V = \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi (d/2)^{3}$, so $V \propto d^{3}$ ($V$ depends on $d$ cubed).
The percentage uncertainty in $d$ is
Because $V \propto d^{3}$, the percentage uncertainty in $V$ is three times this, about $1.14\%$. The volume is $\tfrac{4}{3}\pi(2.63)^{3} \approx 76.2\ \text{cm}^{3}$. So the absolute uncertainty is $0.0114 \times 76.2 \approx 0.87\ \text{cm}^{3}$. The final answer is $V = (76.2 \pm 0.9)\ \text{cm}^{3}$.
Do two values agree? Practical questions often ask whether two calculated values of a constant support a suggested relationship. Do not just say "they are close". Work out the percentage difference 百分比差异 between them,
and compare it with the percentage uncertainty in $k$ (or with the criterion the question gives, often 10%). If the difference is smaller, the two values agree within the uncertainty and the relationship is supported. If it is larger, they do not.
When you write a calculated quantity, give it the same number of significant figures 有效数字 as the least precise measurement you used — usually two or three in this syllabus. Too many significant figures makes the answer look more exact than it really is. Too few loses useful information.
Practical papers ask you to justify the number you chose. Name the raw readings: "$a$ is given to 2 significant figures because $L$ and $t$ were each measured to 2 significant figures." The phrase "because of the raw data" on its own does not earn the mark. In a table of results, judge each row from the least precise raw reading in that row; do not force every row to the same number of figures.
Paper 5 asks you to carry uncertainties through a graph:
A percentage uncertainty in the gradient then flows into any quantity you calculate from it, using the rules above.
A systematic error shifts every reading the same way and survives repetition; a random error scatters the readings and shrinks when you average. · Kesalahan sistematis menggeser setiap pembacaan dengan jumlah yang sama dan tetap ada meskipun diulang; kesalahan acak menyebarkan pembacaan dan mengecil saat dirata-ratakan.
| English | Bahasa Indonesia |
|---|---|
| diameter/daɪˈæmɪtə/ | diameter |
| uncertainty/ʌnˈsɜːtənti/ | ketidakpastian |
| measurement/ˈmeʒəmənt/ | pengukuran |
| random error/ˈrændəm ˈerə/ | kesalahan acak |
| systematic error/ˌsɪstəˈmætɪk ˈerə/ | kesalahan sistematis |
| zero error/ˈzɪərəʊ ˈerə/ | kesalahan nol |
| calibration/ˌkælɪˈbreɪʃn/ | kalibrasi |
| parallax/ˈpærəlæks/ | paralaks |
| accuracy/ˈækjʊrəsi/ | akurasi |
| precision/prɪˈsɪʒn/ | presisi |
| instrument/ˈɪnstrəmənt/ | alat ukur |
| mean/miːn/ | rata-rata |
| line of best fit/laɪn ɒv best fɪt/ | garis tren terbaik |
| intercept/ˌɪntəˈsept/ | intersep |
| division/dɪˈvɪʒn/ | pembagian |
| reaction time/rɪˈækʃn taɪm/ | waktu reaksi |
| half the range/hɑːf ðə reɪndʒ/ | setengah dari jangkauan |
| derived quantity/dɪˈraɪvd ˈkwɒntɪti/ | kuantitas turunan |
| absolute uncertainty/ˈæbsəluːt ʌnˈsɜːtənti/ | ketidakpastian absolut |
| percentage uncertainty/pəˈsentɪdʒ ʌnˈsɜːtənti/ | ketidakpastian persentase |
| volume/ˈvɒljuːm/ | volume |
| percentage difference/pəˈsentɪdʒ ˈdɪfrəns/ | selisih persentase |
| significant figures/sɪɡˈnɪfɪkənt ˈfɪɡəz/ | angka bererti |
| same number of significant figures/seɪm ˈnʌmbə ɒv sɪɡˈnɪfɪkənt ˈfɪɡəz/ | jumlah angka penting yang sama |
| error bar/ˈerə bɑː/ | batang kesalahan |
| worst acceptable line/wɜːst əkˈseptəbl laɪn/ | garis terbaik yang dapat diterima |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A scalar 标量 has size only. A vector 矢量 has both size and direction.
Examples from the syllabus:
Quick test: if it makes sense to ask "in which direction?", the quantity is a vector. You cannot ask "in which direction is the temperature?", so temperature is a scalar. You can ask "in which direction is the velocity?", so velocity is a vector.
A vector is drawn as an arrow: the direction of the arrow gives the direction of the quantity, and the length of the arrow (drawn to scale) gives the magnitude.
To add two coplanar 共面 vectors (vectors in the same flat plane), draw them tip to tail. The resultant 合矢量 goes from the tail of the first arrow to the tip of the second.
To find $\vec{X} - \vec{Y}$, add the reverse of $\vec{Y}$: $\vec{X} + (-\vec{Y})$. The reverse of $\vec{Y}$ has the same size as $\vec{Y}$ but points the opposite way.
If the two vectors are at right angles (90°), the size of the resultant is
and its direction comes from $\tan\theta = Y / X$.
Worked example. A swimmer heads north at $1.2\ \text{m s}^{-1}$ across a river that flows east at $0.5\ \text{m s}^{-1}$. Find the size and direction of the resultant velocity.
The two velocities are perpendicular, so
at an angle $\tan\theta = 0.5/1.2$, giving $\theta \approx 23°$ east of north.
If two vectors have the same size $F$ with an angle $2\alpha$ between them, the resultant has size $2F\cos\alpha$ and lies along the line that cuts the angle in half.
Worked example. Two forces of $6.0\ \text{N}$ act on an object with $60°$ between them. Find the resultant.
Here $2\alpha = 60°$, so $\alpha = 30°$ and $R = 2 \times 6.0 \times \cos 30° = 10.4\ \text{N}$, along the line halfway between the two forces. Resolving gives the same answer: each force has a component $6.0\cos 30°$ along that line, and their components at right angles to it cancel.
For two vectors at any other angle, either make a scale drawing 按比例作图 of the vector triangle and measure the resultant, or resolve both vectors into perpendicular components, add the components, and recombine with $\sqrt{X^{2} + Y^{2}}$.
Worked example. A ball moving east at $5.0\ \text{m s}^{-1}$ is hit so that it moves north at $5.0\ \text{m s}^{-1}$. Find its change in velocity.
A change in velocity is a vector subtraction: $\Delta\vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}}$. Draw $5.0\ \text{m s}^{-1}$ north, then add the reverse of the initial velocity, $5.0\ \text{m s}^{-1}$ west. The two are perpendicular, so $|\Delta\vec{v}| = \sqrt{5.0^{2} + 5.0^{2}} = 7.1\ \text{m s}^{-1}$, pointing north-west. The speed did not change, but the velocity did. That is why a force must have acted on the ball (topic 3).
Any vector can be split into two perpendicular 垂直 (at right angles) components 分量. Usually these are horizontal 水平 and vertical 竖直, or along and across a surface. For a vector $\vec{v}$ at angle $\theta$ to the horizontal:
Choose the directions that make the problem easiest. On a slope (an inclined plane 斜面), split the weight into one part along the slope and one part at right angles to it:
where $\theta$ is the angle of the slope to the horizontal.
You split a vector into components whenever you need to know how much of it acts in one direction. For example: the part of a force that acts along a slope, or the horizontal and vertical parts of a ball's velocity after it is thrown.
resultant = a + b · resultan = a + b
Vectors add tip to tail. The resultant runs from the start of the first arrow to the tip of the second, and its length is found from a scale drawing or by Pythagoras, never by adding the two magnitudes. · Vektor dijumlahkan ujung ke pangkal. Vektor resultan membentang dari pangkal anak panah pertama hingga ujung anak panah kedua, dan panjangnya ditentukan dari gambar berskala atau menggunakan teorema Pythagoras, bukan dengan menjumlahkan kedua magnitudenya.
| English | Bahasa Indonesia |
|---|---|
| vertical/ˈvɜːtɪkl/ | vertikal |
| scalar/ˈskeɪlə/ | skalar |
| vector/ˈvektə/ | vektor |
| component/kəmˈpəʊnənt/ | komponen |
| electric charge/ɪˈlektrɪk tʃɑːdʒ/ | muatan listrik |
| displacement/dɪˈspleɪsmənt/ | pindahan |
| momentum/məʊˈmentəm/ | momentum |
| coplanar/ˌkəʊˈpleɪnə/ | koplanar |
| resultant/rɪˈzʌltənt/ | resultan |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | tegak lurus |
| scale drawing/skeɪl ˈdrɔːɪŋ/ | gambar skala |
| horizontal/ˌhɒrɪˈzɒntl/ | horizontal |
| inclined plane/ɪnˈklaɪnd pleɪn/ | bidang miring |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| physical quantity | a numerical magnitude together with a unit |
| homogeneous equation | an equation in which every term has the same base units |
| systematic error | an error that shifts every reading in the same direction by the same amount; not reduced by repeating |
| random error | an error that scatters readings above and below the true value; reduced by repeating and averaging |
| zero error | the reading an instrument shows when the true value is zero |
| precision | how close repeated readings are to each other |
| accuracy | how close a reading, or the mean of several readings, is to the true value |
| absolute uncertainty | the range within which the true value is expected to lie, in the units of the quantity |
| percentage uncertainty | the absolute uncertainty divided by the value, multiplied by 100% |
| scalar | a quantity with magnitude only |
| vector | a quantity with magnitude and direction |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
These five quantities come up in almost every kinematics 运动学 question. Learn the exact words — the examiner gives marks for precise wording.
The unit of speed and velocity is $\text{m s}^{-1}$; the unit of acceleration is $\text{m s}^{-2}$.
Two more phrases the examiner uses: uniform acceleration 匀加速 means constant acceleration (a straight line on a velocity–time graph), and the acceleration of free fall 自由落体加速度 $g$ is the acceleration of an object falling freely in a uniform gravitational field 匀强重力场 when air resistance is negligible 可忽略的 (small enough to ignore).
A common mistake: deceleration 减速度 just means acceleration in the opposite direction to the velocity. It is not a separate quantity.
Speeds are sometimes given in $\text{km h}^{-1}$. Convert before you calculate: $85\ \text{km h}^{-1} = 85 \times 1000 / 3600 = 23.6\ \text{m s}^{-1}$.
Lima besaran ini muncul dalam hampir setiap soal kinematika. Hafalkan kata-kata yang tepat — penguji memberikan nilai untuk wording yang presisi.
Satuan kecepatan dan kecepatan (velocity) adalah $\text{m s}^{-1}$; satuan percepatan adalah $\text{m s}^{-2}$.
Dua frasa lagi yang digunakan penguji: percepatan seragam berarti percepatan konstan (garis lurus pada grafik kecepatan–waktu), dan percepatan jatuh bebas $g$ adalah percepatan benda yang jatuh bebas dalam medan gravitasi seragam ketika hambatan udara dapat diabaikan (cukup kecil untuk diabaikan).
Kesalahan umum: perlambatan hanya berarti percepatan dalam arah berlawanan dengan kecepatan. Ini bukan besaran terpisah.
Kecepatan kadang diberikan dalam $\text{km h}^{-1}$. Konversikan sebelum menghitung: $85\ \text{km h}^{-1} = 85 \times 1000 / 3600 = 23.6\ \text{m s}^{-1}$.
v = u + at
On a speed–time graph the gradient is the acceleration and the area underneath is the distance travelled. · Pada grafik kecepatan–waktu, gradien adalah percepatan dan luas di bawahnya adalah jarak tempuh.
| English | Bahasa Indonesia |
|---|---|
| projectile/prəˈdʒektaɪl/ | proyektil |
| speed/spiːd/ | kecepatan (besaran skalar) |
| distance/ˈdɪstəns/ | jarak tempuh |
| kinematics/ˌkɪnɪˈmætɪks/ | kinematika |
| scalar/ˈskeɪlə/ | skalar |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
Many marks come from reading or drawing motion graphs, the graphical methods 图像法 the syllabus names. Two graphs matter.
The gradient 斜率 (steepness) of a displacement–time graph at a point gives the velocity at that moment.
The gradient of a velocity–time graph gives the acceleration at that moment.
The area between the line and the time axis gives the displacement in that time.
To find the displacement, split the area into triangles and rectangles, or count grid squares. Area of a triangle is $\tfrac{1}{2} \times \text{base} \times \text{height}$; area of a rectangle is $\text{base} \times \text{height}$.
Worked example. The graph shows the velocity of a lift. Find the acceleration in each stage and the total distance travelled.
When the graph is a curve, the area is still the displacement: count the squares under the curve (half a square or more counts as one) and multiply by the value of one square.
Banyak nilai diperoleh dari membaca atau menggambar grafik gerak, metode grafis yang disebut dalam silabus. Dua grafik sangat penting.
Gradien (kemiringan) dari grafik perpindahan–waktu pada suatu titik memberikan kecepatan pada momen itu.
Gradien dari grafik kecepatan–waktu memberikan percepatan pada momen itu.
Luas antara garis dan sumbu waktu memberikan perpindahan dalam waktu tersebut.
Untuk mencari sesaran, bahagikan kawasan kepada segitiga dan segi empat tepat, atau hitung petak grid. Kawasan segitiga ialah $\tfrac{1}{2} \times \text{base} \times \text{height}$; kawasan segi empat tepat ialah $\text{base} \times \text{height}$.
Contoh berkerja. Graf menunjukkan kelajuan lift. Cari pecutan dalam setiap peringkat dan jarak total yang dilalui.
Apabila graf adalah lengkung, kawasan itu masih merupakan sesaran: hitung petak di bawah lengkung (separuh petak atau lebih dikira sebagai satu) dan darabkan dengan nilai satu petak.
Change the start velocity u and the acceleration a. The gradient of the line is the acceleration; the area under it is the displacement. · Ubah kecepatan awal u dan percepatan a. Gradien garis menunjukkan percepatan; luas di bawahnya menunjukkan perpindahan.
| English | Bahasa Indonesia |
|---|---|
| displacement/dɪˈspleɪsmənt/ | pindahan |
| vector/ˈvektə/ | vektor |
| acceleration/əkˌseləˈreɪʃn/ | percepatan |
| uniform acceleration/ˈjuːnɪfɔːm əkˌseləˈreɪʃn/ | percepatan seragam |
| acceleration of free fall/əkˌseləˈreɪʃn ɒv friː fɔːl/ | percepatan jatuh bebas |
For motion in a straight line with uniform acceleration, we use five symbols: starting velocity $u$, final velocity $v$, acceleration $a$, displacement $s$, and time $t$. Four equations link them:
Each equation uses four of the five symbols. To pick the right one: write down what you know and what you want, then choose the equation with exactly those four.
Worked example. A car accelerates uniformly from $8\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over a distance of $56\ \text{m}$. Find its acceleration.
We know $u$, $v$ and $s$ and want $a$, so use $v^{2} = u^{2} + 2as$:
You should be able to get these from the definitions of velocity and acceleration. Think of the velocity–time graph of the motion: a straight line from $u$ at $t = 0$ to $v$ at time $t$.
A "derive" question wants exactly these steps, each starting from a definition or an equation already established, with no numbers.
If a question asks "which equation can be found using only the gradient of a velocity–time graph?", the answer is $v = u + at$ (the gradient is the acceleration).
"Show that" questions. When the question gives the answer ("show that the height is 3.2 km"), the marks are for the method. State the equation, substitute every value with its unit, and give the result to one more significant figure than the value quoted ($3.24\ \text{km}$, which rounds to $3.2\ \text{km}$). Reaching the quoted value proves nothing on its own.
Pick a positive direction at the start and keep it. Anything pointing the other way gets a minus sign. For a ball thrown straight up, if "up" is positive: $u$ is positive, $a = -g$ (gravity 重力 pulls down), and at the highest point the displacement is positive but the velocity is zero.
Bagi gerakan dalam garis lurus dengan pecutan seragam, kita menggunakan lima simbol: kelajuan permulaan $u$, kelajuan akhir $v$, pecutan $a$, sesaran $s$, dan masa $t$. Empat persamaan menghubungkannya:
Setiap persamaan menggunakan empat daripada lima simbol. Untuk memilih yang betul: tulis apa yang anda tahu dan apa yang anda mahukan, kemudian pilih persamaan yang mempunyai tepat empat simbol tersebut.
Contoh berkerja. Sebuah kereta memecut secara seragam dari $8\ \text{m s}^{-1}$ kepada $20\ \text{m s}^{-1}$ merentasi jarak $56\ \text{m}$. Cari pecutannya.
Kita tahu $u$, $v$ dan $s$ dan mahu $a$, jadi gunakan $v^{2} = u^{2} + 2as$:
Anda sepatutnya boleh mendapatkan ini daripada definisi kelajuan dan pecutan. Bayangkan graf kelajuan–masa bagi gerakan: garis lurus dari $u$ pada $t = 0$ ke $v$ pada masa $t$.
Soalan "terbitkan" memerlukan langkah-langkah ini tepat, masing-masing bermula dengan definisi atau persamaan yang telah ditetapkan, tanpa nombor.
Jika soalan bertanya "persamaan manakah boleh didapati hanya menggunakan kecerunan graf kelajuan–masa?", jawapannya ialah $v = u + at$ (kecerunan adalah pecutan).
Soalan "Buktikan bahawa". Apabila soalan memberikan jawapan ("buktikan bahawa ketinggian ialah 3.2 km"), markah diberikan untuk kaedah. Nyatakan persamaan, gantikan setiap nilai dengan unitnya, dan berikan hasil dengan satu angka bererti lebih banyak daripada nilai yang dikutip ($3.24\ \text{km}$, yang dibundarkan kepada $3.2\ \text{km}$). Mencapai nilai yang dikutip tidak membuktikan apa-apa sendiri.
Pilih arah positif pada permulaan dan kekalkan ia. Segala sesuatu yang mengarah ke arah lain mendapat tanda negatif. Bagi bola yang dilempar lurus ke atas, jika "atas" adalah positif: $u$ adalah positif, $a = -g$ (graviti menarik ke bawah), dan pada titik tertinggi sesaran adalah positif tetapi kelajuan ialah sifar.
| English | Bahasa Indonesia |
|---|---|
| parabola/pəˈræbələ/ | parabola |
| gravity/ˈɡrævɪti/ | gravitasi |
| symmetric/sɪˈmetrɪk/ | simetris |
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ | kecepatan terminal |
When air resistance 空气阻力 can be ignored, an object in free fall 自由落体 has a constant acceleration $g \approx 9.81\ \text{m s}^{-2}$ downwards. This is the same for every mass.
For a ball dropped from rest and falling a distance $h$:
For a ball thrown straight up with speed $u$:
Worked example. A ball is thrown straight up at $20\ \text{m s}^{-1}$. Find the greatest height it reaches (take $g = 9.81\ \text{m s}^{-2}$).
At the highest point $v = 0$, so from $h = u^{2}/(2g)$:
The equations above assume the only force is the weight. With air resistance, the resultant force on a falling object is smaller than its weight, so the acceleration is less than $g$; and because air resistance grows with speed, the acceleration keeps decreasing as the object speeds up. On a velocity–time graph the line curves, its gradient falling towards zero as the object approaches terminal velocity 收尾速度 (topic 3). For a projectile, air resistance shortens the range and lowers the maximum height, and the path is no longer a symmetrical parabola 抛物线: the object comes down more steeply than it went up. A "state and explain" question wants the force first, then its effect on the acceleration, then the effect on the motion.
A common method: drop an object from rest, then measure the distance $h$ it falls and the time $t$ it takes. Then
Repeat for several heights and plot $h$ against $t^{2}$. The gradient of the best straight line is $g/2$, so $g$ is twice the gradient. Repeating reduces random error 随机误差. An electronic timer — using light gates 光电门, or a switch the ball hits — removes reaction-time 反应时间 error.
Apabila rintangan udara boleh diabaikan, objek dalam jatuh bebas mempunyai pecutan malar $g \approx 9.81\ \text{m s}^{-2}$ ke bawah. Ini sama untuk setiap jisim.
Bagi bola yang dijatuhkan dari keadaan rehat dan jatuh sejauh $h$:
Bagi bola yang dilempar lurus ke atas dengan kelajuan $u$:
Contoh berkerja. Sebuah bola dilempar lurus ke atas pada $20\ \text{m s}^{-1}$. Cari ketinggian maksimum yang dicapai (ambil $g = 9.81\ \text{m s}^{-2}$).
Pada titik tertinggi $v = 0$, jadi dari $h = u^{2}/(2g)$:
Persamaan di atas mengandaikan daya tunggal ialah berat. Dengan rintangan udara, daya resultan pada objek jatuh adalah kurang daripada beratnya, jadi pecutan adalah kurang daripada $g$; dan kerana rintangan udara bertambah dengan kelajuan, pecutan terus berkurang semasa objek mempercepatkan diri. Pada graf kelajuan–masa, garis melengkung, kecerunannya menurun menuju sifar apabila objek mendekati kelajuan terminal (topik 3). Bagi peluru, rintangan udara memendekkan julat dan menurunkan ketinggian maksimum, dan laluan tidak lagi menjadi parabol simetri: objek datang turun lebih curam daripada bagaimana ia naik. Soalan "nyatakan dan terangkan" memerlukan daya dahulu, kemudian kesannya terhadap pecutan, kemudian kesannya terhadap gerakan.
Kaedah biasa: jatuhkan objek dari keadaan rehat, kemudian ukur jarak $h$ yang jatuh dan masa $t$ yang diambil. Kemudian
Ulangi untuk beberapa ketinggian dan plot $h$ terhadap $t^{2}$. Gradien garis lurus terbaik adalah $g/2$, sehingga $g$ adalah dua kali gradien. Pengulangan mengurangi kesalahan acak. Timer elektronik — menggunakan pintu cahaya, atau saklar yang dipukul bola — menghilangkan kesalahan waktu reaksi.
| English | Bahasa Indonesia |
|---|---|
| free fall/friː fɔːl/ | jatuh bebas |
| uniform gravitational field/ˈjuːnɪfɔːm ˌɡrævɪˈteɪʃənl fiːld/ | medan gravitasi seragam |
| air resistance/eə rɪˈzɪstəns/ | hambatan udara |
| negligible/ˈneɡlɪdʒəbl/ | dapat diabaikan |
| deceleration/dɪˌseləˈreɪʃn/ | perlambatan |
| graphical methods/ˈɡræfɪkl ˈmeθədz/ | metode grafik |
| gradient/ˈɡreɪdɪənt/ | gradien |
| tangent/ˈtændʒənt/ | tangens |
| trapezium/trəˈpiːzɪəm/ | trapesium |
When an object moves at constant velocity in one direction (say horizontal 水平) and speeds up in a direction at right angles to it (say vertical 竖直, under gravity), the two motions do not affect each other. Treat each direction on its own, with its own SUVAT equation.
An object thrown horizontally with speed $u_{\text{H}}$ from height $h$, with air resistance ignored:
The time to reach the ground depends only on the height $h$, not on $u_{\text{H}}$. Solve $h = \tfrac{1}{2} g t^{2}$ for $t$; then the horizontal range 射程 is $u_{\text{H}} t$.
Worked example. A ball is thrown horizontally at $15\ \text{m s}^{-1}$ from the top of a cliff $20\ \text{m}$ high. Find the time it takes to land and how far from the base it lands (take $g = 9.81\ \text{m s}^{-2}$).
Vertical motion gives the time: from $h = \tfrac{1}{2}g t^{2}$,
Horizontal motion then gives the range: $x = u_{\text{H}} t = 15 \times 2.0 \approx 30\ \text{m}.$
The horizontal-velocity graph is a flat line at $u_{\text{H}}$. The vertical-velocity graph is a straight line from the origin with gradient $g$.
Comparing two objects. If object A is dropped and object B is thrown horizontally from the same height at the same moment, both reach the ground at the same time: their vertical motions are identical. If instead B is thrown with an upward component, it takes longer, because it must first rise and come back to the start height before it falls the same distance. Its speed on landing is still found from energy: the kinetic energy gained equals the potential energy lost, whatever the direction of the throw, so a ball launched at any angle with the same speed from the same height lands at the same speed (topic 5).
A projectile 抛体 thrown at speed $u$ at angle $\theta$ above the horizontal:
At the highest point, $v_{\text{V}} = 0$, but $v_{\text{H}}$ is still $u\cos\theta$. The time to the top is $t_{\text{up}} = u\sin\theta / g$; the total flight time (back to the start height) is $2t_{\text{up}}$.
Sketching the velocity graphs. Examiners often ask you to sketch $v_{\text{H}}$ and $v_{\text{V}}$ against time on the same axes, taking upwards as positive. The horizontal component is a flat line at $u\cos\theta$ for the whole flight. The vertical component is a straight line of gradient $-g$: it starts at $+u\sin\theta$, crosses zero at the top of the flight, and reaches $-u\sin\theta$ on landing at the start height. Label each line, mark the times where the lines start and stop, and make the crossing sit exactly at $t_{\text{up}}$.
Worked example. A ball leaves the ground at $18\ \text{m s}^{-1}$ at $60°$ to the horizontal. Show that it reaches its maximum height at $t = 1.6\ \text{s}$.
$u_{\text{V}} = 18 \sin 60° = 15.6\ \text{m s}^{-1}$. At the top $v_{\text{V}} = 0$, so from $v = u + at$ with $a = -9.81\ \text{m s}^{-2}$: $t = 15.6 / 9.81 = 1.59\ \text{s} \approx 1.6\ \text{s}$.
When a ball bounces, its velocity–time graph is a set of straight sloping lines (constant $g$) with a sudden jump at each bounce (the velocity flips direction, and gets smaller if some energy 能量 is lost). Add up the times and the distances across the bounces.
Read such a graph carefully: the gradient of every sloping line is the same $g$; the ball is at its highest point wherever the line crosses the time axis; and the height of each bounce comes from the area of the triangle above (or below) the axis for that flight.
Ketika benda bergerak dengan kecepatan konstan dalam satu arah (misalnya horizontal) dan percepatan dalam arah tegak lurus dengannya (misalnya vertikal, akibat gravitasi), kedua gerak tersebut tidak saling mempengaruhi. Perlakukan setiap arah secara terpisah, dengan persamaan SUVAT masing-masing.
Benda yang dilempar horizontal dengan kecepatan $u_{\text{H}}$ dari ketinggian $h$, dengan pengabaian hambatan udara:
Waktu untuk mencapai tanah bergantung hanya pada ketinggian $h$, bukan pada $u_{\text{H}}$. Selesaikan $h = \tfrac{1}{2} g t^{2}$ untuk $t$; kemudian jangkauan horizontal adalah $u_{\text{H}} t$.
Contoh terpecahkan. Sebuah bola dilempar horizontal pada $15\ \text{m s}^{-1}$ dari puncak tebing setinggi $20\ \text{m}$. Tentukan waktu yang dibutuhkan untuk mendarat dan seberapa jauh dari dasar tebing ia mendarat (ambil $g = 9.81\ \text{m s}^{-2}$).
Gerak vertikal memberikan waktu: dari $h = \tfrac{1}{2}g t^{2}$,
Gerak horizontal kemudian memberikan jangkauan: $x = u_{\text{H}} t = 15 \times 2.0 \approx 30\ \text{m}.$
Graf kecepatan horizontal adalah garis datar pada $u_{\text{H}}$. Grafik kecepatan vertikal adalah garis lurus dari titik asal dengan gradien $g$.
Membandingkan dua benda. Jika benda A dijatuhkan dan benda B dilempar horizontal dari ketinggian yang sama pada saat yang sama, keduanya mencapai tanah pada waktu yang sama: gerak vertikal mereka identik. Jika sebaliknya B dilempar dengan komponen ke atas, ia membutuhkan waktu lebih lama, karena harus naik terlebih dahulu dan kembali ke ketinggian awal sebelum jatuh sejauh itu. Kecepatannya saat mendarat tetap ditemukan dari energi: energi kinetik yang didapat sama dengan energi potensial yang hilang, apa pun arah lemparan, sehingga bola yang diluncurkan dengan sudut dan kecepatan yang sama dari ketinggian yang sama akan mendarat dengan kecepatan yang sama (topik 5).
Peluru yang dilempar dengan kecepatan $u$ pada sudut $\theta$ di atas horizontal:
Di titik tertinggi, $v_{\text{V}} = 0$, tetapi $v_{\text{H}}$ masih $u\cos\theta$. Waktu untuk mencapai puncak adalah $t_{\text{up}} = u\sin\theta / g$; total waktu penerbangan (kembali ke ketinggian awal) adalah $2t_{\text{up}}$.
Menggambar grafik kecepatan. Penguji sering meminta Anda untuk menggambar $v_{\text{H}}$ dan $v_{\text{V}}$ terhadap waktu pada sumbu yang sama, dengan mengambil arah ke atas sebagai positif. Komponen horizontal adalah garis datar pada $u\cos\theta$ sepanjang penerbangan. Komponen vertikal adalah garis lurus bergradien $-g$: dimulai dari $+u\sin\theta$, memotong nol di puncak penerbangan, dan mencapai $-u\sin\theta$ saat mendarat pada ketinggian awal. Beri label setiap garis, tandai waktu di mana garis dimulai dan berakhir, dan pastikan perpotongan tepat berada di $t_{\text{up}}$.
Contoh terpecahkan. Sebuah bola meninggalkan tanah pada $18\ \text{m s}^{-1}$ pada $60°$ terhadap horizontal. Tunjukkan bahwa bola tersebut mencapai ketinggian maksimumnya pada $t = 1.6\ \text{s}$.
$u_{\text{V}} = 18 \sin 60° = 15.6\ \text{m s}^{-1}$. Di puncak $v_{\text{V}} = 0$, jadi dari $v = u + at$ dengan $a = -9.81\ \text{m s}^{-2}$: $t = 15.6 / 9.81 = 1.59\ \text{s} \approx 1.6\ \text{s}$.
Ketika bola memantul, grafik kecepatan–waktunya terdiri dari serangkaian garis miring (percepatan konstan $g$) dengan loncatan mendadak di setiap pantulan (kecepatan membalik arah, dan menjadi lebih kecil jika ada energi yang hilang). Jumlahkan waktu dan jarak lintas pantulan.
Baca grafik seperti ini dengan teliti: gradien setiap garis miring adalah sama $g$; bola berada pada titik tertingginya di mana pun garis memotong sumbu waktu; dan ketinggian setiap pantulan berasal dari luas segitiga di atas (atau di bawah) sumbu untuk penerbangan tersebut.
Fire the ball, then change the angle and speed. The horizontal motion is steady while gravity pulls it down — together they trace a parabola. Find the angle for the longest range, and try the Moon. · Tembak bola, lalu ubah sudut dan kecepatan. Gerak horizontal tetap konstan sementara gravitasi menariknya ke bawah — bersama-sama mereka melacak parabola. Temukan sudut untuk jangkauan terjauh, dan coba di Bulan.
| English | Bahasa Indonesia |
|---|---|
| range/reɪndʒ/ | jangkauannya |
| random error/ˈrændəm ˈerə/ | kesalahan acak |
| light gate/laɪt ɡeɪt/ | pintu cahaya |
| reaction-time/rɪˈækʃn taɪm/ | waktu reaksi |
| horizontal/ˌhɒrɪˈzɒntl/ | horizontal |
| vertical/ˈvɜːtɪkl/ | vertikal |
| component/kəmˈpəʊnənt/ | komponen |
| energy/ˈenədʒi/ | tenaga |
When two objects move along the same line in different ways, write a displacement equation for each. Use the same start time and the same positive direction. Then set the two displacements equal (or set their difference to a given gap).
For a goods train at constant velocity $u_{\text{G}}$ and an express train starting from rest with acceleration $a$, both passing the same point at $t = 0$:
They are level again when $s_{\text{G}} = s_{\text{E}}$, giving $t = 2 u_{\text{G}} / a$.
Ketika dua benda bergerak sepanjang garis yang sama dengan cara berbeda, tulis persamaan perpindahan untuk masing-masing. Gunakan waktu awal yang sama dan arah positif yang sama. Kemudian samakan kedua perpindahan tersebut (atau samakan selisihnya dengan jarak tertentu).
Untuk kereta barang berkecepatan konstan $u_{\text{G}}$ dan kereta ekspres yang mulai diam dengan percepatan $a$, keduanya melewati titik yang sama pada $t = 0$:
Keduanya sejajar kembali ketika $s_{\text{G}} = s_{\text{E}}$, menghasilkan $t = 2 u_{\text{G}} / a$.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| distance | the total length of the path travelled (a scalar) |
| displacement | the distance moved in a stated direction, from start to finish (a vector) |
| speed | the rate of change of distance with time |
| velocity | the rate of change of displacement with time |
| acceleration | the rate of change of velocity with time |
| uniform acceleration | acceleration that is constant in magnitude and direction |
| acceleration of free fall | the acceleration of an object falling freely under gravity alone, with air resistance negligible |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| jarak | total panjang lintasan yang ditempuh (skalar) |
| perpindahan | jarak yang berpindah dalam arah yang ditentukan, dari awal hingga akhir (vektor) |
| kecepatan | laju perubahan jarak terhadap waktu |
| kecepatan vektor | laju perubahan perpindahan terhadap waktu |
| percepatan | laju perubahan kecepatan terhadap waktu |
| percepatan seragam | percepatan yang konstan dalam besaran dan arah |
| percepatan jatuh bebas | percepatan benda yang jatuh bebas hanya di bawah gravitasi, dengan hambatan udara dapat diabaikan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Mass 质量 tells you how hard it is to change an object's motion. The larger the mass, the larger the force 力 needed to give it a certain acceleration 加速度. Mass is measured in kilograms ($\text{kg}$) and is a scalar 标量.
Linear momentum 动量 is the product of mass and velocity:
Momentum is a vector 矢量 — it points the same way as the velocity 速度. Its unit is $\text{kg m s}^{-1}$, which is the same as $\text{N s}$.
Newton's second law, in its general form: the resultant force 合力 on an object equals the rate of change of its momentum.
When the mass is constant this becomes $F = ma$, because $\Delta p = m\,\Delta v$ and $\Delta v / \Delta t = a$. Cambridge questions often want you to use $F = \Delta p / \Delta t$ directly for a collision 碰撞 or an impulse 冲量: the average force equals the change in momentum divided by the contact time.
A ball hits a wall with momentum $p_{1}$ and bounces back with momentum $p_{2}$ in the opposite direction. The change in momentum is $\Delta p = p_{2} - p_{1}$ (give each direction the correct sign). The average force is $\Delta p / \Delta t$, where $\Delta t$ is the contact time.
Worked example. A $0.20\ \text{kg}$ ball hits a wall at $8.0\ \text{m s}^{-1}$ and bounces straight back at $6.0\ \text{m s}^{-1}$. The contact lasts $0.050\ \text{s}$. Find the average force on the ball.
Take the rebound direction as positive, so $u = -8.0\ \text{m s}^{-1}$ and $v = +6.0\ \text{m s}^{-1}$:
When you know the momentum but not the speed, the change in kinetic energy 动能 is
This comes from $E_{\text{k}} = \tfrac{1}{2} m v^{2} = p^{2} / (2m)$.
Because $F = \Delta p / \Delta t$, the gradient of a momentum–time graph is the resultant force, just as the gradient of a velocity–time graph is the acceleration. A straight line means a constant resultant force; a horizontal line means zero resultant force; a line sloping down means a force acting against the motion.
Worked example. The graph shows the momentum of a motorcycle. Find the resultant force on it during the first $10\ \text{s}$.
Gradient $= \Delta p / \Delta t = 5000 / 10 = 500\ \text{N}$. Between $10\ \text{s}$ and $16\ \text{s}$ the momentum is constant, so the resultant force is zero: the driving force just balances the resistive forces. From $16\ \text{s}$ to $20\ \text{s}$ the momentum falls by $2000\ \text{kg m s}^{-1}$, so the resultant force is $-500\ \text{N}$, acting backwards.
If a question tells you the speed is changing while the resultant force stays constant, that force cannot be air resistance: drag depends on speed, so it would change as the speed changes, and drag always acts against the motion.
The resultant of the forces, divided by the mass, gives the acceleration. · Resultan gaya, dibagi massa, menghasilkan percepatan.
F = ma (resultant) · F = ma (resultan)
The resultant force sets the acceleration; balanced forces ⇒ none. · Gaya resultan menentukan percepatan; gaya seimbang ⇒ tidak ada percepatan.
| English | Bahasa Indonesia |
|---|---|
| mass/mæs/ | massa |
| force/fɔːs/ | gaya |
| acceleration/əkˌseləˈreɪʃn/ | percepatan |
| scalar/ˈskeɪlə/ | skalar |
| linear momentum/ˈlɪnɪə məʊˈmentəm/ | momentum linear |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
| vector/ˈvektə/ | vektor |
| resultant force/rɪˈzʌltənt fɔːs/ | resultan gaya |
| impulse/ˈɪmpʌls/ | impuls |
| collision/kəˈlɪʒn/ | tabrakan |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
An object stays at rest, or keeps moving at constant velocity in a straight line, unless a resultant external force 外力 acts on it. In short: zero resultant force means zero acceleration.
The resultant force on an object equals its rate of change of momentum, and acts in the same direction as that change. In SI units,
Acceleration and resultant force always point the same way.
When body A pushes on body B, body B pushes back on body A with an equal and opposite force. The two forces:
A common trap: the weight 重力 of a block on a table and the normal contact force 支持力 from the table are not a third-law pair (they act on the same object and are different types). The third-law partner of the block's weight is the pull the block makes on the Earth. The third-law partner of the table's contact force is the push the block makes on the table.
To name a third-law partner, swap the two bodies in the sentence. "The back wheel of a bicycle pushes backwards on the road" pairs with "the road pushes forwards on the back wheel", and that forward push is what drives the bicycle. "The Earth pulls the Moon" pairs with "the Moon pulls the Earth", with the same size of force.
For a rocket: the thrust 推力 on the rocket and the force on the gases are a third-law pair (the engine pushes the gas down, the gas pushes the engine up). Weight and air resistance are not part of this pair.
| English | Bahasa Indonesia |
|---|---|
| thrust/θrʌst/ | dorongan |
| external force/ekˈstɜːnl fɔːs/ | gaya eksternal |
| weight/weɪt/ | berat |
| normal contact force/ˈnɔːml ˈkɒntækt fɔːs/ | gaya kontak normal |
| electrostatic/ɪˌlektrəʊˈstætɪk/ | elektrostatik |
Weight is the force on an object from a gravitational field 重力场. Near the Earth's surface,
where $g \approx 9.81\ \text{m s}^{-2}$ is the acceleration of free fall. Weight is a vector that points towards the centre of the Earth. Do not mix it up with mass: mass is the same everywhere, but weight changes with place.
On the Moon $g$ is about $1.6\ \text{m s}^{-2}$, so an astronaut's mass is unchanged but their weight is about one sixth of its value on Earth. When a multiple-choice question asks which statement describes weight, the answer is the one that says "the force on the object due to a gravitational field", not "mass times acceleration" and not "the resultant force".
| English | Bahasa Indonesia |
|---|---|
| gravitational field/ˌɡrævɪˈteɪʃənl fiːld/ | medan gravitasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A friction 摩擦力 force between two solid surfaces acts along the surface and opposes the sliding. A viscous 黏性 or drag 阻力 force is the resistive force from a fluid 流体 (a liquid or gas) on an object moving through it; air resistance is the case for air. You do not need to use any coefficient 系数 of friction or viscosity.
A simple model: the drag gets bigger as the speed gets bigger. At zero speed, the drag is zero. Drag also grows with the cross-sectional area 横截面积 of the object and with the density of the fluid; a parachute works by making the area, and so the drag, much larger. These frictional forces and drag forces are all resistive forces: they act against the direction of motion and transfer kinetic energy to thermal energy.
For an object dropped from rest and falling through air:
On a velocity–time graph, the line starts straight with gradient $g$, then bends and flattens at the terminal velocity. This shape (fast start, then slowing acceleration, then constant speed) is how "falling with air resistance" differs from "free fall in a vacuum" in a uniform gravitational field 匀强重力场. On an acceleration–time graph the acceleration starts at $g$ and decreases, curving down to zero.
Two spheres of the same size but different density have different terminal velocities: the denser sphere has the larger weight, so it needs a larger drag to balance it, and drag only reaches that value at a higher speed. That is also why a parachute changes everything: when it opens, the drag suddenly exceeds the weight, the resultant force is upwards, and the skydiver decelerates (the acceleration is upwards while the velocity is still downwards) until the drag has fallen back to equal the weight at a new, much lower terminal velocity.
A ball falling through a liquid has three forces on it: its weight downwards, the upthrust 浮力 upwards, and the viscous drag upwards. The upthrust comes from hydrostatic pressure 流体静压强: the pressure in a liquid increases with depth, so the liquid pushes harder on the bottom of the ball than on the top, and the resultant of these pushes is upwards. Its size is the weight of the liquid the ball displaces, $U = \rho_{\text{liquid}} V g$ (topic 4). The drag grows with speed. When weight $=$ upthrust $+$ drag, the resultant force is zero and the ball falls at its terminal speed. A "draw labelled arrows" question wants exactly these three, with the upthrust and drag both up and the weight down.
Worked example. A steel ball of radius $2.0\ \text{mm}$ and weight $2.56 \times 10^{-3}\ \text{N}$ falls through oil of density $850\ \text{kg m}^{-3}$. The viscous drag on it is $F = 6\pi \eta r v$, where $\eta$ is the viscosity 黏度 of the oil and $v$ the speed. (a) Find the SI base units of $\eta$. (b) Show that the upthrust on the ball is $2.8 \times 10^{-4}\ \text{N}$. (c) Taking $\eta = 0.20$ in SI units, find the terminal speed.
(a) $\eta = F / (6\pi r v)$, so its units are $\text{N} / (\text{m} \cdot \text{m s}^{-1}) = \text{kg m s}^{-2} / (\text{m}^{2}\ \text{s}^{-1}) = \text{kg m}^{-1}\ \text{s}^{-1}$.
(b) $V = \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi (2.0 \times 10^{-3})^{3} = 3.35 \times 10^{-8}\ \text{m}^{3}$, so $U = 850 \times 3.35 \times 10^{-8} \times 9.81 = 2.79 \times 10^{-4}\ \text{N} \approx 2.8 \times 10^{-4}\ \text{N}$.
(c) At the terminal speed, drag $= W - U = 2.56 \times 10^{-3} - 2.8 \times 10^{-4} = 2.28 \times 10^{-3}\ \text{N}$. So $v = F / (6\pi \eta r) = 2.28 \times 10^{-3} / (6\pi \times 0.20 \times 2.0 \times 10^{-3}) = 0.30\ \text{m s}^{-1}$.
At terminal velocity, a parachutist 跳伞者 has constant kinetic energy. But the gravitational potential energy 重力势能 keeps falling as they go down. Where does it go? Almost all of it turns into thermal energy 热能 of the air around them. It does not become kinetic energy of the parachutist — that stays constant.
A vehicle at constant speed on a flat road has zero resultant force. The forward driving force 驱动力 is equal and opposite to the total resistive force (friction, air resistance, and rolling resistance). At higher speed the drag is larger, so the driving force must be larger too — and so the power 功率 must be larger.
On a slope the weight has a component along the road. Going up at constant speed, the driving force must balance the resistive force and the part of the weight that pulls down the slope, $W\sin\theta$; going down, that same component helps the motion.
Worked example. A car of mass $1500\ \text{kg}$ climbs a road inclined at $12°$ to the horizontal at a constant $30\ \text{m s}^{-1}$. The total resistive force is $1600\ \text{N}$. Find the driving force and the useful output power of the engine.
Constant speed means zero resultant force along the slope: $D = 1600 + 1500 \times 9.81 \times \sin 12° = 1600 + 3060 = 4660\ \text{N} \approx 4700\ \text{N}$. Power is force times velocity (topic 5): $P = Dv = 4660 \times 30 = 1.4 \times 10^{5}\ \text{W}$.
Friction is what brakes a car. Set a speed and brake — the car keeps moving while the driver reacts, then friction slows it. Double the speed and watch the braking distance quadruple. · Gesekanlah yang memperlambat mobil. Tetapkan kecepatan dan rem — mobil terus bergerak sementara pengemudi bereaksi, lalu gesekan memperlambatnya. Gandakan kecepatan dan perhatikan jarak pengereman menjadi empat kali lipat.
Jump and watch the air-resistance arrow grow until it balances the weight — then the speed is constant. Open the parachute and the much bigger drag drops the diver to a slow, safe terminal velocity. · Lompat dan lihat panang hambatan udara membesar hingga menyeimbangkan berat — lalu kecepatannya tetap. Buka payung terpal dan hambat alir yang jauh lebih besar menurunkan pendakian ke kecepatan terminal yang lambat dan aman.
| English | Bahasa Indonesia |
|---|---|
| driving force/ˈdraɪvɪŋ fɔːs/ | tenaga pendorong |
| drag/dræɡ/ | menggeser |
| friction/ˈfrɪkʃn/ | gesekan |
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ | kecepatan terminal |
| viscous/ˈvɪskəs/ | viskos |
| fluid/ˈfluːɪd/ | fluida |
| coefficient/ˌkəʊɪˈfɪʃənt/ | koefisien |
| viscosity/vɪˈskɒsɪti/ | viskositas |
| cross-sectional area/krɒs ˈsekʃənl ˈeərɪə/ | luas penampang |
| thermal energy/ˈθɜːml ˈenədʒi/ | energi termal |
| uniform gravitational field/ˈjuːnɪfɔːm ˌɡrævɪˈteɪʃənl fiːld/ | medan gravitasi seragam |
| upthrust/ˈʌpθrʌst/ | gaya apung |
| hydrostatic pressure/ˌhaɪdrəˈstætɪk ˈpreʃə/ | tekanan hidrostatik |
| parachutist/ˈpærətʃuːtɪst/ | penurun payung |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | energi potensial gravitasi |
| power/ˈpaʊə/ | kuasa |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
For a system with no resultant external force, the total momentum stays constant. This is conservation of momentum 动量守恒. Stated for the two marks the examiner gives: the total momentum of a system of objects remains constant provided no resultant external force acts on the system. Both halves are needed: "total momentum is constant" alone scores one.
It always holds when there is no outside resultant force — in collisions, explosions 爆炸, and recoil 反冲. In two dimensions, momentum is conserved along each direction on its own.
Worked example. A $2.0\ \text{kg}$ trolley and a $3.0\ \text{kg}$ trolley are held together against a compressed spring, then released from rest. The $2.0\ \text{kg}$ trolley flies off at $6.0\ \text{m s}^{-1}$. Find the speed of the other trolley.
The total momentum stays zero (it started at rest), so
in the opposite direction.
In every collision, momentum is conserved (if there is no outside resultant force).
An elastic collision 弹性碰撞 is one where the total kinetic energy is also conserved. A quick test: in an elastic collision, the relative speed 相对速率 of approach equals the relative speed of separation.
In an inelastic collision 非弹性碰撞, momentum is conserved but kinetic energy goes down — some becomes thermal, sound, or deformation 形变 energy. If the two objects stick together, the collision is perfectly inelastic.
For two objects with masses $m_{1}, m_{2}$ and starting velocities $u_{1}, u_{2}$ that hit head-on 正面, write
Use signed velocities (positive in one chosen direction). If the collision is elastic, add the relative-speed equation
or, the same thing, $\tfrac{1}{2} m_{1} u_{1}^{2} + \tfrac{1}{2} m_{2} u_{2}^{2} = \tfrac{1}{2} m_{1} v_{1}^{2} + \tfrac{1}{2} m_{2} v_{2}^{2}$. That gives two equations for two unknowns.
Worked example. A $1500\ \text{kg}$ car moving at $12\ \text{m s}^{-1}$ runs into a stationary $1000\ \text{kg}$ car and they lock together. Find their common velocity just after the collision.
Momentum is conserved (the cars stick, so $v_{1} = v_{2} = v$):
Questions often go on to ask what fraction of the kinetic energy is transferred to other forms. Before: $E_{\text{k}} = \tfrac{1}{2} \times 1500 \times 12^{2} = 1.08 \times 10^{5}\ \text{J}$. After: $\tfrac{1}{2} \times 2500 \times 7.2^{2} = 6.48 \times 10^{4}\ \text{J}$. So $4.3 \times 10^{4}\ \text{J}$, which is $40\%$ of the original kinetic energy, becomes thermal energy, sound and deformation of the cars. Momentum is conserved; kinetic energy is not.
Worked example. A stationary nucleus of mass $222\,u$ decays by emitting an alpha particle α粒子 of mass $4\,u$ at $1.6 \times 10^{7}\ \text{m s}^{-1}$. Find the speed of the remaining nucleus.
Before the decay the total momentum is zero, so afterwards the two momenta are equal and opposite: $218\,u \times v = 4\,u \times 1.6 \times 10^{7}$, giving $v = 2.9 \times 10^{5}\ \text{m s}^{-1}$ in the opposite direction to the alpha particle. The unit $u$ cancels, so its value is never needed. In a decay the momentum, the charge (proton number), the nucleon number and the total mass–energy are all conserved (topic 11).
A useful result for a head-on elastic collision of mass $m$ with a stationary 静止 mass $M$:
If the objects move in two dimensions, split the velocities into perpendicular 垂直 components 分量 and apply conservation of momentum along each direction on its own. For a collision where the objects hit at an angle, choose one axis along the first object's motion and one across it. The total momentum is conserved along each axis.
Worked example. On a frictionless surface, ball A has momentum $4.0\ \text{N s}$ due east and ball B has momentum $3.0\ \text{N s}$ due north. They collide and stick together. Find the momentum of the combined object.
Momentum is conserved along each axis: $4.0\ \text{N s}$ east and $3.0\ \text{N s}$ north after the collision, exactly as before. The total momentum is the vector sum, $\sqrt{4.0^{2} + 3.0^{2}} = 5.0\ \text{N s}$ at $\tan^{-1}(3.0 / 4.0) = 37°$ north of east; the combined object moves that way at $5.0\ \text{N s}$ divided by the total mass. A multiple-choice diagram of two momentum arrows is asking exactly this: add the arrows tip to tail.
A rocket pushes out gas at velocity $u$ (relative to itself) at a mass-flow rate 质量流率 $\dot m$ (kg per second). It feels a thrust
which comes from $F = \Delta p / \Delta t$. The momentum given to the gas each second equals the thrust on the rocket (Newton's third law: the rocket pushes the gas one way, the gas pushes the rocket the other way). An engine ejecting $90\ \text{kg}$ of gas per second at $190\ \text{m s}^{-1}$ produces a thrust of $90 \times 190 = 1.7 \times 10^{4}\ \text{N}$.
Worked example. A rocket of weight $W$ leaves the ground with an initial acceleration $a$. What thrust does its engine produce?
Take upwards as positive. The resultant force is thrust minus weight, so $T - W = ma$. With $m = W / g$, $T = W + Wa / g = W(1 + a / g)$. A rocket of weight $2.0 \times 10^{6}\ \text{N}$ accelerating at $4.0\ \text{m s}^{-2}$ needs a thrust of $2.0 \times 10^{6} \times (1 + 4.0 / 9.81) = 2.8 \times 10^{6}\ \text{N}$. The thrust must exceed the weight before the rocket moves at all.
Set the masses and speeds, then collide them. Total momentum is conserved — the total before equals the total after. · Tetapkan massa dan kecepatan, lalu tabrakkan keduanya. Momentum total kekal — total sebelum tabrakan sama dengan total setelah tabrakan.
| English | Bahasa Indonesia |
|---|---|
| component/kəmˈpəʊnənt/ | komponen |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | tegak lurus |
| conservation of momentum/ˌkɒnsəˈveɪʃn ɒv məʊˈmentəm/ | kekekalan momentum |
| explosion/ekˈspləʊʒn/ | peledakan |
| recoil/rɪˈkɔɪl/ | recoil |
| inelastic collision/ɪnɪˈlæstɪk kəˈlɪʒn/ | tabrakan inelastis |
| elastic collision/ɪˈlæstɪk kəˈlɪʒn/ | tabrakan elastis |
| relative speed/ˈrelətɪv spiːd/ | kecepatan relatif |
| deformation/ˌdiːfɔːˈmeɪʃn/ | deformasi |
| head-on/hed ɒn/ | tabrakan kepala-kepalan |
| stationary/ˈsteɪʃənəri/ | stasioner |
| alpha particle/ˈælfə ˈpɑːtɪkl/ | partikel alfa |
| mass-flow rate/mæs fləʊ reɪt/ | laju aliran massa |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| mass | the property of an object that resists a change in its motion |
| linear momentum | the product of an object's mass and its velocity |
| force | the rate of change of momentum |
| Newton's first law | an object remains at rest or moves at constant velocity unless acted on by a resultant force |
| Newton's second law | the resultant force on an object is proportional to the rate of change of its momentum, and acts in the direction of the change |
| Newton's third law | when two bodies interact, the force on one is equal in magnitude and opposite in direction to the force on the other |
| weight | the force on an object due to a gravitational field; equal to the product of its mass and the acceleration of free fall |
| principle of conservation of momentum | the total momentum of a system of objects remains constant provided no resultant external force acts on the system |
| elastic collision | a collision in which total kinetic energy is conserved (and the relative speed of approach equals the relative speed of separation) |
| inelastic collision | a collision in which momentum is conserved but some kinetic energy is transferred to other forms |
| terminal velocity | the constant velocity reached when the resistive force on a moving object equals the force driving it, so the resultant force is zero |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The weight 重力 of a large object can be treated as acting at one single point, called the centre of gravity 重心 (the same as the centre of mass 质心 in a uniform gravitational field). For a uniform, regular shape — a rectangle, a sphere, a uniform rod — the centre of gravity is at the middle.
When you draw a free-body diagram 受力图, always put the weight arrow at the centre of gravity.
For an irregular shape, find the centre of gravity by experiment. Hang the object freely from a pin: in equilibrium it settles with its centre of gravity vertically below the pin, because only then does the weight have no moment about the pin. A plumb line 铅垂线 hung from the same pin marks that vertical. Hang the object from a second point and draw the second line; the centre of gravity is where the two lines cross. The same fact answers the exam's "explain why the sheet hangs like this": its centre of gravity is directly below the support, so the weight's line of action passes through the pivot and there is no resultant torque.
For a non-uniform 非均匀 bar, the centre of gravity is not at the middle. Find it by balancing the bar on a pivot, or by taking moments: if the bar's weight $W$ balances a known weight, the distance of its centre of gravity from the pivot is the one moment that makes the two sides equal.
The moment 力矩 of a force about a point is
where $F$ is the size of the force 力 and $d$ is the perpendicular distance 垂直距离 from the point to the line of action 作用线 of the force. Unit: $\text{N m}$.
If the force acts at angle $\theta$ to a lever arm 力臂 of length $r$ from the pivot 支点, then $d = r\sin\theta$, so $M = F r\sin\theta$. Only the part of the force perpendicular 垂直 to the lever arm makes it turn.
Worked example. A force of $50\ \text{N}$ is applied to the end of a spanner $0.40\ \text{m}$ long, at $30°$ to the spanner. Find the moment of the force about the nut.
A moment is either clockwise 顺时针 or anticlockwise 逆时针 about the chosen point.
A couple 力偶 is a pair of forces that are:
A couple makes the body turn only — its resultant force is zero, so it gives no straight-line acceleration.
The torque 力偶矩 of a couple is the turning effect it makes:
where $F$ is the size of one force and $d$ is the perpendicular distance between the two lines of action. Unit: $\text{N m}$. The torque is the same about any point — a special property of couples.
A common multiple-choice trap: two equal forces in the same direction are not a couple (they have a resultant force and cause translation 平动). A couple needs equal size and opposite direction.
Watch the distance in the formula. For a couple of two forces $F$ each acting a distance $d$ from a central pivot, on opposite sides, the perpendicular distance between the two lines of action is $2d$, so the torque is $F \times 2d = 2Fd$. Taking moments about the pivot gives the same answer, $Fd + Fd$. The two-mark definition wants both parts: the product of one of the forces and the perpendicular distance between the lines of action of the forces.
Berat benda besar dapat dianggap bekerja pada satu titik tunggal, disebut pusat gravitasi (sama dengan pusat massa dalam medan gravitasi seragam). Untuk bentuk seragam dan teratur — persegi panjang, bola, batang seragam — pusat gravitasinya berada di tengah.
Saat menggambar diagram benda-bebas, selalu letakkan panah berat di pusat gravitasi.
Untuk bentuk tidak beraturan, tentukan pusat gravitasi melalui eksperimen. Gantungkan benda secara bebas dari sebuah pin: dalam kesetimbangan, benda akan stabil dengan pusat gravitasinya secara vertikal tepat di bawah pin, karena hanya dengan demikian berat tidak memiliki momen terhadap pin. Sebuah benang plumb yang digantung dari pin yang sama menandai garis vertikal tersebut. Gantungkan benda dari titik kedua dan gambar garis kedua; pusat gravitasi berada di persimpangan kedua garis. Fakta yang sama menjawab soal ujian "jelaskan mengapa lembaran menggantung seperti itu": pusat gravitasinya berada tepat di bawah penyangga, sehingga garis kerja berat melewati poros dan tidak ada torsi resultan.
Untuk batang tidak seragam, pusat gravitasinya tidak berada di tengah. Tentukan dengan menyeimbangkan batang pada poros, atau dengan menghitung momen: jika berat batang $W$ menyeimbangkan berat yang diketahui, jarak pusat gravitasinya dari poros adalah momen yang membuat kedua sisi setara.
Momen gaya terhadap suatu titik adalah
di mana $F$ adalah besaran gaya dan $d$ adalah jarak tegak lurus dari titik ke garis kerja gaya. Satuan: $\text{N m}$.
Jika gaya bekerja pada sudut $\theta$ terhadap tongkat pengungkit sepanjang $r$ dari poros, maka $d = r\sin\theta$, sehingga $M = F r\sin\theta$. Hanya komponen gaya yang tegak lurus terhadap tongkat pengungkit yang membuatnya berputar.
Contoh terpecahkan. Gaya sebesar $50\ \text{N}$ diterapkan pada ujung kunci pas sepanjang $0.40\ \text{m}$, pada sudut $30°$ terhadap kunci pas. Hitunglah momen gaya terhadap mur.
Momen bisa searah jarum jam atau berlawanan arah jarum jam terhadap titik yang dipilih.
Sepasang gaya adalah sepasang gaya yang:
Sepasang gaya menyebabkan benda hanya berputar — resultan gayanya nol, sehingga tidak menghasilkan percepatan linear.
Torsi dari sepasang gaya adalah efek putarannya:
di mana $F$ adalah besaran salah satu gaya dan $d$ adalah jarak tegak lurus antara kedua garis kerja. Satuan: $\text{N m}$. Torsinya sama terhadap titik manapun — sifat khusus dari pasangan gaya.
*Pasangan gaya: dua gaya yang sama besar dan berlawanan, terpisah jarak, menghasilkan torsi
Jebakan pilihan ganda yang umum: dua gaya yang sama besar dalam arah yang sama bukan merupakan pasangan gaya (karena memiliki resultan gaya dan menyebabkan translasi). Pasangan gaya memerlukan besaran yang sama dan arah berlawanan.
Perhatikan jarak dalam rumus. Untuk pasangan gaya dari dua gaya $F$ masing-masing bekerja pada jarak $d$ dari engsel pusat, di sisi yang berlawanan, jarak tegak lurus antara kedua garis kerja adalah $2d$, sehingga torsinya adalah $F \times 2d = 2Fd$. Mengambil momen tentang engsel memberikan jawaban yang sama, $Fd + Fd$. Definisi bernilai dua poin membutuhkan kedua bagian tersebut: hasil kali salah satu gaya dan jarak tegak lurus antara garis kerja kedua gaya.
Put a weight on each side and slide it in or out. A small weight far from the pivot can balance a big weight close in — the beam is level when force × distance matches on both sides. · Letakkan beban di masing-masing sisi dan geser masuk atau keluar. Beban kecil yang jauh dari tumpu dapat menyeimbangkan beban besar yang dekat — batang mendatar ketika gaya × jarak seimbang di kedua sisi.
| English | Bahasa Indonesia |
|---|---|
| force/fɔːs/ | gaya |
| moment/ˈməʊmənt/ | momen |
| centre of gravity/ˈsentə ɒv ˈɡrævɪti/ | pusat gravitasi |
| weight/weɪt/ | berat |
| centre of mass/ˈsentə ɒv mæs/ | pusat massa |
| free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ | diagram benda bebas |
| torque/tɔːk/ | tork |
| non-uniform/nɒn ˈjuːnɪfɔːm/ | tidak seragam |
| perpendicular distance/ˌpɜːpənˈdɪkjʊlə ˈdɪstəns/ | jarak tegak lurus |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | tegak lurus |
| lever arm/ˈliːvə ɑːm/ | lengan tuas |
| horizontal/ˌhɒrɪˈzɒntl/ | horizontal |
| couple/ˈkʌpl/ | kopel |
| resultant force/rɪˈzʌltənt fɔːs/ | resultan gaya |
| translation/trænˈsleɪʃn/ | terjemahan |
| angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ | pecutan sudut |
| tension/ˈtenʃn/ | tegangan |
| component/kəmˈpəʊnənt/ | komponen |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A body is in equilibrium 平衡 when:
Both must hold. A body with no resultant force can still be turning; a body with no resultant moment can still be moving in a straight line.
The classic multiple-choice case: four forces, two equal pairs pointing in opposite directions but not along the same lines. The resultant force is zero, so the centre of mass does not accelerate, but the pairs form couples with a resultant torque, so the object spins faster and faster while its centre of mass moves at constant velocity. Only when both conditions hold is the body in equilibrium.
For a body that is not turning, the total clockwise moment about any point equals the total anticlockwise moment about the same point. This is the principle of moments 力矩原理.
To solve a balance problem:
A ruler balanced on a pivot with masses on each side is solved this way. For a heavy uniform rod, remember to include its weight acting at its centre of gravity. Stated for the marks: for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.
Worked example. A uniform beam of length $4.0\ \text{m}$ is pivoted at its centre. A $40\ \text{N}$ weight hangs $1.5\ \text{m}$ from the pivot on one side. How far from the pivot, on the other side, must a $30\ \text{N}$ weight hang to balance the beam?
The beam's own weight acts at the centre (the pivot), so it has no moment. Setting clockwise $=$ anticlockwise moments:
Worked example. A uniform rod of length $L$ and weight $W = 28\ \text{N}$ is hinged to a wall at one end and held horizontal by a wire from its other end. The wire makes $40°$ with the rod. Find the tension $T$ in the wire.
Take moments about the hinge, so the unknown hinge force drops out. The weight acts at the centre, a distance $L/2$ from the hinge: its moment is $W \times L/2$, clockwise. The tension acts at the end, but not at right angles to the rod, so its moment is $T\sin 40° \times L$, anticlockwise (only the component of $T$ perpendicular to the rod turns it). Equate them:
The length $L$ cancels, which is why the question need not give it. If a load hangs from the rod as well, add its moment on the clockwise side. This is the shape of almost every moments question: choose the pivot where the unknown force acts, use the perpendicular component of any angled force, and let unknown lengths cancel.
Worked example. A uniform beam of weight $200\ \text{N}$ and length $9.0\ \text{m}$ rests on two supports, X at $1.0\ \text{m}$ from the left end and Y at $6.5\ \text{m}$. A person of weight $700\ \text{N}$ stands $1.7\ \text{m}$ from Y, beyond it. Find the force from each support.
Two unknown forces, $R_{\text{X}}$ and $R_{\text{Y}}$, so take moments about X to remove $R_{\text{X}}$. The beam's weight acts at its centre, $3.5\ \text{m}$ from X; Y is $5.5\ \text{m}$ from X; the person is $7.2\ \text{m}$ from X:
Then the resultant force is zero: $R_{\text{X}} + R_{\text{Y}} = 200 + 700$, so $R_{\text{X}} = 900 - 1040 = -140\ \text{N}$. A negative answer means the beam would lift off X: with the person that far past Y, X must hold the beam down, not up. Questions often ask where a person can stand before the beam tips; that is the position at which $R_{\text{X}}$ becomes zero.
Three forces in the same plane that are in equilibrium can be drawn as a closed vector triangle 矢量三角形 — drawn tip-to-tail, the three arrows come back to the start. This is a drawing method instead of splitting into components 分量.
Use the sine rule 正弦定理 or the cosine rule 余弦定理 on the triangle to find unknown sizes or directions, or draw the triangle to scale on graph paper.
You can also split each force into horizontal 水平 and vertical 竖直 components and set $\sum F_{x} = 0$ and $\sum F_{y} = 0$.
To draw the triangle for the marks: draw the weight first (vertical, to scale), then add the other two forces tip to tail in their real directions, so the third arrow closes the triangle; label every side with its force and mark the angles. For a block resting on a slope the three forces are the weight (vertical), the normal contact force (perpendicular to the slope) and friction (along the slope), so the triangle is right-angled with the weight as the hypotenuse. For a picture hanging from a string over a pin, the two equal tensions make an isosceles triangle with the weight. The test of any answer is that the triangle closes; if it does not, the object is not in equilibrium.
Worked example. A pulley of negligible weight is held by a spring. A single cable passes under the pulley, and each side of the cable makes $30°$ with the vertical. The tension in the cable is $T = 60\ \text{N}$. Find the force from the spring and the extension of the spring, given its spring constant is $2000\ \text{N m}^{-1}$.
Because it is one continuous cable, the tension is the same on both sides. The two tensions pull down and outwards; their horizontal components cancel and their vertical components add, so the spring must pull up with $F = 2T\cos 30° = 2 \times 60 \times 0.866 = 104\ \text{N}$. From Hooke's law (topic 6), $x = F / k = 104 / 2000 = 0.052\ \text{m}$. If the angle to the vertical grows, $\cos\theta$ falls, so the spring force and extension fall: the cable is pulling more sideways and less upwards.
Sebuah benda berada dalam keseimbangan ketika:
Keduanya harus terpenuhi. Benda tanpa resultan gaya masih bisa berputar; benda tanpa resultan momen masih bisa bergerak lurus.
Kasus pilihan ganda klasik: empat gaya, dua pasangan sama besar mengarah ke arah berlawanan tetapi tidak pada garis kerja yang sama. Resultan gayanya nol, sehingga pusat massa tidak mengalami percepatan, namun pasangan-pasangan tersebut membentuk pasangan gaya dengan resultan torsi, sehingga benda berputar semakin cepat sementara pusat massanya bergerak dengan kecepatan konstan. Hanya ketika kedua syarat terpenuhi benda tersebut berada dalam keseimbangan.
Untuk benda yang tidak berputar, total momen searah jarum jam tentang titik mana pun sama dengan total momen berlawanan arah jarum jam tentang titik yang sama. Ini disebut prinsip momen.
Untuk menyelesaikan masalah kesetimbangan:
Penggaris yang seimbang di atas engsel dengan massa di kedua sisinya diselesaikan dengan cara ini. Untuk batang seragam yang berat, ingatlah untuk menyertakan beratnya yang bekerja di pusat gravitasinya. Ditulis untuk nilai soal: untuk benda dalam keseimbangan, jumlah momen searah jarum jam tentang titik mana pun sama dengan jumlah momen berlawanan arah jarum jam tentang titik yang sama.
Contoh pengerjaan. Balok seragam sepanjang $4.0\ \text{m}$ dipivot di tengahnya. Beban $40\ \text{N}$ digantung $1.5\ \text{m}$ dari pivot di satu sisi. Berapa jauh dari pivot, di sisi lainnya, beban $30\ \text{N}$ harus digantung untuk menyeimbangkan balok?
Berat batang sendiri bekerja di tengah (di engsel), sehingga tidak memiliki momen. Menyetel momen searah jarum jam $=$ momen berlawanan arah jarum jam:
Contoh terpecahkan. Batang seragam sepanjang $L$ dan berat $W = 28\ \text{N}$ disambung ke dinding di satu ujung dan dijaga tetap horizontal oleh kawat dari ujung lainnya. Kawat tersebut membentuk sudut $40°$ terhadap batang. Tentukan tegangan $T$ dalam kawat tersebut.
Ambil momen tentang engsel, sehingga gaya engsel yang tak diketahui hilang. Berat bekerja di tengah, sejauh $L/2$ dari engsel: momennya adalah $W \times L/2$, searah jarum jam. Tegangan bekerja di ujung, tetapi tidak tegak lurus terhadap batang, sehingga momennya adalah $T\sin 40° \times L$, berlawanan arah jarum jam (hanya komponen dari $T$ yang tegak lurus terhadap batang yang memutarkannya). Samakan keduanya:
Panjang $L$ saling menghilangkan, itulah sebabnya pertanyaan tidak perlu memberikannya. Jika ada beban yang digantung dari batang juga, tambahkan momennya di sisi searah jarum jam. Ini adalah pola hampir setiap soal momen: pilih engsel tempat gaya tak diketahui bekerja, gunakan komponen tegak lurus dari gaya miring apa pun, dan biarkan panjang tak diketahui saling menghilangkan.
Contoh terpecahkan. Balok seragam beratnya $200\ \text{N}$ dan panjangnya $9.0\ \text{m}$ bertumpu pada dua penyangga, X pada $1.0\ \text{m}$ dari ujung kiri dan Y pada $6.5\ \text{m}$. Seseorang beratnya $700\ \text{N}$ berdiri $1.7\ \text{m}$ dari Y, melewati posisinya. Tentukan gaya dari masing-masing penyangga.
Dua gaya tak diketahui, $R_{\text{X}}$ dan $R_{\text{Y}}$, jadi ambil momen terhadap X untuk menghilangkan $R_{\text{X}}$. Berat balok bekerja di tengahnya, $3.5\ \text{m}$ dari X; Y terletak $5.5\ \text{m}$ dari X; orang tersebut terletak $7.2\ \text{m}$ dari X:
Kemudian resultan gayanya nol: $R_{\text{X}} + R_{\text{Y}} = 200 + 700$, sehingga $R_{\text{X}} = 900 - 1040 = -140\ \text{N}$. Jawaban negatif berarti balok akan terangkat dari X: dengan orang itu sedemikian jauh melewati Y, X harus menahan balok ke bawah, bukan ke atas. Soal sering bertanya di mana seseorang bisa berdiri sebelum balok terbalik; itu adalah posisi di mana $R_{\text{X}}$ menjadi nol.
Tiga gaya dalam bidang yang sama dan berada dalam kesetimbangan dapat digambarkan sebagai segitiga vektor tertutup — digambar ujung ke ekor, ketiga panah kembali ke titik awal. Ini adalah metode gambar alih-alih memecahnya menjadi komponen.
Gunakan aturan sinus atau aturan kosinus pada segitiga untuk mencari besaran atau arah yang tidak diketahui, atau gambarlah segitiga dengan skala pada kertas grafik.
Anda juga dapat memecah setiap gaya menjadi komponen horizontal dan vertikal serta menetapkan $\sum F_{x} = 0$ dan $\sum F_{y} = 0$.
Untuk menggambar segitiga untuk soal: gambarlah berat terlebih dahulu (vertikal, sesuai skala), lalu tambahkan dua gaya lainnya ujung ke ekor dalam arah aslinya, sehingga panah ketiga menutup segitiga; beri label setiap sisi dengan gayanya dan tandai sudutnya. Untuk balok yang diam di atas bidang miring, ketiga gayanya adalah berat (vertikal), gaya kontak normal (tegak lurus terhadap bidang miring) dan gesekan (sejajar dengan bidang miring), sehingga segitiganya siku-siku dengan berat sebagai hipotenusa. Untuk lukisan yang tergantung pada tali di atas paku, dua tegangan yang sama membentuk segitiga sama kaki dengan berat. Uji dari setiap jawaban adalah bahwa segitiga tersebut tertutup; jika tidak, benda tersebut tidak dalam kesetimbangan.
Contoh terpecahkan. Katrol dengan berat yang dapat diabaikan dipegang oleh sebuah pegas. Sebuah kabel tunggal melewati bawah katrol, dan masing-masing sisi kabel membuat sudut $30°$ terhadap vertikal. Tegangan dalam kabel adalah $T = 60\ \text{N}$. Tentukan gaya dari pegas dan perpanjangan pegas, given konstanta pegasnya adalah $2000\ \text{N m}^{-1}$.
Karena ini adalah kabel yang berkelanjutan, tegangan sama di kedua sisi. Dua tegangan menarik ke bawah dan ke luar; komponen horizontalnya saling meniadakan dan komponen vertikalnya menjumlah, sehingga pegas harus menarik ke atas dengan $F = 2T\cos 30° = 2 \times 60 \times 0.866 = 104\ \text{N}$. Dari hukum Hooke (topik 6), $x = F / k = 104 / 2000 = 0.052\ \text{m}$. Jika sudut terhadap vertikal membesar, $\cos\theta$ turun, sehingga gaya pegas dan perpanjangannya turun: kabel menarik lebih banyak ke samping dan kurang ke atas.
When forces are balanced the resultant is zero — the vectors form a closed loop. Drag the arrows to keep them cancelling. · Ketika gaya seimbang, resultannya nol — vektor-vektornya membentuk loop tertutup. Seret panah agar tetap saling meniadakan.
| English | Bahasa Indonesia |
|---|---|
| principle of moments/ˈprɪnsɪpl ɒv ˈməʊmənts/ | prinsip momen |
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | kesetimbangan |
| plumb line/plʌm laɪn/ | garis plumb |
| vertical/ˈvɜːtɪkl/ | vertikal |
| line of action/laɪn ɒv ˈækʃn/ | garis aksi |
| pivot/ˈpɪvət/ | engsel |
| clockwise/ˈklɒkwaɪz/ | searah jarum jam |
| anticlockwise/ˌæntɪˈklɒkwaɪz/ | berlawanan arah jarum jam |
| vector triangle/ˈvektə ˈtraɪæŋɡl/ | segitiga vektor |
| closed vector triangle/kləʊzd ˈvektə ˈtraɪæŋɡl/ | segitiga vektor tertutup |
| sine rule/saɪn ruːl/ | aturan sinus |
| cosine rule/ˈkəʊsaɪn ruːl/ | aturan kosinus |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Density 密度 is the mass per unit volume:
Unit: $\text{kg m}^{-3}$ (or $\text{g cm}^{-3}$; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$). Density is a scalar 标量.
Some useful densities to know:
Measuring density is a practical question in disguise. For a cuboid, measure the mass on a balance and the three side lengths with a rule or calipers, then $\rho = m / (xyz)$. Because density is a product and quotient, the percentage uncertainty in $\rho$ is the sum of the percentage uncertainties in $m$, $x$, $y$ and $z$ (topic 1). A zero error on the balance, or a rule whose end is worn, is a systematic error; repeating the readings does not remove it, but subtracting the zero reading does.
Worked example. A block has mass $(120 \pm 1)\ \text{g}$ and sides $(4.0 \pm 0.1)\ \text{cm}$, $(3.0 \pm 0.1)\ \text{cm}$ and $(2.0 \pm 0.1)\ \text{cm}$. Find its density and the percentage uncertainty.
$\rho = 120 / (4.0 \times 3.0 \times 2.0) = 5.0\ \text{g cm}^{-3} = 5000\ \text{kg m}^{-3}$. The percentage uncertainties are $0.8\%$, $2.5\%$, $3.3\%$ and $5.0\%$, so the density is uncertain by $12\%$, about $\pm 600\ \text{kg m}^{-3}$. The shortest side contributes most, which is why a question asks which measurement to improve first.
Kepadatan adalah massa per satuan volume:
Satuan: $\text{kg m}^{-3}$ (atau $\text{g cm}^{-3}$; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$). Kepadatan adalah besaran skalar.
Beberapa kepadatan yang berguna untuk diingat:
Mengukur kepadatan adalah soal praktik yang disembunyikan. Untuk balok persegi panjang, ukur massanya menggunakan neraca dan tiga panjang sisinya dengan penggaris atau jangka sorong, lalu $\rho = m / (xyz)$. Karena kepadatan adalah hasil kali dan hasil bagi, ketidakpastian persentase dalam $\rho$ adalah jumlah dari ketidakpastian persentase dalam $m$, $x$, $y$ dan $z$ (topik 1). Kesalahan nol pada neraca, atau penggaris yang ujungnya aus, adalah kesalahan sistematis; mengulang pembacaan tidak menghapusnya, tetapi mengurangi pembacaan nol dapat melakukannya.
Contoh terpecahkan. Sebuah balok memiliki massa $(120 \pm 1)\ \text{g}$ dan sisi-sisi $(4.0 \pm 0.1)\ \text{cm}$, $(3.0 \pm 0.1)\ \text{cm}$ dan $(2.0 \pm 0.1)\ \text{cm}$. Temukan kepadatannya dan ketidakpastian persentasenya.
$\rho = 120 / (4.0 \times 3.0 \times 2.0) = 5.0\ \text{g cm}^{-3} = 5000\ \text{kg m}^{-3}$. Ketidakpastian persentasenya adalah $0.8\%$, $2.5\%$, $3.3\%$ dan $5.0\%$, sehingga kepadatannya tidak pasti sebesar $12\%$, sekitar $\pm 600\ \text{kg m}^{-3}$. Sisi terpendek memberikan kontribusi terbesar, itulah sebabnya pertanyaan meminta pengukuran mana yang perlu diperbaiki terlebih dahulu.
| English | Bahasa Indonesia |
|---|---|
| density/ˈdensɪti/ | densitas |
Pressure 压强 is the force per unit area, where the force acts at right angles to the surface:
Unit: $\text{Pa} = \text{N m}^{-2}$. Pressure is a scalar.
The force is the one pressing at right angles to the surface, and the area is the area of contact. A tree trunk of weight $9.0\ \text{kN}$ standing on a post of diameter $0.28\ \text{m}$ presses with $p = F / A = 9000 / (\pi \times 0.14^{2}) = 1.5 \times 10^{5}\ \text{Pa}$; the same weight on a wider post gives a smaller pressure. Convert diameters to radii, and centimetres to metres, before squaring.
Take a column of fluid 流体 with density $\rho$, cross-sectional area 横截面积 $A$ and height $\Delta h$. Its weight is
This weight presses down on the area $A$ at the bottom, so the extra pressure at the bottom compared with the top is
This is the hydrostatic pressure 流体静压强 equation. It depends only on the density of the fluid and the depth 深度 — the shape of the container does not matter.
Worked example. Find the extra pressure due to the water at the bottom of a swimming pool $2.5\ \text{m}$ deep. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)
For a submarine at depth $h$ below the surface, the pressure from the water is $\rho_{\text{seawater}}\, g\, h$. For the total pressure, add the atmospheric pressure 大气压强 at the surface (about $1.0 \times 10^{5}\ \text{Pa}$).
When an object is submerged 浸没 in a fluid, the pressure at the bottom of the object is greater than the pressure at the top (by $\rho g \Delta h$, where $\Delta h$ is the object's height). This difference gives a net upward force called the upthrust 浮力.
For an object of volume $V$ (the volume of fluid displaced 排开), the upthrust is
This is Archimedes' principle 阿基米德原理: the upthrust on a body in a fluid equals the weight of the fluid it pushes aside.
Worked example. A metal block of volume $2.0 \times 10^{-3}\ \text{m}^{3}$ is fully submerged in water. Find the upthrust on it. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)
For a fully submerged object, $V$ is its full volume. For a floating 漂浮 object, $V$ is only the volume below the surface — the object floats when the upthrust on the part below the surface equals its weight.
Whether an object floats at all depends on densities: for a fully submerged object the ratio of upthrust to weight is $\rho_{\text{fluid}} V g / \rho_{\text{object}} V g = \rho_{\text{fluid}} / \rho_{\text{object}}$, so it floats if it is less dense than the fluid and sinks if it is denser. Halving the object's density doubles that ratio; changing its volume or the depth changes nothing.
Worked example. A cylinder of mass $11\ \text{kg}$ and diameter $0.78\ \text{m}$ floats upright in water of density $990\ \text{kg m}^{-3}$. Find the depth $y$ of its base below the surface.
Floating means upthrust $=$ weight. The submerged volume is the base area times the depth, $A y$, with $A = \pi \times 0.39^{2} = 0.478\ \text{m}^{2}$:
Each extra newton of load on the cylinder needs an extra $1 / (\rho g A)$ of depth, so the depth rises in a straight line with the added weight, wherever on the cylinder it is placed.
A block held under water by a string tied to the bottom of the container is in equilibrium under three vertical forces: weight (down), tension 张力 (down), upthrust (up). Set $F_{\text{upthrust}} = W + T$ to find the tension.
A submerged block hanging from a newton meter 弹簧测力计 reads less than its weight in air, because of the upthrust: reading $= W - F_{\text{upthrust}}$.
Worked example. A cylinder of weight $25.0\ \text{N}$ hangs from a newton meter fully submerged in water, and the meter reads $10.0\ \text{N}$. Find the volume of the cylinder.
The upthrust is the missing $15.0\ \text{N}$, and upthrust $= \rho g V$, so $V = 15.0 / (1000 \times 9.81) = 1.53 \times 10^{-3}\ \text{m}^{3}$. The same method finds the density of the cylinder: $25.0 / (9.81 \times 1.53 \times 10^{-3}) = 1670\ \text{kg m}^{-3}$.
The upthrust depends on the fluid density and the displaced volume, not on the object's material or depth (for an incompressible 不可压缩 fluid). On a planet with smaller $g$, the upthrust is smaller in the same ratio as the weight, so a floating object still floats with the same fraction below the surface.
Tekanan adalah gaya per satuan luas, di mana gaya tersebut bekerja tegak lurus terhadap permukaan:
Satuan: $\text{Pa} = \text{N m}^{-2}$. Tekanan adalah besaran skalar.
Gaya adalah gaya yang menekan tegak lurus terhadap permukaan, dan luas adalah area kontak. Batang pohon dengan berat $9.0\ \text{kN}$ berdiri di atas tiang berdiameter $0.28\ \text{m}$ menekan dengan $p = F / A = 9000 / (\pi \times 0.14^{2}) = 1.5 \times 10^{5}\ \text{Pa}$; berat yang sama di atas tiang yang lebih lebar menghasilkan tekanan yang lebih kecil. Ubah diameter menjadi jari-jari, dan sentimeter menjadi meter, sebelum dikuadratkan.
Ambil kolom fluida dengan kepadatan $\rho$, luas penampang $A$ dan tinggi $\Delta h$. Beratnya adalah
Berat ini menekan ke bawah pada area $A$ di bagian bawah, sehingga tambahan tekanan di bagian bawah dibandingkan dengan bagian atas adalah
Ini adalah persamaan tekanan hidrostatik. Ini hanya bergantung pada kepadatan fluida dan kedalaman — bentuk wadah tidak berpengaruh.
Contoh terpecahkan. Temukan tambahan tekanan akibat air di dasar kolam renang $2.5\ \text{m}$ dalam. (Kepadatan air $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)
Untuk kapal selam pada kedalaman $h$ di bawah permukaan, tekanan dari air adalah $\rho_{\text{seawater}}\, g\, h$. Untuk tekanan total, tambahkan tekanan atmosfer di permukaan (sekitar $1.0 \times 10^{5}\ \text{Pa}$).
Ketika suatu benda tercelup dalam fluida, tekanan di bagian bawah benda lebih besar daripada tekanan di bagian atas (selisih $\rho g \Delta h$, di mana $\Delta h$ adalah tinggi benda). Perbedaan ini menghasilkan gaya bersih ke atas yang disebut gaya apung.
Untuk benda dengan volume $V$ (volume fluida yang dipindahkan), gaya apungnya adalah
Ini adalah prinsip Archimedes: gaya apung pada sebuah benda dalam fluida sama dengan berat fluida yang dipindahkannya.
Contoh terpecahkan. Sebuah balok logam dengan volume $2.0 \times 10^{-3}\ \text{m}^{3}$ dicelupkan sepenuhnya dalam air. Tentukan gaya apung yang bekerja padanya. (Kerapatan air $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)
Untuk benda yang tercelup sepenuhnya, $V$ adalah volumenya seluruhnya. Untuk benda yang mengapung, $V$ hanyalah volume di bawah permukaan — benda mengapung ketika gaya apung pada bagian di bawah permukaan sama dengan beratnya.
Apakah suatu benda dapat mengapung atau tidak bergantung pada kerapatan: untuk benda yang tercelup sepenuhnya rasio antara gaya apung terhadap berat adalah $\rho_{\text{fluid}} V g / \rho_{\text{object}} V g = \rho_{\text{fluid}} / \rho_{\text{object}}$, sehingga benda tersebut akan mengapung jika kerapatannya lebih rendah dari fluida dan tenggelam jika lebih tinggi. Mengurangi setengah kerapatan benda menggandakan rasio tersebut; mengubah volumenya atau kedalaman tidak mengubah apa-apa.
Contoh terpecahkan. Sebuah silinder dengan massa $11\ \text{kg}$ dan diameter $0.78\ \text{m}$ mengapung tegak di dalam air dengan kerapatan $990\ \text{kg m}^{-3}$. Tentukan kedalaman $y$ dasar silinder di bawah permukaan.
Mengapung berarti gaya apung $=$ berat. Volume yang tercelup adalah luas alas dikalikan kedalaman, $A y$, dengan $A = \pi \times 0.39^{2} = 0.478\ \text{m}^{2}$:
Setiap新增 newton beban pada silinder memerlukan tambahan $1 / (\rho g A)$ kedalaman, sehingga kedalaman meningkat secara linear seiring bertambahnya beban, di mana pun beban tersebut diletakkan pada silinder.
Sebuah balok yang diikat di bawah air oleh tali yang diikatkan ke dasar wadah berada dalam kesetimbangan di bawah tiga gaya vertikal: berat (ke bawah), tegangan (ke bawah), gaya apung (ke atas). Tetapkan $F_{\text{upthrust}} = W + T$ untuk menemukan tegangan.
Sebuah balok tercelup yang digantung pada neraca pegas menunjukkan bacaan kurang dari beratnya di udara, karena adanya gaya apung: bacaan $= W - F_{\text{upthrust}}$.
Contoh terpecahkan. Sebuah silinder dengan berat $25.0\ \text{N}$ digantung pada neraca pegas dan dicelupkan sepenuhnya dalam air, dan meter menunjukkan $10.0\ \text{N}$. Tentukan volume silinder tersebut.
Gaya apung adalah kekurangan $15.0\ \text{N}$, dan gaya apung $= \rho g V$, sehingga $V = 15.0 / (1000 \times 9.81) = 1.53 \times 10^{-3}\ \text{m}^{3}$. Metode yang sama digunakan untuk mencari kerapatan silinder: $25.0 / (9.81 \times 1.53 \times 10^{-3}) = 1670\ \text{kg m}^{-3}$.
Gaya apung bergantung pada kerapatan fluida dan volume yang dipindahkan, bukan pada material benda atau kedalaman (untuk fluida tak termampatkan). Di planet dengan gravitasi $g$ yang lebih kecil, gaya apung juga lebih kecil dalam rasio yang sama dengan berat, sehingga benda yang mengapung tetap mengapung dengan fraksi yang sama di bawah permukaan.
p = ρg·h
Pressure is proportional to depth — the gradient is ρg. · Tekanan berbanding lurus dengan kedalaman — gradiennya adalah ρg.
| English | Bahasa Indonesia |
|---|---|
| floating/ˈfləʊtɪŋ/ | mengambang |
| scalar/ˈskeɪlə/ | skalar |
| pressure/ˈpreʃə/ | tekanan |
| atmospheric pressure/ˌætməsˈferɪk ˈpreʃə/ | tekanan atmosfer |
| hydrostatic pressure/ˌhaɪdrəˈstætɪk ˈpreʃə/ | tekanan hidrostatik |
| fluid/ˈfluːɪd/ | fluida |
| cross-sectional area/krɒs ˈsekʃənl ˈeərɪə/ | luas penampang |
| depth/depθ/ | kedalaman |
| upthrust/ˈʌpθrʌst/ | gaya apung |
| Archimedes' principle/ˌɑːkɪˈmiːdiːz ˈprɪnsɪpl/ | Prinsip Archimedes |
| submerged/səbˈmɜːdʒd/ | tercelup |
| displaced/dɪˈspleɪst/ | dipindahkan |
| newton meter/ˈnjuːtn ˈmiːtə/ | newton meter |
| incompressible/ˌɪnkəmˈpresəbl/ | tak termampatkan |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| centre of gravity | the point at which the whole weight of an object may be considered to act |
| moment of a force | the product of the force and the perpendicular distance from the point to the line of action of the force |
| couple | a pair of equal and opposite forces whose lines of action do not coincide, which produces rotation only |
| torque of a couple | the product of one of the forces and the perpendicular distance between the lines of action of the forces |
| principle of moments | for a body in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point |
| equilibrium | the state of a body on which the resultant force and the resultant torque are both zero |
| density | mass per unit volume |
| pressure | force per unit area, where the force acts at right angles to the surface |
| upthrust | the upward force on a body in a fluid caused by the pressure being greater on its lower surface than on its upper surface |
| Archimedes' principle | the upthrust on a body in a fluid is equal to the weight of the fluid displaced by the body |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| titik berat | titik di mana seluruh berat benda dapat dianggap bekerja |
| momen gaya | hasil kali gaya dan jarak tegak lurus dari titik ke garis kerja gaya |
| sepasang gaya | sepasang gaya yang sama besar dan berlawanan arah whose lines of action do not coincide, which produces rotation only |
| torsi dari sepasang gaya | hasil kali salah satu gaya dan jarak tegak lurus antara garis kerja kedua gaya tersebut |
| prinsip momen | untuk benda dalam kesetimbangan, jumlah momen searah jarum jam tentang titik manapun sama dengan jumlah momen berlawanan arah jarum jam tentang titik itu |
| kesetimbangan | keadaan benda di mana resultan gaya dan resultan torsi keduanya nol |
| kerapatan | massa per satuan volume |
| tekanan | gaya per satuan area, di mana gaya bekerja tegak lurus terhadap permukaan |
| gaya apung | gaya ke atas pada benda dalam fluida yang disebabkan oleh tekanan yang lebih besar pada permukaannya yang bawah dibandingkan permukaannya yang atas |
| prinsip Archimedes | gaya apung pada benda dalam fluida sama dengan berat fluida yang dipindahkan oleh benda tersebut |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Work 功 is done when a force moves its point of contact along the line of the force. For the one-mark definition write: work done is the product of the force and the distance moved in the direction of the force. The work done by a constant force $F$ that causes a displacement 位移 $s$ is
where $\theta$ is the angle between the force and the displacement. Only the component 分量 of the force along the displacement does work.
Worked example. A child pulls a sledge $5.0\ \text{m}$ across the snow with a rope, using a force of $20\ \text{N}$ at $60°$ to the ground. Find the work done by the rope.
Unit: $\text{J} = \text{N m}$. Work is a scalar 标量.
Special cases:
For an object moving up a slope at angle $\alpha$ to the horizontal 水平, the work done against gravity in rising a height $h$ is $mgh$, while the work done by a horizontal push over the slope length $L$ uses $\cos\alpha$.
Two multiple-choice tests of the definition: no work is done on an object that slides at constant velocity along a frictionless surface (no force along the motion), or on one that is simply held still (no distance moved); work is done when a force lifts it. And when a box is pushed at constant velocity across a rough floor, the work done by the push is $F \times d$ for the whole distance, however the path is split into stages.
Worked example. A block of mass $4.9\ \text{kg}$ is pushed at constant velocity up a rough slope of length $8.0\ \text{m}$ that rises $1.5\ \text{m}$, by a force of $12\ \text{N}$ acting along the slope. Find the work done against friction.
Constant velocity means no change in kinetic energy, so the work done by the push is shared between the gain in gravitational potential energy and the work done against friction. Work by the push: $12 \times 8.0 = 96\ \text{J}$. Gain in GPE, using the vertical rise: $4.9 \times 9.81 \times 1.5 = 72\ \text{J}$. So the work done against friction is $96 - 72 = 24\ \text{J}$, and the friction force is $24 / 8.0 = 3.0\ \text{N}$. Using the slope length in $mgh$ is the commonest error on this question.
Energy is never made or destroyed — it only changes from one form to another, or moves from one object to another. In a closed system 封闭系统, the total energy stays constant. This is conservation of energy 能量守恒, stated for the marks as: energy cannot be created or destroyed; it can only be transferred from one form to another (so the total energy of a closed system is constant).
When you write an energy equation, list every form the energy starts as and ends as. Common forms in this syllabus: kinetic, gravitational potential, elastic potential energy 弹性势能, electrical 电能, thermal, sound, chemical energy 化学能.
A ball rolling down a frictionless 无摩擦 ramp 斜坡 turns gravitational potential energy 重力势能 into kinetic energy 动能: $mgh = \tfrac{1}{2} m v^{2}$, so $v = \sqrt{2gh}$. With friction, some of this energy becomes thermal energy 热能 of the ramp and the air.
Worked example. A ball is released from rest at the top of a smooth ramp $1.2\ \text{m}$ high. Find its speed at the bottom (take $g = 9.81\ \text{m s}^{-2}$).
All the gravitational potential energy becomes kinetic energy, so $v = \sqrt{2gh}$ (the mass cancels):
The same idea reads a graph. For a ball falling from height $H$ with air resistance negligible, the GPE falls in a straight line with height ($mgh$), the KE rises in a straight line ($mg(H - h)$), and the two add to the constant $mgH$ at every height.
The energy chain can pass through a spring. A block that hits a spring with kinetic energy $110\ \text{J}$ while sliding a little downhill, so that its GPE falls by a further $20\ \text{J}$ before the spring stops it, stores $110 + 20 = 130\ \text{J}$ of elastic potential energy at the greatest compression, if nothing is lost to friction. The spring's force–compression graph then gives the compression: the energy stored is the area under the line, $\tfrac{1}{2} F_{0} x_{0}$ for a spring that obeys Hooke's law (topic 6), so a maximum force of $2600\ \text{N}$ means $x_{0} = 2 \times 130 / 2600 = 0.10\ \text{m}$. At that instant the resultant force on the block is the spring force minus the component of the weight down the slope, and its acceleration follows from $F = ma$ (topic 3).
The efficiency 效率 of a system is
The same idea with power:
Efficiency is always less than 100% in a real system, because some input energy becomes "useless" forms — usually thermal energy. If a total energy $E$ is supplied and an amount $Q$ is wasted, the useful output is $E - Q$ and the efficiency is $(E - Q) / E$; a motor rated at $1.7\ \text{kW}$ input with efficiency $53\%$ delivers $0.53 \times 1.7 = 0.90\ \text{kW}$ of useful output.
For an electric motor lifting a load with efficiency $\eta$ at voltage 电压 $V$ and current 电流 $I$, the useful output power is $\eta V I$. From this you can find a force, a lifting speed, or a tension 张力.
Worked example. An electric motor lifts a $50\ \text{kg}$ load at a steady $0.40\ \text{m s}^{-1}$ while drawing $250\ \text{W}$ of electrical power. Find its efficiency (take $g = 9.81\ \text{m s}^{-2}$).
The useful output power is $P = mgv = 50 \times 9.81 \times 0.40 = 196\ \text{W}$, so
Worked example. A motor raises a block with a force of $240\ \text{N}$ through a vertical distance of $12\ \text{m}$ in $60\ \text{s}$. The input power to the motor is $900\ \text{W}$. Find the useful output power and the efficiency.
Work done on the block: $W = Fd = 240 \times 12 = 2880\ \text{J}$. Useful output power: $P = W / t = 2880 / 60 = 48\ \text{W}$. Efficiency: $48 / 900 \times 100\% = 5.3\%$. Most of the input goes into thermal energy in the motor and its cable. Turned round, a lift motor of useful output $20\ \text{kW}$ raising $1500\ \text{kg}$ through $20\ \text{m}$ takes $t = mgh / P = 1500 \times 9.81 \times 20 / 20\,000 = 15\ \text{s}$.
Usaha dilakukan ketika sebuah gaya menggeser titik tangkapnya sepanjang garis kerja gayanya. Untuk definisi bernilai satu poin, tuliskan: usaha adalah hasil kali gaya dan jarak yang ditempuh searah dengan gaya. Usaha yang dilakukan oleh gaya konstan $F$ yang menyebabkan perpindahan $s$ adalah
di mana $\theta$ adalah sudut antara gaya dan perpindahan. Hanya komponen gaya yang sejajar dengan perpindahan yang melakukan usaha.
Contoh terlatih. Seorang anak menarik sled $5.0\ \text{m}$ melintasi salju dengan tali, menggunakan gaya sebesar $20\ \text{N}$ pada sudut $60°$ terhadap tanah. Hitunglah usaha yang dilakukan oleh tali tersebut.
Satuan: $\text{J} = \text{N m}$. Usaha adalah besaran skalar.
Kasus khusus:
Untuk benda bergerak menaiki lereng dengan sudut $\alpha$ terhadap horizontal, usaha melawan gravitasi saat naik setinggi $h$ adalah $mgh$, sedangkan usaha dari dorongan horizontal sepanjang panjang lereng $L$ menggunakan $\cos\alpha$.
Dua uji pilihan ganda mengenai definisi: tidak ada usaha yang dilakukan pada benda yang meluncur dengan kecepatan konstan di atas permukaan tanpa gesekan (tidak ada gaya searah gerak), atau pada benda yang sekadar dipegang diam (tidak ada jarak yang ditempuh); usaha dilakukan ketika gaya mengangkatnya. Dan ketika kotak didorong dengan kecepatan konstan melintasi lantai kasar, usaha yang dilakukan oleh dorongan adalah $F \times d$ untuk seluruh jarak, terlepas dari bagaimana lintasan dibagi menjadi tahap-tahap.
Contoh terlatih. Sebuah balok bermassa $4.9\ \text{kg}$ didorong dengan kecepatan konstan menaiki lereng kasar sepanjang $8.0\ \text{m}$ yang memiliki ketinggian vertikal $1.5\ \text{m}$, dengan gaya sebesar $12\ \text{N}$ yang bekerja sepanjang lereng. Hitunglah usaha yang dilakukan melawan gesekan.
Kecepatan konstan berarti tidak ada perubahan energi kinetik, sehingga usaha dari dorongan terbagi antara peningkatan energi potensial gravitasi dan usaha yang dilakukan melawan gesekan. Usaha oleh dorongan: $12 \times 8.0 = 96\ \text{J}$. Peningkatan GPE, menggunakan kenaikan vertikal: $4.9 \times 9.81 \times 1.5 = 72\ \text{J}$. Jadi usaha yang dilakukan melawan gesekan adalah $96 - 72 = 24\ \text{J}$, dan gaya gesek adalah $24 / 8.0 = 3.0\ \text{N}$. Menggunakan panjang lereng dalam $mgh$ adalah kesalahan paling umum pada soal ini.
Energi tidak pernah diciptakan atau dimusnahkan — ia hanya berubah dari satu bentuk ke bentuk lain, atau berpindah dari satu benda ke benda lain. Dalam sistem tertutup, total energi tetap konstan. Ini adalah kekekalan energi, dinyatakan untuk nilai poin sebagai: energi tidak dapat diciptakan atau dimusnahkan; ia hanya dapat ditransfer dari satu bentuk ke bentuk lain (sehingga total energi sistem tertutup adalah konstan).
Saat Anda menulis persamaan energi, daftar setiap bentuk energi mulai dari dan berakhir pada. Bentuk-bentuk umum dalam silabus ini: kinetik, potensial gravitasi, energi potensial elastis, listrik, termal, bunyi, energi kimia.
Bola yang bergulir menuruni ramp tanpa gesekan mengubah energi potensial gravitasi menjadi energi kinetik: $mgh = \tfrac{1}{2} m v^{2}$, sehingga $v = \sqrt{2gh}$. Dengan adanya gesekan, sebagian energi ini menjadi energi termal ramp dan udara.
Contoh terlatih. Sebuah bola dilepas dari keadaan diam di puncak ramp licin $1.2\ \text{m}$ tingginya. Hitunglah kecepatannya di bagian bawah (ambil $g = 9.81\ \text{m s}^{-2}$).
Semua energi potensial gravitasi berubah menjadi energi kinetik, sehingga $v = \sqrt{2gh}$ (massa saling menghilangkan):
Ide yang sama dapat dibaca melalui grafik. Untuk bola jatuh dari ketinggian $H$ dengan hambatan udara diabaikan, GPE turun secara linear terhadap ketinggian ($mgh$), KE naik secara linear ($mg(H - h)$), dan keduanya menjumlahkan ke nilai konstan $mgH$ pada setiap ketinggian.
Rantai energi dapat melewati pegas. Sebuah balok yang menabrak pegas dengan energi kinetik $110\ \text{J}$ sambil meluncur sedikit ke bawah, sehingga GPE-nya turun sebesar $20\ \text{J}$ lebih lanjut sebelum pegas menghentikannya, menyimpan $110 + 20 = 130\ \text{J}$ energi potensial elastis pada kompresi terbesar, jika tidak ada energi yang hilang karena gesekan. Grafik gaya–kompresi pegas kemudian memberikan kompresi: energi yang tersimpan adalah luas di bawah garis tersebut, $\tfrac{1}{2} F_{0} x_{0}$ untuk pegas yang mematuhi Hukum Hooke (topik 6), sehingga gaya maksimum $2600\ \text{N}$ berarti $x_{0} = 2 \times 130 / 2600 = 0.10\ \text{m}$. Pada saat itu, resultan gaya pada balok adalah gaya pegas dikurangi komponen berat sepanjang kemiringan, dan percepatannya mengikuti $F = ma$ (topik 3).
Efisiensi suatu sistem adalah
Konsep yang sama dengan daya:
Efisiensi selalu kurang dari 100% dalam sistem nyata, karena sebagian energi masukan berubah menjadi bentuk "tidak berguna" — biasanya energi termal. Jika total energi $E$ disuplai dan jumlah $Q$ terbuang, output yang berguna adalah $E - Q$ dan efisiensinya adalah $(E - Q) / E$; sebuah motor dengan rating masukan $1.7\ \text{kW}$ dan efisiensi $53\%$ menghasilkan output yang berguna $0.53 \times 1.7 = 0.90\ \text{kW}$.
Untuk motor listrik yang mengangkat beban dengan efisiensi $\eta$ pada tegangan $V$ dan arus $I$, daya output yang berguna adalah $\eta V I$. Dari sini Anda dapat menemukan gaya, kecepatan angkat, atau tegangan tali.
Contoh terpecahkan. Motor listrik mengangkat beban $50\ \text{kg}$ dengan kecepatan konstan $0.40\ \text{m s}^{-1}$ sementara menyerap daya listrik $250\ \text{W}$. Tentukan efisiensinya (gunakan $g = 9.81\ \text{m s}^{-2}$).
Daya output yang berguna adalah $P = mgv = 50 \times 9.81 \times 0.40 = 196\ \text{W}$, sehingga
Contoh terpecahkan. Motor mengangkat balok dengan gaya $240\ \text{N}$ melalui jarak vertikal $12\ \text{m}$ dalam waktu $60\ \text{s}$. Daya masukan ke motor adalah $900\ \text{W}$. Tentukan daya output yang berguna dan efisiensinya.
Usaha yang dilakukan pada balok: $W = Fd = 240 \times 12 = 2880\ \text{J}$. Daya output yang berguna: $P = W / t = 2880 / 60 = 48\ \text{W}$. Efisiensi: $48 / 900 \times 100\% = 5.3\%$. Sebagian besar masukan masuk ke energi termal di dalam motor dan kabelnya. Dibalik, motor lift dengan output berguna $20\ \text{kW}$ yang mengangkat $1500\ \text{kg}$ melalui $20\ \text{m}$ membutuhkan $t = mgh / P = 1500 \times 9.81 \times 20 / 20\,000 = 15\ \text{s}$.
The input energy divides into useful work and wasted energy; efficiency = useful ÷ input, and useful + wasted always equals the input. · Energi input terbagi menjadi usaha berguna dan energi terbuang; efisiensi = berguna ÷ input, dan berguna + terbuang selalu sama dengan input.
PE + KE = constant · PE + KE = konstan
Work transfers energy; as it falls, PE becomes KE with the total fixed. · Usaha mentransfer energi; saat jatuh, PE berubah menjadi KE dengan total tetap.
| English | Bahasa Indonesia |
|---|---|
| work/wɜːk/ | usaha |
| energy/ˈenədʒi/ | tenaga |
| power/ˈpaʊə/ | kuasa |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| electrical/ɪˈlektrɪkl/ | listrik |
| electrical energy/ɪˈlektrɪkl ˈenədʒi/ | energi listrik |
| displacement/dɪˈspleɪsmənt/ | pindahan |
| component/kəmˈpəʊnənt/ | komponen |
| horizontal/ˌhɒrɪˈzɒntl/ | horizontal |
| scalar/ˈskeɪlə/ | skalar |
| normal contact force/ˈnɔːml ˈkɒntækt fɔːs/ | gaya kontak normal |
| friction/ˈfrɪkʃn/ | gesekan |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
| frictionless/ˈfrɪkʃənləs/ | tanpa geseran |
| mass/mæs/ | massa |
| weight/weɪt/ | berat |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | energi potensial gravitasi |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | hukum kekekalan energi |
| closed system/kləʊzd ˈsɪstəm/ | sistem tertutup |
| elastic potential energy/ɪˈlæstɪk pəˈtenʃl ˈenədʒi/ | energi potensial elastis |
| chemical energy/ˈkemɪkl ˈenədʒi/ | energi kimia |
| ramp/ræmp/ | landai |
| thermal energy/ˈθɜːml ˈenədʒi/ | energi termal |
| spring/sprɪŋ/ | pegas |
| compression/kəmˈpreʃn/ | pemampatan |
| efficiency/ɪˈfɪʃənsi/ | efisiensi |
| voltage/ˈvəʊltɪdʒ/ | tegangan |
| current/ˈkʌrənt/ | arus |
| tension/ˈtenʃn/ | tegangan |
| drag/dræɡ/ | menggeser |
Power 功率 is the rate of doing work, or the rate of transferring energy:
Unit: $\text{W} = \text{J s}^{-1}$. Power is a scalar.
For an object moving at velocity 速度 $v$ with a force $F$ along the direction of motion, in a short time $\Delta t$ the displacement is $v\,\Delta t$ and the work done is $F v\,\Delta t$. Dividing by $\Delta t$:
This is one of the most useful results in mechanics. A "derive $P = Fv$" answer needs exactly those three lines: $P = W / t$, $W = Fs$ for a force along the motion, and $s / t = v$; then use it to solve problems in which a vehicle or a load moves at constant speed.
Worked example. A car travels at a steady $25\ \text{m s}^{-1}$ against a total resistive force of $600\ \text{N}$. Find the output power of its engine.
At constant speed the driving force equals the resistive force, so
Daya adalah laju melakukan usaha, atau laju mentransfer energi:
Unit: $\text{W} = \text{J s}^{-1}$. Daya adalah besaran skalar.
Untuk benda yang bergerak dengan kecepatan $v$ dengan gaya $F$ searah arah gerak, dalam waktu singkat $\Delta t$ perpindahannya adalah $v\,\Delta t$ dan usaha yang dilakukan adalah $F v\,\Delta t$. Dibagi dengan $\Delta t$:
Ini adalah salah satu hasil paling berguna dalam mekanika. Jawaban "turunkan $P = Fv$" memerlukan tepat tiga baris tersebut: $P = W / t$, $W = Fs$ untuk gaya searah gerak, dan $s / t = v$; lalu gunakan untuk menyelesaikan masalah di mana kendaraan atau beban bergerak dengan kecepatan konstan.
Contoh terpecahkan. Mobil berjalan dengan kecepatan konstan $25\ \text{m s}^{-1}$ melawan total gaya hambat $600\ \text{N}$. Tentukan daya output mesinnya.
Pada kecepatan konstan, gaya penggerak sama dengan gaya hambat, sehingga
*Pembangkit hidrolistrik dinilai dalam megawatt: energi air yang disimpan, dibagi dengan waktu yang dibutuhkan untuk melepaskannya
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
In a uniform gravitational field (close to a planet's surface), the change in gravitational potential energy of mass 质量 $m$ rising or falling through a height $\Delta h$ is
The work done against gravity to raise a mass $m$ slowly (no change in kinetic energy) through height $\Delta h$ equals the gravitational potential energy gained:
So $\Delta E_{\text{P}} = mg \Delta h$. To use it you need $m$ and $\Delta h$ (and $g$). You do not need speed or time. The two-mark derivation wants the same four lines, with the condition stated: the block is raised at constant speed, so the lifting force equals the weight $mg$, and the work done by that force over the vertical distance $\Delta h$ is $mg\Delta h$, which is the gain in gravitational potential energy. A $100\ \text{g}$ object falling $10\ \text{m}$ loses $0.100 \times 9.81 \times 10 = 9.8\ \text{J}$, about $10\ \text{J}$; the mass must be in kilograms.
*Pertukaran energi dalam bandul
Dalam medan gravitasi seragam (dekat permukaan planet), perubahan energi potensial gravitasi dari massa $m$ naik atau turun melalui ketinggian $\Delta h$ adalah
Usaha yang dilakukan melawan gravitasi untuk menaikkan massa $m$ perlahan (tanpa perubahan energi kinetik) melalui ketinggian $\Delta h$ sama dengan energi potensial gravitasi yang diperoleh:
Jadi $\Delta E_{\text{P}} = mg \Delta h$. Untuk menggunakannya, Anda perlu $m$ dan $\Delta h$ (dan $g$). Anda tidak perlu kecepatan atau waktu. Penurunan nilai dua poin menginginkan empat baris yang sama itu, dengan kondisi yang dinyatakan: balok diangkat dengan kecepatan konstan, sehingga gaya angkat sama dengan berat $mg$, dan usaha yang dilakukan oleh gaya tersebut selama jarak vertikal $\Delta h$ adalah $mg\Delta h$, yang merupakan peningkatan energi potensial gravitasi. Benda $100\ \text{g}$ yang jatuh $10\ \text{m}$ kehilangan $0.100 \times 9.81 \times 10 = 9.8\ \text{J}$, sekitar $10\ \text{J}$; massa harus dalam kilogram.
The kinetic energy of an object of mass $m$ moving at speed $v$ is
Apply a resultant force $F$ to a mass $m$ that starts at rest. It speeds up evenly from $0$ to $v$ over a displacement $s$. From $v^{2} = u^{2} + 2as$ with $u = 0$,
The work done on the mass is
All this work becomes kinetic energy, so $E_{\text{K}} = \tfrac{1}{2} m v^{2}$. In the exam, state the assumptions as you go: the object starts from rest, the force is constant, so the acceleration is uniform, and $F = ma$ is Newton's second law.
Because $E_{\text{K}} \propto v^{2}$, quadrupling the speed multiplies the kinetic energy by sixteen: an object with $1500\ \text{J}$ at $10\ \text{m s}^{-1}$ has $24\,000\ \text{J}$ at $40\ \text{m s}^{-1}$. The same square is why a stopping force is found from energy: a ball of mass $1.2\ \text{kg}$ moving at $3.0\ \text{m s}^{-1}$ that is stopped by a cushion over $0.020\ \text{m}$ loses $\tfrac{1}{2} \times 1.2 \times 3.0^{2} = 5.4\ \text{J}$, so the average force on it is $5.4 / 0.020 = 270\ \text{N}$: the work done by the cushion equals the kinetic energy lost.
Combining $p = mv$ and $E_{\text{K}} = \tfrac{1}{2} m v^{2}$:
This is handy when the momentum 动量 is given but not the velocity. For a momentum change from $p_{1}$ to $p_{2}$ at constant mass, the change in kinetic energy is $(p_{2}^{2} - p_{1}^{2}) / (2m)$.
Energi kinetik dari benda bermassa $m$ bergerak dengan kecepatan $v$ adalah
Terapkan resultan gaya $F$ pada massa $m$ yang dimulai dari keadaan diam. Benda melaju merata dari $0$ ke $v$ selama perpindahan $s$. Dari $v^{2} = u^{2} + 2as$ dengan $u = 0$,
Usaha yang dilakukan pada massa tersebut adalah
Semua usaha ini menjadi energi kinetik, jadi $E_{\text{K}} = \tfrac{1}{2} m v^{2}$. Dalam ujian, nyatakan asumsi saat Anda berjalan: benda dimulai dari keadaan diam, gayanya konstan, sehingga percepatannya seragam, dan $F = ma$ adalah hukum kedua Newton.
Karena $E_{\text{K}} \propto v^{2}$, menggandakan empat kali kecepatan mengalikan energi kinetik dengan enam belas: benda dengan $1500\ \text{J}$ pada $10\ \text{m s}^{-1}$ memiliki $24\,000\ \text{J}$ pada $40\ \text{m s}^{-1}$. Kuadrat yang sama inilah alasan mengapa gaya pengereman ditemukan dari energi: bola bermassa $1.2\ \text{kg}$ bergerak pada $3.0\ \text{m s}^{-1}$ yang dihentikan oleh bantal di $0.020\ \text{m}$ kehilangan $\tfrac{1}{2} \times 1.2 \times 3.0^{2} = 5.4\ \text{J}$, sehingga gaya rata-rata pada bola adalah $5.4 / 0.020 = 270\ \text{N}$: usaha yang dilakukan bantal sama dengan energi kinetik yang hilang.
Menggabungkan $p = mv$ dan $E_{\text{K}} = \tfrac{1}{2} m v^{2}$:
Ini berguna ketika momentum diberikan tetapi tidak kecepatannya. Untuk perubahan momentum dari $p_{1}$ ke $p_{2}$ pada massa konstan, perubahan energi kinetik adalah $(p_{2}^{2} - p_{1}^{2}) / (2m)$.
PE + KE = constant · PE + KE = konstan
Potential energy turns into kinetic energy — the total never changes. · Potensial energi berubah menjadi kinetik energi — totalnya tidak pernah berubah.
| English | Bahasa Indonesia |
|---|---|
| momentum/məʊˈmentəm/ | momentum |
| spring constant/sprɪŋ ˈkɒnstənt/ | konstanta pegas |
| projectile/prəˈdʒektaɪl/ | proyektil |
| bungee jumper/ˈbʌndʒiː ˈdʒʌmpə/ | pembawa bungee |
A useful plan for problems that mix forces and energy:
Examples:
Rencana berguna untuk masalah yang mencampurkan gaya dan energi:
Contoh-contoh:
Drop the mass and watch GPE turn into KE. With no friction the total energy stays the same — that's the energy method. · Jatuhkan massa dan perhatikan GPE berubah menjadi KE. Tanpa gesekan, energi total tetap konstan — inilah metode energi.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| work done by a force | the product of the force and the distance moved in the direction of the force |
| principle of conservation of energy | energy cannot be created or destroyed, only transferred from one form to another, so the total energy of a closed system is constant |
| efficiency | the ratio of useful energy (or power) output from a system to the total energy (or power) input, usually as a percentage |
| power | the work done per unit time, or the rate at which energy is transferred |
| gravitational potential energy | the energy an object has because of its position in a gravitational field |
| kinetic energy | the energy an object has because of its motion |
| elastic potential energy | the energy stored in an object that has been stretched or compressed |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| usaha oleh gaya | hasil kali gaya dan jarak yang ditempuh searah dengan gaya |
| prinsip kekekalan energi | energi tidak dapat diciptakan atau dimusnahkan, hanya ditransfer dari satu bentuk ke bentuk lain, sehingga total energi sistem tertutup adalah konstan |
| efisiensi | rasio energi (atau daya) output yang berguna dari suatu sistem terhadap total energi (atau daya) input, biasanya dalam persentase |
| daya | usaha per satuan waktu, atau laju transfer energi |
| energi potensial gravitasi | energi yang dimiliki benda karena posisinya dalam medan gravitasi |
| energi kinetik | energi yang dimiliki benda karena geraknya |
| energi potensial elastis | energi yang disimpan dalam benda yang telah diregangkan atau ditekan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
When a force acts on a solid along its length, the object changes shape (deformation 形变). Two cases (treated as one-dimensional here):
The applied force is the load 负载. The change from the natural length is the extension (or compression).
Both are measured from the natural, unstretched length, never from the loaded length: on a diagram of a spring under a box, the distance the spring has shortened is its compression. A spring's own mass is usually said to be negligible 可忽略的, so the only forces on it are the load and the tension it provides.
Apabila daya bertindak pada padatan sepanjang panjangnya, objek berubah bentuk (ubah bentuk). Dua kes (dipertimbangkan sebagai satu dimensi di sini):
Daya yang diaplikasikan ialah beban. Perubahan daripada panjang semula jadi ialah panjangan (atau pemampatan).
Keduanya diukur dari panjang semula jadi yang tidak diregangkan, bukan dari panjang yang dibebani: pada rajah spring di bawah kotak, jarak spring telah memendek ialah pemampatannya. Jisim spring sendiri biasanya dikatakan negligible, jadi hanya beban dan ketegangan yang dilakukannya bertindak ke atasnya.
Hang a load on the real spring: up to the elastic limit the extension is proportional to the force; beyond it the spring is stretched for good. · Gantungkan beban pada pegas nyata: hingga batas elastis, peregangan sebanding dengan gaya; melebihi itu, pegas meregang permanen.
| English | Bahasa Indonesia |
|---|---|
| tensile/ˈtensaɪl/ | tarik |
| negligible/ˈneɡlɪdʒəbl/ | dapat diabaikan |
| spring constant/sprɪŋ ˈkɒnstənt/ | konstanta pegas |
| limit of proportionality/ˈlɪmɪt ɒv prəˌpɔːʃəˈnælɪti/ | batas proporsionalitas |
For many materials at small extensions, the extension is proportional to the load — this is Hooke's law 胡克定律. The constant that links them is the spring constant 劲度系数 $k$:
Unit of $k$: $\text{N m}^{-1}$.
Worked example. A spring stretches by $4.0\ \text{cm}$ when a $2.0\ \text{N}$ load is hung from it. Find its spring constant.
Converting the extension to metres ($4.0\ \text{cm} = 0.040\ \text{m}$):
For the one-mark "state Hooke's law", write it in words: the extension is directly proportional to the applied force (load), provided the limit of proportionality is not exceeded. The spring constant is the force per unit extension; the multiple-choice distractors are "extension per unit force" (that is $1/k$) and $\tfrac{1}{2}Fx$ (the energy stored).
Reading a graph:
A common trap: if a graph plots length $L$ against force, you can still find the spring constant from the gradient (since $L = L_{0} + F/k$, the gradient is $1/k$ — read it off carefully).
Worked example. A spring has length $90\ \text{mm}$ under a force of $0.80\ \text{N}$ and $115\ \text{mm}$ under $1.30\ \text{N}$. Find the spring constant and the unstretched length.
The extra force $0.50\ \text{N}$ produces the extra extension $25\ \text{mm} = 0.025\ \text{m}$, so $k = 0.50 / 0.025 = 20\ \text{N m}^{-1}$. At $0.80\ \text{N}$ the extension is $x = F/k = 0.80 / 20 = 0.040\ \text{m}$, so the unstretched length is $90 - 40 = 50\ \text{mm}$. On a graph of length against force the intercept on the length axis is $L_{0}$, the gradient is $1/k$, and the work done in stretching the spring is the triangle between the line and the level $L = L_{0}$, not the whole area down to the force axis: the rectangle below $L_{0}$ is the unstretched length multiplied by the force, which means nothing.
Hooke's law only holds up to the limit of proportionality 比例极限. Past this point the $F$ against $x$ line curves and is no longer straight. The material may still be elastic 弹性 (it returns to its first length when you remove the load) a little further, then it becomes plastic 塑性.
You may need to combine spring constants:
Worked example. Springs of spring constant $6.0\ \text{N cm}^{-1}$ and $4.0\ \text{N cm}^{-1}$ are joined end to end and a load of $80\ \text{N}$ hangs from them. Find the total extension.
Each spring carries the full $80\ \text{N}$, so the extensions are $80/6.0 = 13.3\ \text{cm}$ and $80/4.0 = 20\ \text{cm}$: a total of $33\ \text{cm}$ (the same as $k = 2.4\ \text{N cm}^{-1}$ from $1/k = 1/6.0 + 1/4.0$). Three identical springs give the largest combined constant when all three are in parallel ($3k$) and the smallest when all are in series ($k/3$). When a load is shared by two identical springs in parallel, each carries half the force and so stores a quarter of the energy of one spring holding the whole load; the pair together stores half as much as the single spring did. A spring constant in $\text{N cm}^{-1}$ is $100$ times smaller than the same constant in $\text{N m}^{-1}$; give the unit the question asks for.
Untuk banyak bahan pada panjangan kecil, panjangan berkadar terus dengan beban — ini adalah Hukum Hooke. Pemalar yang menghubungkannya ialah pemalar spring $k$:
Unit bagi $k$: $\text{N m}^{-1}$.
Contoh kerja. Sebuah spring meregang sebanyak $4.0\ \text{cm}$ apabila beban $2.0\ \text{N}$ digantung daripadanya. Cari pemalar springnya.
Menukar panjangan kepada meter ($4.0\ \text{cm} = 0.040\ \text{m}$):
Bagai markah "nyatakan Hukum Hooke", tulis dalam kata-kata: panjangan berkadar terus dengan daya yang diaplikasikan (beban), dengan syarat had perkadaran tidak dilampaui. Pemalar spring ialah daya per unit panjangan; pilihan jawapan salah pelbagai ialah "panjangan per unit daya" (iaitu $1/k$) dan $\tfrac{1}{2}Fx$ (tenaga yang disimpan).
Membaca graf:
Jebakan biasa: jika graf memaparkan panjang $L$ terhadap daya, anda masih boleh mencari pemalar spring dari kecerunan (kerana $L = L_{0} + F/k$, kecerunan adalah $1/k$ — baca dengan teliti).
Contoh kerja. Sebuah spring mempunyai panjang $90\ \text{mm}$ di bawah daya $0.80\ \text{N}$ dan $115\ \text{mm}$ di bawah $1.30\ \text{N}$. Cari pemalar spring dan panjang tidak diregangkan.
Daya tambahan $0.50\ \text{N}$ menghasilkan panjangan tambahan $25\ \text{mm} = 0.025\ \text{m}$, jadi $k = 0.50 / 0.025 = 20\ \text{N m}^{-1}$. Pada $0.80\ \text{N}$ panjangan adalah $x = F/k = 0.80 / 20 = 0.040\ \text{m}$, jadi panjang tidak diregangkan ialah $90 - 40 = 50\ \text{mm}$. Pada graf panjang terhadap daya, pintasan pada paksi panjang ialah $L_{0}$, kecerunan ialah $1/k$, dan kerja dilakukan dalam meregangkan spring ialah segitiga antara garis dan aras $L = L_{0}$, bukan keseluruhan kawasan hingga paksi daya: segi empat tepat di bawah $L_{0}$ ialah panjang tidak diregangan didarabkan dengan daya, yang tiada maksud apa-apa.
Hukum Hooke hanya sah sehingga had perkadaran. Melepasi titik ini, garis $F$ terhadap $x$ melengkung dan tidak lagi lurus. Bahan mungkin masih anjalik (kembali ke panjang pertama apabila beban ditanggalkan) sedikit lagi, kemudian menjadi plastik.
Anda mungkin perlu menggabungkan pemalar spring:
Contoh kerja. Spring dengan pemalar spring $6.0\ \text{N cm}^{-1}$ dan $4.0\ \text{N cm}^{-1}$ disambung hujung ke hujung dan beban $80\ \text{N}$ digantung daripadanya. Cari panjangan total.
Setiap pegas membawa $80\ \text{N}$ penuh, sehingga perpanjangan adalah $80/6.0 = 13.3\ \text{cm}$ dan $80/4.0 = 20\ \text{cm}$: total $33\ \text{cm}$ (sama dengan $k = 2.4\ \text{N cm}^{-1}$ dari $1/k = 1/6.0 + 1/4.0$). Tiga pegas identik memberikan konstanta gabungan terbesar ketika ketiganya sejajar ($3k$) dan terkecil ketika semuanya seri ($k/3$). Ketika beban dibagi oleh dua pegas identik yang sejajar, masing-masing menahan setengah gaya dan menyimpan seperempat energi dari satu pegas yang menahan seluruh beban; pasangan tersebut bersama-sama menyimpan setengah jumlah energi dari pegas tunggal. Konstanta pegas dalam $\text{N cm}^{-1}$ adalah $100$ kali lebih kecil daripada konstanta yang sama dalam $\text{N m}^{-1}$; berikan satuan yang diminta soal.
F = k·x
Force is proportional to extension — the gradient is the spring constant k. · Gaya berbanding lurus dengan regangan — gradiennya adalah konstanta pegas k.
F = kx
Up to the limit, extension is proportional to force — the gradient is the spring constant k. · Hingga batas, peregangan berbanding lurus dengan gaya — gradien adalah konstanta pegas k.
F = ax
Drag the spring constant. Force is proportional to extension — a straight line through the origin whose gradient is the spring constant. · Geser konstanta pegas. Gaya berbanding lurus dengan pertambahan panjang — garis lurus melalui titik asal yang gradiennya adalah konstanta pegas.
| English | Bahasa Indonesia |
|---|---|
| extension/ekˈstenʃn/ | ekstensi |
| compression/kəmˈpreʃn/ | pemampatan |
| load/ləʊd/ | beban |
For a wire of uniform cross-section under a tensile load:
For the one-mark definitions: stress is the force per unit cross-sectional area (the force acting normally to the area); strain is the extension per unit original length. The unit of stress, $\text{Pa} = \text{N m}^{-2}$, is $\text{kg m}^{-1}\ \text{s}^{-2}$ in base units. The area of a round wire comes from its diameter $d$: $A = \pi d^{2}/4 = \pi r^{2}$, and using $d$ in place of $r$ makes the area four times too big. Convert $\text{mm}^{2}$ to $\text{m}^{2}$ with $10^{-6}$.
Worked example. A tensile force of $18\ \text{N}$ acts on a wire of cross-sectional area $3.2\ \text{mm}^{2}$. Find the stress.
$\sigma = F/A = 18 / (3.2 \times 10^{-6}) = 5.6 \times 10^{6}\ \text{Pa}$. The same force acts along the whole wire, so a bolt whose diameter is $2d$ at one end and $d$ at the other has four times the stress at the narrow end ($\sigma \propto 1/d^{2}$), and a wire of three times the radius carries one ninth of the stress.
The Young modulus 杨氏模量 is the ratio of stress to strain in the Hooke's-law region:
Unit: $\text{Pa}$ (about $10^{11}$ for metals; e.g. steel $\approx 2.0 \times 10^{11}$ Pa).
Worked example. A steel wire of length $2.0\ \text{m}$ and cross-sectional area $1.5 \times 10^{-7}\ \text{m}^{2}$ stretches by $1.0\ \text{mm}$ under a load of $15\ \text{N}$. Find the Young modulus.
The Young modulus is a property of the material — it does not depend on the wire's shape or size. The spring constant $k$ depends on both the material and the size: $k = EA/L_{0}$.
For the one-mark definition: the Young modulus is the ratio of stress to strain (for a material deformed within its limit of proportionality). Because it belongs to the material, a thicker wire of the same metal has the same Young modulus: it stretches less under the same load because its area is larger, not because the metal is stiffer. A "show that $k = EA/L_{0}$" question wants two lines: $k = F/x$ and $E = FL_{0}/(Ax)$, so $E = kL_{0}/A$. Two things follow: a wire's spring constant stays constant while Hooke's law holds, whatever the force, and a wire of the same metal with twice the diameter has four times the spring constant.
Worked example. A copper wire of length $1.7\ \text{m}$ and diameter $0.64\ \text{mm}$ has Young modulus $1.2 \times 10^{11}\ \text{Pa}$. Find its spring constant.
$A = \pi (0.32 \times 10^{-3})^{2} = 3.2 \times 10^{-7}\ \text{m}^{2}$, so $k = EA/L_{0} = 1.2 \times 10^{11} \times 3.2 \times 10^{-7} / 1.7 = 2.3 \times 10^{4}\ \text{N m}^{-1}$.
On a stress–strain graph the gradient of the straight part is $E$, so of two materials drawn on the same axes the steeper line is the stiffer material, and a wire with double the Young modulus is drawn as a line through the origin with twice the gradient. On a force–extension graph the gradient is $EA/L_{0}$ and the Young modulus is the gradient multiplied by $L_{0}/A$; only a stress–strain graph has a gradient equal to $E$ itself.
A standard setup:
Why a long, thin wire? To make the extension big enough to measure well. Why repeat readings and measure $d$ at several places? To reduce random error 随机误差 and check the wire is uniform.
A "describe an experiment" answer earns its marks for the quantities measured and how ($L_{0}$ with a metre rule; $d$ with a micrometer at several points; the load from the masses or a newton meter; the extension from a marker read against a fixed scale, or a vernier scale), the graph ($F$ against $x$, gradient $EA/L_{0}$, or stress against strain, gradient $E$) and the precautions: safety goggles in case the wire snaps, a small load first to straighten kinks, and readings taken on unloading as well as loading to check the wire stayed elastic.
Untuk kawat dengan penampang seragam di bawah beban tarik:
Untuk definisi bernilai satu poin: tegangan adalah gaya per satuan luas penampang (gaya yang bekerja tegak lurus terhadap area); regangan adalah perpanjangan per satuan panjang awal. Satuan tegangan, $\text{Pa} = \text{N m}^{-2}$, adalah $\text{kg m}^{-1}\ \text{s}^{-2}$ dalam satuan dasar. Luas kawat bulat berasal dari diameternya $d$: $A = \pi d^{2}/4 = \pi r^{2}$, dan menggunakan $d$ sebagai pengganti $r$ membuat luas empat kali terlalu besar. Konversi $\text{mm}^{2}$ ke $\text{m}^{2}$ dengan $10^{-6}$.
Contoh terpecahkan. Gaya tarik sebesar $18\ \text{N}$作用于 pada kawat dengan luas penampang $3.2\ \text{mm}^{2}$. Temukan tegangannya.
$\sigma = F/A = 18 / (3.2 \times 10^{-6}) = 5.6 \times 10^{6}\ \text{Pa}$. Gaya yang sama bekerja sepanjang kawat, sehingga baut dengan diameter $2d$ di satu ujung dan $d$ di ujung lainnya memiliki empat kali tegangan di ujung sempit ($\sigma \propto 1/d^{2}$), dan kawat dengan tiga kali jari-jari membawa satu per sembilan tegangan.
Modulus Young adalah rasio tegangan terhadap regangan di daerah hukum Hooke:
Satuan: $\text{Pa}$ (sekitar $10^{11}$ untuk logam; misalnya baja $\approx 2.0 \times 10^{11}$ Pa).
Contoh terpecahkan. Kawat baja sepanjang $2.0\ \text{m}$ dan luas penampang $1.5 \times 10^{-7}\ \text{m}^{2}$ meregang sebesar $1.0\ \text{mm}$ di bawah beban $15\ \text{N}$. Tentukan modulus Young.
Modulus Young adalah sifat bahan—tidak bergantung pada bentuk atau ukuran kawat. Konstanta pegas $k$ bergantung pada kedua bahan dan ukuran: $k = EA/L_{0}$.
Untuk definisi satu nilai: modulus Young adalah rasio tegangan terhadap regangan (untuk material yang terdeformasi dalam batas proporsionalnya). Karena ini merupakan sifat material, kawat yang lebih tebal dari logam yang sama memiliki modulus Young yang sama; ia meregang lebih sedikit di bawah beban yang sama karena luas penampangnya lebih besar, bukan karena logamnya lebih kaku. Soal "buktikan bahwa $k = EA/L_{0}$" membutuhkan dua baris: $k = F/x$ dan $E = FL_{0}/(Ax)$, sehingga $E = kL_{0}/A$. Dua hal berikut berlaku: konstanta pegas kawat tetap konstan selama hukum Hooke berlaku, apa pun gayanya, dan kawat dari logam yang sama dengan diameter dua kali lipat memiliki konstanta pegas empat kali lipatnya.
Contoh terpecahkan. Kawat tembaga sepanjang $1.7\ \text{m}$ dan diameter $0.64\ \text{mm}$ memiliki modulus Young $1.2 \times 10^{11}\ \text{Pa}$. Tentukan konstanta pegasnya.
$A = \pi (0.32 \times 10^{-3})^{2} = 3.2 \times 10^{-7}\ \text{m}^{2}$, sehingga $k = EA/L_{0} = 1.2 \times 10^{11} \times 3.2 \times 10^{-7} / 1.7 = 2.3 \times 10^{4}\ \text{N m}^{-1}$.
Pada grafik tegangan-regangan, gradien bagian lurus adalah $E$, jadi dari dua material yang digambar pada sumbu yang sama, garis yang lebih curam adalah material yang lebih kaku, dan kawat dengan modulus Young dua kali lipat digambarkan sebagai garis melalui titik asal dengan gradien dua kali lipat. Pada grafik gaya-perpanjangan, gradiennya adalah $EA/L_{0}$ dan modulus Young adalah gradien dikalikan dengan $L_{0}/A$; hanya grafik tegangan-regangan yang memiliki gradien sama dengan $E$ itu sendiri.
Pengaturan standar:
Mengapa kawat panjang dan tipis? Untuk membuat perpanjangan cukup besar untuk diukur dengan baik. Mengapa mengulang pembacaan dan mengukur $d$ di beberapa tempat? Untuk mengurangi kesalahan acak dan memeriksa apakah kawat seragam.
Jawaban "deskripsikan sebuah eksperimen" mendapatkan nilainya untuk besaran yang diukur dan bagaimana ($L_{0}$ dengan penggaris meter; $d$ dengan mikrometer pada beberapa titik; beban dari massa atau newton meter; perpanjangan dari penanda dibaca terhadap skala tetap, atau skala vernier), grafik ($F$ terhadap $x$, gradien $EA/L_{0}$, atau tegangan terhadap regangan, gradien $E$) dan tindakan pencegahan: kacamata pelindung jika kawat patah, beban kecil terlebih dahulu untuk meluruskan bengkokan, dan pembacaan diambil saat pembongkaran maupun penambahan beban untuk memeriksa apakah kawat tetap elastis.
| English | Bahasa Indonesia |
|---|---|
| stress/stres/ | tegangan |
| strain/streɪn/ | regangan |
| Young modulus/jʌŋ ˈmɒdjʊləs/ | Modulus Young |
| cross-sectional area/krɒs ˈsekʃənl ˈeərɪə/ | luas penampang |
| diameter/daɪˈæmɪtə/ | diameter |
| pulley/ˈpʊli/ | katrol |
| micrometer/maɪˈkrɒmɪtə/ | mikrometer |
| random error/ˈrændəm ˈerə/ | kesalahan acak |
| thermal energy/ˈθɜːml ˈenədʒi/ | energi termal |
| rubber band/ˈrʌbə bænd/ | karet gelang |
| elastic hysteresis/ɪˈlæstɪk ˌhɪstəˈriːsɪs/ | histeresis elastis |
| elastic potential energy/ɪˈlæstɪk pəˈtenʃl ˈenədʒi/ | energi potensial elastis |
| trapezia/trəˈpiːzɪə/ | trapesium |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
As the load grows:
Hooke's law only holds in the straight, elastic region.
In words: elastic deformation means the object returns to its original length (or shape) when the load is removed; plastic deformation means it does not, and a permanent extension remains; the elastic limit is the maximum load (or extension) for which the deformation is still elastic; the limit of proportionality is the point beyond which the extension is no longer proportional to the load. On a graph the limit of proportionality is where the line stops being straight. The elastic limit lies a little beyond it and cannot be read from a loading line alone. So in a "which statement must be correct" item, the end of the straight part is the limit of proportionality, but whether a later point is the elastic limit or the breaking point cannot be told from the shape of the loading line.
On a force–extension graph for a material taken into the plastic region and then unloaded, the loading line and the unloading line are different. The unloading line is parallel to the first Hooke line but shifted to the right (the permanent extension left when the load reaches zero). The area between the loading and unloading lines is the energy turned into thermal energy 热能 in the material.
A rubber band 橡皮筋 also has different loading and unloading curves, but it returns to its original length: its deformation is elastic. The area between the two curves is again energy dissipated as thermal energy (elastic hysteresis 弹性滞后), and the energy recovered on unloading is the area under the lower curve. The two-mark "explain why the work done in stretching the wire is not equal to the energy recovered when the force is removed" answer says that the wire was taken past its elastic limit, so part of the deformation is plastic and a permanent extension remains, and that part of the work done became thermal energy in the wire.
Saat beban bertambah:
Hukum Hooke hanya berlaku di daerah elastis yang lurus.
Dalam kata-kata: deformasi elastis berarti objek kembali ke panjang aslinya (atau bentuknya) ketika beban dihilangkan; deformasi plastis berarti tidak, dan perpanjangan permanen tersisa; batas elastis adalah beban maksimum (atau perpanjangan) untuk mana deformasi masih elastis; batas proporsionalitas adalah titik di mana perpanjangan tidak lagi berbanding lurus dengan beban. Pada grafik, batas proporsionalitas adalah tempat garis berhenti menjadi lurus. Batas elastis terletak sedikit setelahnya dan tidak dapat dibaca hanya dari garis pembebanan. Jadi dalam soal "pernyataan mana yang pasti benar", akhir bagian garis lurus adalah batas proporsionalitas, tetapi apakah titik selanjutnya adalah batas elastis atau titik patah tidak dapat ditentukan hanya dari bentuk garis pembebanan.
*Mesin uji (tarik) universal modern meregangkan sampel dan mencatat gaya dan perpanjangan
Pada grafik gaya–perpanjangan untuk suatu material yang ditarik ke daerah plastis lalu dibongkar, garis pembebanan dan garis pembongkaran berbeda. Garis pembongkaran sejajar dengan garis Hooke pertama tetapi bergeser ke kanan (perpanjangan permanen yang tersisa saat beban mencapai nol). Area antara garis pembebanan dan pembongkaran adalah energi yang diubah menjadi energi termal di dalam material.
Karet gelang juga memiliki kurva pembebanan dan pembongkaran yang berbeda, namun kembali ke panjang aslinya: deformasinya elastis. Area antara kedua kurva tersebut kembali merupakan energi yang terdisipasi sebagai energi termal (histeresis elastis), dan energi yang dipulihkan saat pembongkaran adalah area di bawah kurva bawah. Jawaban dua poin "jelaskan mengapa usaha yang dilakukan dalam meregangkan kawat tidak sama dengan energi yang dipulihkan ketika gaya dihilangkan" menyatakan bahwa kawat telah ditarik melebihi batas elastisnya, sehingga sebagian deformasi bersifat plastis dan perpanjangan permanen tersisa, serta sebagian usaha yang dilakukan berubah menjadi energi termal di dalam kawat.
*Karet gelang kembali ke panjang aslinya, namun area antara pembebanan dan pembongkaran adalah energi yang hilang sebagai energi termal
| English | Bahasa Indonesia |
|---|---|
| deformation/ˌdiːfɔːˈmeɪʃn/ | deformasi |
| Hooke's law/hʊks lɔː/ | Hukum Hooke |
| elastic/ɪˈlæstɪk/ | anjal |
| elastic limit/ɪˈlæstɪk ˈlɪmɪt/ | batas elastis |
| plastic/ˈplæstɪk/ | plastis |
| series/ˈsɪəriːz/ | seri |
| parallel/ˈpærəlel/ | paralel |
The work done in stretching a material from $0$ to extension $x$, as the load grows from $0$ to $F$, is the area under the force–extension graph.
Why the area: work done is force multiplied by the distance moved, but here the force grows as the material stretches, so the work is the sum of $F\,\Delta x$ over many small extensions, which is the area under the line. Within the limit of proportionality the force rises uniformly from $0$ to $F$, so the average force is $\tfrac{1}{2}F$ and the work is $\tfrac{1}{2}Fx$. On a force–extension graph the gradient is the spring constant.
When Hooke's law holds, the $F$ against $x$ graph is a straight line through the origin. The area under it from $0$ to $x$ is a triangle:
This is the elastic potential energy 弹性势能 stored in a spring or wire stretched within its limit of proportionality. An equal form:
Worked example. A spring of spring constant $50\ \text{N m}^{-1}$ is stretched by $0.20\ \text{m}$, within its limit of proportionality. Find the elastic potential energy stored.
Worked example. A wire of spring constant $2.0 \times 10^{4}\ \text{N m}^{-1}$ is already extended by $2.0\ \text{mm}$. Find the work done to increase its extension to $3.0\ \text{mm}$.
The stored energy rises from $\tfrac{1}{2} k x_{1}^{2}$ to $\tfrac{1}{2} k x_{2}^{2}$: $W = \tfrac{1}{2} \times 2.0 \times 10^{4} \times \left[(3.0 \times 10^{-3})^{2} - (2.0 \times 10^{-3})^{2}\right] = 0.050\ \text{J}$. This is the trapezium under the line between the two extensions, not $\tfrac{1}{2} k (x_{2} - x_{1})^{2}$. Because $E_{\text{P}} \propto x^{2}$ at fixed $k$, doubling the extension of a wire stores four times the energy ($0.65\ \text{J}$ becomes $2.6\ \text{J}$), and a stored energy gives the extension as $x = \sqrt{2E_{\text{P}}/k}$: a spring of $k = 400\ \text{N m}^{-1}$ storing $0.32\ \text{J}$ is compressed by $\sqrt{2 \times 0.32 / 400} = 0.040\ \text{m}$. For two wires joined end to end the tension is the same in both, so the total energy stored is $\tfrac{1}{2}F x_{1} + \tfrac{1}{2}F x_{2}$ with each wire's own extension.
For a graph that is not a straight line (a stretched rubber band, or a spring past its limit of proportionality), find the area by counting grid squares or by using trapezia 梯形. The same idea holds: the area under the force–extension graph is the work done on the material. To estimate the work done up to the breaking point, count the squares under the whole curve (part squares as halves) and multiply by the energy one square represents, the force step multiplied by the extension step; saying that the area was found by counting squares is the "explain your reasoning" mark.
A common multiple-choice case: two materials are stretched by the same force, or by the same extension. Using $E_{\text{P}} = \tfrac{1}{2} F x$:
A sketch of $E_{\text{P}}$ against extension or compression is a curve through the origin that gets steeper ($E_{\text{P}} \propto x^{2}$), not a straight line. Of two wires of the same length and area under the same load, the one with the smaller Young modulus extends more and so stores more energy.
When a stretched spring is released onto a mass, the elastic potential energy becomes kinetic energy 动能 (and gravitational potential energy if the mass rises). Set $\tfrac{1}{2} k x^{2}$ equal to $\tfrac{1}{2} m v^{2}$ (plus any $mgh$) to find the speed or height.
Usaha yang dilakukan dalam meregangkan material dari $0$ ke perpanjangan $x$, saat beban bertambah dari $0$ ke $F$, adalah area di bawah grafik gaya–perpanjangan.
*Usaha yang dilakukan meregangkan material adalah area di bawah grafik gaya–perpanjangan
Mengapa area: usaha adalah gaya dikali jarak tempuh, namun di sini gaya bertambah saat material meregang, sehingga usahanya adalah jumlah $F\,\Delta x$ atas banyak perpanjangan kecil, yang merupakan area di bawah garis. Dalam batas proporsionalitas, gaya naik secara seragam dari $0$ ke $F$, sehingga rata-rata gaya adalah $\tfrac{1}{2}F$ dan usahanya adalah $\tfrac{1}{2}Fx$. Pada grafik gaya–perpanjangan, gradien adalah konstanta pegas.
Ketika hukum Hooke berlaku, grafik $F$ terhadap $x$ adalah garis lurus melalui titik asal. Area di bawahnya dari $0$ ke $x$ berbentuk segitiga:
Ini adalah energi potensial elastis yang tersimpan dalam pegas atau kawat yang diregangkan dalam batas proporsionalitasnya. Bentuk setara:
Contoh dikerjakan. Sebuah pegas dengan konstanta pegas $50\ \text{N m}^{-1}$ diregangkan oleh $0.20\ \text{m}$, dalam batas proporsionalitasnya. Temukan energi potensial elastis yang tersimpan.
Contoh terpecahkan. Sebuah kawat dengan konstanta pegas $2.0 \times 10^{4}\ \text{N m}^{-1}$ sudah diperpanjang sejauh $2.0\ \text{mm}$. Hitung usaha yang diperlukan untuk memperpanjangnya menjadi $3.0\ \text{mm}$.
Energi tersimpan naik dari $\tfrac{1}{2} k x_{1}^{2}$ ke $\tfrac{1}{2} k x_{2}^{2}$: $W = \tfrac{1}{2} \times 2.0 \times 10^{4} \times \left[(3.0 \times 10^{-3})^{2} - (2.0 \times 10^{-3})^{2}\right] = 0.050\ \text{J}$. Ini adalah trapesium di bawah garis antara kedua perpanjangan tersebut, bukan $\tfrac{1}{2} k (x_{2} - x_{1})^{2}$. Karena $E_{\text{P}} \propto x^{2}$ pada $k$ tetap, menggandakan perpanjangan kawat menyimpan empat kali energi ($0.65\ \text{J}$ menjadi $2.6\ \text{J}$), dan energi tersimpan memberikan perpanjangan sebagai $x = \sqrt{2E_{\text{P}}/k}$: sebuah pegas dengan $k = 400\ \text{N m}^{-1}$ yang menyimpan $0.32\ \text{J}$ terkompresi sejauh $\sqrt{2 \times 0.32 / 400} = 0.040\ \text{m}$. Untuk dua kawat yang disambungkan ujung demi ujung, tegangan sama di keduanya, sehingga total energi tersimpan adalah $\tfrac{1}{2}F x_{1} + \tfrac{1}{2}F x_{2}$ dengan perpanjangan masing-masing kawat.
Untuk grafik yang bukan garis lurus (karet yang diregangkan, atau pegas melewati batas proporsionalnya), hitung luas dengan menghitung kotak grid atau menggunakan trapesia. Prinsip yang sama berlaku: luas di bawah grafik gaya–perpanjangan adalah usaha yang dilakukan pada bahan tersebut. Untuk memperkirakan usaha hingga titik putus, hitung kotak di bawah seluruh kurva (kotak parsial dihitung sebagai setengah) dan kalikan dengan energi yang diwakili satu kotak, yaitu langkah gaya dikali langkah perpanjangan; menyatakan bahwa luas ditemukan dengan menghitung kotak adalah poin "jelaskan alasanmu".
Kasus pilihan ganda umum: dua bahan diregangkan oleh gaya yang sama, atau oleh perpanjangan yang sama. Menggunakan $E_{\text{P}} = \tfrac{1}{2} F x$:
Sketsa $E_{\text{P}}$ terhadap perpanjangan atau kompresi adalah kurva melalui titik asal yang semakin curam ($E_{\text{P}} \propto x^{2}$), bukan garis lurus. Dari dua kawat dengan panjang dan luas penampang yang sama di bawah beban yang sama, yang memiliki modulus Young lebih kecil mengalami perpanjangan lebih besar dan sehingga menyimpan lebih banyak energi.
Ketika pegas yang diregangkan dilepaskan ke atas suatu massa, energi potensial elastis berubah menjadi energi kinetik (dan energi potensial gravitasi jika massa naik). Set $\tfrac{1}{2} k x^{2}$ sama dengan $\tfrac{1}{2} m v^{2}$ (ditambah apa pun $mgh$) untuk menemukan kecepatan atau ketinggian.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| Hooke's law | the extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded |
| spring constant | the force per unit extension |
| limit of proportionality | the point beyond which the extension is no longer proportional to the applied force |
| elastic limit | the maximum force (or extension) for which the material returns to its original length when the force is removed |
| elastic deformation | the material returns to its original length (shape) when the force is removed |
| plastic deformation | the material does not return to its original length when the force is removed; a permanent extension remains |
| stress | the force per unit cross-sectional area |
| strain | the extension per unit original length |
| Young modulus | the ratio of stress to strain (within the limit of proportionality) |
| elastic potential energy | the energy stored in an object because it has been stretched or compressed |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| Hukum Hooke | perpanjangan berbanding lurus dengan gaya yang diberikan, asalkan batas proporsionalitas tidak dilampaui |
| Konstanta pegas | gaya per satuan perpanjangan |
| Batas proporsionalitas | titik di mana perpanjangan不再 lagi berbanding lurus dengan gaya yang diberikan |
| Batas elastis | gaya maksimum (atau perpanjangan) di mana bahan kembali ke panjang aslinya ketika gaya dilepas |
| Deformasi elastis | bahan kembali ke panjang aslinya (bentuk) ketika gaya dilepas |
| Deformasi plastis | bahan tidak kembali ke panjang aslinya ketika gaya dilepas; perpanjangan permanen tersisa |
| Tegangan | gaya per satuan luas penampang melintang |
| Regangan | perpanjangan per satuan panjang awal |
| Modulus Young | rasio tegangan terhadap regangan (dalam batas proporsionalitas) |
| Energi potensial elastis | energi yang tersimpan dalam benda karena telah diregangkan atau dikompresi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A wave 波 carries energy 能量 from one place to another without moving matter overall. The particles of the medium 介质 oscillate 振动 about fixed rest positions; only the disturbance (and its energy) propagates 传播. Examples: a transverse wave 横波 on a rope, a longitudinal wave 纵波 on a slinky spring, ripples on water, and sound in air. A wave that travels and carries energy is a progressive wave 行波.
Wave motion 波动 is a series of oscillations of the particles of a medium, each passing the disturbance on to the next: shake one end of a rope, push one end of a spring, or touch the surface of the water in a ripple tank 水波槽, and the oscillation travels away from where it started. For the one-mark definition, a progressive wave transfers energy from one place to another without transferring matter; the particles only oscillate about their rest positions.
Two points one wavelength apart are in phase (phase difference 0 or $2\pi$). Two points half a wavelength apart are exactly out of phase (phase difference $\pi$).
To find the phase difference between two points from a displacement–distance graph, divide their separation by the wavelength and multiply by $360°$ (or $2\pi$): two points $0.50\ \text{m}$ apart on a wave of wavelength $2.0\ \text{m}$ ($v = 600\ \text{m s}^{-1}$, $f = 300\ \text{Hz}$) differ in phase by $0.25 \times 360° = 90°$. Only the fraction of a cycle matters, so $450°$ is the same as $90°$. From two displacement–time graphs, read the time shift between the peaks as a fraction of the period. Which way is a point moving at the instant a graph shows? The whole profile moves along, so each point takes up the displacement that its neighbour on the side the wave comes from has now; a point at a crest or a trough is momentarily at rest.
Worked example. Points R and T on a string are $0.62\ \text{cm}$ apart and a quarter of a cycle out of phase. The wave speed is $0.27\ \text{m s}^{-1}$. Find the frequency.
A quarter of a cycle is a quarter of a wavelength, so $\lambda = 4 \times 0.62 = 2.5\ \text{cm}$ and $f = v / \lambda = 0.27 / 0.025 = 11\ \text{Hz}$.
In one period $T$, the wave moves forward by one wavelength $\lambda$. So speed $= \text{distance} / \text{time} = \lambda / T = \lambda f$:
This comes straight from the definitions of speed, frequency and wavelength, and works for every progressive wave. For the two-mark derivation write both steps: in one period $T$ the wave travels one wavelength $\lambda$, so $v = \lambda / T$; and $f = 1/T$, so $v = f\lambda$. Because $v$ is fixed by the medium, a higher frequency means a shorter wavelength. A period question is the same equation the other way round: light of wavelength $460\ \text{nm}$ has $f = c / \lambda = 6.5 \times 10^{14}\ \text{Hz}$ and $T = 1/f = 1.5 \times 10^{-15}\ \text{s}$.
A cathode-ray oscilloscope 示波器 (CRO) draws a voltage signal — for sound, the output of a microphone — against time. Two controls matter:
If the time-base is $5\ \text{ms}/\text{div}$ and one full cycle takes $4$ divisions, then $T = 4 \times 5\ \text{ms} = 20\ \text{ms}$ and $f = 50\ \text{Hz}$.
The time-base is the time represented by one division (or one centimetre) across the screen, in $\text{s}\ \text{div}^{-1}$. To set it for a given signal, work backwards: a $2000\ \text{Hz}$ sound has $T = 0.50\ \text{ms}$, so for one cycle to span $2.5\ \text{cm}$ the time-base must be $0.50 / 2.5 = 0.20\ \text{ms cm}^{-1}$. For the amplitude, multiply the peak height in divisions by the y-gain: $2.0$ divisions at $3.5\ \text{mV cm}^{-1}$ is $7.0\ \text{mV}$. When the intensity of the sound is reduced to a quarter at the same frequency, the trace keeps its period and its peaks halve in height, because $I \propto A^{2}$.
A wave carries energy. The intensity 强度 at a point is the power 功率 passing through unit area at right angles to the direction of travel:
Unit: $\text{W m}^{-2}$.
Intensity is proportional to the square of the amplitude:
For a point source 点源 sending out energy equally in all directions, the wavefronts 波前 are spheres; the surface area at distance $r$ is $4\pi r^{2}$, so
Doubling the distance cuts the intensity to a quarter, which means the amplitude is halved (since $I \propto A^{2}$).
Worked example. A lamp emits $60\ \text{W}$ of light equally in all directions. Find the intensity of the light $2.0\ \text{m}$ away.
Worked example. Light of power $750\ \text{W}$ falls at right angles on a square solar panel of side $1.2\ \text{m}$. Find the intensity.
$I = P/A = 750 / 1.2^{2} = 520\ \text{W m}^{-2}$. A smaller panel in the same light receives the same intensity but less power, in proportion to its area. The other way round, a magnifying glass of radius $r$ collects a power $I \times \pi r^{2}$ and concentrates it on a small spot.
Two rules the multiple-choice questions turn on. Because $I \propto A^{2}$, the ratio of two intensities is the square of the ratio of the amplitudes: waves of amplitude $3.0\ \text{cm}$ and $2.0\ \text{cm}$ have intensities in the ratio $2.25$, and when an amplitude falls to a half the intensity falls to a quarter. For a point source, $I \propto 1/r^{2}$ means $A \propto 1/r$, so a sketch of $A/A_{0}$ against $d/x_{0}$ is a curve falling as $1/d$, not a straight line. The one-line statement the scheme wants: intensity is proportional to the amplitude squared.
y = a sin(bx + c)
A wave: a is amplitude, b sets the wavelength, c the phase. · Sebuah gelombang: a adalah amplitudo, b menetapkan panjang gelombang, c fase.
| English | Bahasa Indonesia |
|---|---|
| progressive wave/prəˈɡresɪv weɪv/ | gelombang merambat |
| wave/weɪv/ | gelombang |
| energy/ˈenədʒi/ | tenaga |
| frequency/ˈfriːkwənsi/ | frekuensi |
| phase difference/feɪz ˈdɪfrəns/ | selisih fase |
| medium/ˈmiːdɪəm/ | medium |
| oscillate/ˈɒsɪleɪt/ | berosilasi |
| propagate/ˈprɒpəɡeɪt/ | merambat |
| transverse wave/trænsˈvɜːs weɪv/ | gelombang melintang |
| longitudinal wave/ˌlɒŋɡɪˈtjuːdɪnl weɪv/ | gelombang longitudinal |
| Wave motion/weɪv ˈməʊʃn/ | Gerak gelombang |
| ripple tank/ˈrɪpl tæŋk/ | bak riak |
| displacement/dɪˈspleɪsmənt/ | pindahan |
| vector/ˈvektə/ | vektor |
| amplitude/ˈæmplɪtjuːd/ | amplitudo |
| wavelength/ˈweɪvleŋθ/ | panjang gelombang |
| in phase/ɪn feɪz/ | sefasa |
| crest/krest/ | puncak |
| period/ˈpɪərɪəd/ | tempoh |
| hertz/hɜːts/ | hertz |
| speed/spiːd/ | kecepatan (besaran skalar) |
| radian/ˈreɪdɪən/ | radian |
| cathode-ray oscilloscope/ˈkæθəʊd reɪ ɒˈsɪləskəʊp/ | osiloskop sinar katoda |
| time-base/taɪm beɪs/ | basis waktu |
| y-gain/waɪ ɡeɪn/ | penguatan y |
| intensity/ɪnˈtensɪti/ | intensitas |
| power/ˈpaʊə/ | kuasa |
| point source/pɔɪnt sɔːs/ | sumber titik |
| wavefront/ˈweɪvfrʌnt/ | gelombang muka |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The particles oscillate perpendicular 垂直 to the direction the energy travels. A wave on a rope, all electromagnetic waves, and S-waves in the Earth are transverse.
The particles oscillate parallel to the direction the energy travels. Sound in any medium, P-waves in the Earth, and the squashes on a slinky are longitudinal. The wave is made of compressions 压缩 (higher pressure, particles close together) and rarefactions 稀疏 (lower pressure, particles spread out).
A graph of particle displacement against position at one moment looks like a sine curve 正弦曲线 for both kinds of wave. The difference: for a transverse wave the displacement axis is the real sideways displacement; for a longitudinal wave it is the small back-and-forth displacement along the direction of travel (positive one way, negative the other).
For a longitudinal wave take displacement to the right as positive. Where the graph crosses zero going from positive to negative, the particles on either side have moved towards that point, so it is a compression; where it crosses from negative to positive they have moved apart, a rarefaction. Two neighbouring compressions are one wavelength apart, and a compression and the next rarefaction are half a wavelength apart. The direction of motion of a particle follows the same rule as for a transverse wave: it moves towards the displacement of its neighbour on the side the wave comes from.
A graph of particle displacement against time at one point in space is also a sine curve for both kinds. Read the period $T$ from this graph.
The two-mark comparison, with reference to the direction of energy transfer: in a transverse wave the oscillations are perpendicular to the direction of energy transfer; in a longitudinal wave they are parallel to it. Both kinds transfer energy without transferring matter, both can be reflected, refracted and diffracted, and both can form stationary waves (topic 8). Only transverse waves can be polarised, and only electromagnetic waves, which are all transverse, can travel through a vacuum; sound needs a medium. Reading the two graphs together: the displacement–distance graph gives $\lambda$, the displacement–time graph gives $T$, and $v = \lambda / T$.
y = a sin(bx + c)
Change the amplitude and wavelength of the wave. · Ubah amplitudo dan panjang gelombang gelombang tersebut.
| English | Bahasa Indonesia |
|---|---|
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | tegak lurus |
| sine curve/saɪn kɜːv/ | kurva sinus |
| compression/kəmˈpreʃn/ | pemampatan |
| rarefaction/ˌreərɪˈfækʃn/ | renggangan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
When the source 波源 of a sound moves relative to a stationary 静止 observer 观察者, the heard frequency is different from the source frequency. This is the Doppler effect 多普勒效应.
The formula (source moving at speed $v_{\text{s}}$ along the line to the observer; wave speed $v$, source frequency $f_{\text{s}}$, heard frequency $f_{\text{o}}$):
Choose the sign to match the physics:
You only need the case of a stationary observer.
Worked example. A car horn at $f_{\text{s}} = 800\ \text{Hz}$ moves at $30\ \text{m s}^{-1}$ towards a still listener. Speed of sound $v = 340\ \text{m s}^{-1}$:
Moving away at the same speed, the listener hears $340 \times 800 / (340 + 30) = 735\ \text{Hz}$: the frequency is constant while the car approaches, drops as it passes, and is constant again, lower, as it recedes. A sketch of observed frequency against time for a source passing at constant speed is two level lines joined by a fall. Working backwards, a source of $1200\ \text{Hz}$ heard as $960\ \text{Hz}$ is moving away ($f_{\text{o}} < f_{\text{s}}$): $960 = 1200 \times 340 / (340 + v_{\text{s}})$ gives $v_{\text{s}} = 85\ \text{m s}^{-1}$. A buzzer swung in a horizontal circle at $25\ \text{m s}^{-1}$ while emitting $846\ \text{Hz}$ is heard between $846 \times 330 / 355 = 786\ \text{Hz}$ (moving directly away) and $846 \times 330 / 305 = 915\ \text{Hz}$ (moving directly towards), once each per revolution. If the period of a CRO trace of the sound rises continuously, the observed frequency is falling: the source is moving away with increasing speed. The Doppler effect happens for all waves, light included (topic 25); only the sound formula is examined here.
Send the source moving and watch the wavefronts bunch up ahead (higher pitch) and stretch out behind — the siren effect, controlled by the source's speed. · Biarkan sumber bergerak dan lihat bagaimana muka gelombang berkumpul di depan (nada lebih tinggi) dan meregang di belakang — efek sirene, dikendalikan oleh kecepatan sumber.
| English | Bahasa Indonesia |
|---|---|
| source/sɔːs/ | sumber |
| stationary/ˈsteɪʃənəri/ | stasioner |
| Doppler effect/ˈdɒplə ɪˈfekt/ | Efek Doppler |
| observer/ɒbˈzɜːvə/ | pengamat |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
All electromagnetic waves 电磁波 (EM waves) are transverse and travel in a vacuum 真空 at the same speed:
The electromagnetic spectrum 电磁波谱 includes radio waves, microwaves 微波, infrared 红外线, visible light, ultraviolet 紫外线, X-rays X射线 and $\gamma$-rays γ射线.
Approximate wavelength ranges in free space (learn the orders of magnitude):
The boundaries between regions are not sharp. Use $c = f\lambda$ to change between wavelength and frequency. Only light with wavelengths $400$–$700\ \text{nm}$ can be seen.
To identify a region, convert to a wavelength with $\lambda = c / f$ and compare with the ranges: $2.1\ \text{cm}$ is a microwave; $138\ \text{pm} = 1.4 \times 10^{-10}\ \text{m}$ is an X-ray; $30\ \text{THz}$ gives $\lambda = 1.0 \times 10^{-5}\ \text{m}$, infrared; $3.0 \times 10^{16}\ \text{Hz}$ gives $1.0 \times 10^{-8}\ \text{m}$, ultraviolet. Visible light spans frequencies of about $4.3 \times 10^{14}$ to $7.5 \times 10^{14}\ \text{Hz}$, so a wave of $5.0 \times 10^{14}\ \text{Hz}$ can be seen and one of wavelength $5.0 \times 10^{-6}\ \text{m}$ (infrared) cannot. Red light has a longer wavelength and a lower frequency than green. A list "in order of increasing wavelength" runs $\gamma$-rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves. A pulse of light reflected from a wall $150\ \text{m}$ away returns after $2 \times 150 / (3.00 \times 10^{8}) = 1.0\ \mu\text{s}$.
Radio waves, visible light and gamma rays are all the same wave — only the wavelength changes, and with it the frequency, photon energy and everyday use. · Gelombang radio, cahaya tampak, dan sinar gamma semuanya merupakan gelombang yang sama — hanya panjang gelombangnya yang berubah, beserta frekuensinya, energi foton, dan penggunaannya sehari-hari.
| English | Bahasa Indonesia |
|---|---|
| electromagnetic wave/ɪˌlektrəʊməɡˈnetɪk weɪv/ | gelombang elektromagnetik |
| vacuum/ˈvækjuːm/ | vakum |
| electromagnetic spectrum/ɪˌlektrəʊməɡˈnetɪk ˈspektrəm/ | spektrum elektromagnetik |
| microwaves/ˈmaɪkrəʊweɪvz/ | gelombang mikro |
| infrared/ˌɪnfrəˈred/ | inframerah |
| ultraviolet/ˌʊltrəˈvaɪəlɪt/ | ultraviolet |
| X-rays/eks reɪz/ | Sinar-X |
| gamma-rays/ˈɡæmə reɪz/ | sinar gamma |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Polarisation 偏振 means making a transverse wave oscillate in one plane only.
So polarisation is a test: if a wave can be polarised, it must be transverse.
Plane-polarised 平面偏振 light of intensity $I_{0}$ passes through a polarising filter 偏振片 whose transmission axis 透光轴 is at angle $\theta$ to the plane of polarisation. The transmitted intensity is given by Malus's law 马吕斯定律:
Worked example. Plane-polarised light of intensity $12\ \text{W m}^{-2}$ meets a polarising filter whose axis is at $30°$ to the plane of polarisation. Find the transmitted intensity.
For two filters in a row, use Malus's law twice with the angle between each pair. Be careful with the angle each time — after the first filter the polarisation is along that filter's axis, and the second filter's angle is measured from there.
(You do not need to work out the effect of a polarising filter on an unpolarised wave.)
When a filter is rotated through $360°$ in front of plane-polarised light, the transmitted intensity varies as $\cos^{2}\theta$: it is a maximum when the transmission axis is parallel to the plane of polarisation ($0°$ and $180°$), zero when perpendicular ($90°$ and $270°$), and never negative. Because $I \propto A^{2}$, the transmitted amplitude is $A_{0}\cos\theta$: at $45°$ the intensity halves and the amplitude falls to $0.71$ of its value. For two filters, the first sets the plane of polarisation along its own axis, so the second filter's angle is measured from that: vertically polarised light through filters at $50°$ and then $80°$ to the vertical keeps $\cos^{2} 50° \times \cos^{2} 30° = 0.31$ of its intensity. The one-mark definitions: polarisation is the oscillation of a transverse wave in a single plane, which contains the direction of travel; sound cannot be polarised because its oscillations are along the direction of travel, so there is no plane to select.
intensity changes with polariser angle · intensitas berubah sesuai sudut polarizer
Rotate a polariser and see why only transverse waves can be polarised. · Putar sebuah polarizer dan lihat mengapa hanya gelombang transversal yang dapat dipolarisasi.
| English | Bahasa Indonesia |
|---|---|
| polarisation/ˌpəʊləraɪˈzeɪʃn/ | polarisasi |
| polarising filter/ˈpəʊləraɪzɪŋ ˈfɪltə/ | filter polarisasi |
| Malus's law/ˈmæləsɪz lɔː/ | hukum Malus |
| plane-polarised/pleɪn ˈpəʊləraɪzd/ | terpolarisasi bidang |
| transmission axis/trænˈsmɪʃn ˈæksɪs/ | sumbu transmisi |
| γ-rays/ˈɡæmə reɪz/ | sinar γ |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| progressive wave | a wave that transfers energy from one place to another without transferring matter |
| transverse wave | a wave whose oscillations are perpendicular to the direction of energy transfer |
| longitudinal wave | a wave whose oscillations are parallel to the direction of energy transfer |
| displacement | the distance of a particle from its equilibrium (rest) position, in a stated direction |
| amplitude | the maximum displacement of a particle from its equilibrium position |
| wavelength | the minimum distance between two points on the wave that are in phase (for example, adjacent crests) |
| period | the time for one complete oscillation of a particle (or for the wave to travel one wavelength) |
| frequency | the number of oscillations per unit time (or the number of wavefronts passing a point per unit time) |
| phase difference | the fraction of a cycle, as an angle, by which one oscillation leads or lags another |
| intensity | the power per unit area, at right angles to the direction of travel |
| Doppler effect | the change in the observed frequency of a wave when the source moves relative to the observer |
| polarisation | the oscillations of a transverse wave are in one plane only, containing the direction of travel |
| time-base | the time represented by one division across the screen of a CRO |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
When two or more waves 波 overlap at a point, the displacement 位移 there is the vector sum 矢量和 of the displacements each wave would make on its own. This is the principle of superposition 叠加.
The waves pass through each other and come out unchanged. Superposition is the base of everything in this topic.
For the two-mark statement: when two or more waves meet at a point, the resultant displacement is the sum of the displacements of the individual waves. It is the displacements that add, not the amplitudes or the intensities, and the principle applies whenever waves of the same type overlap; the waves do not have to be coherent or of equal amplitude.
If two waves of amplitude 振幅 $A_{1}$ and $A_{2}$ meet:
For intensity 强度, $I \propto A^{2}$. Two equal waves meeting in phase give intensity $(2A)^{2} = 4A^{2}$ — four times the intensity of one wave alone.
Because $I \propto A^{2}$, work in amplitudes first. Two waves of intensities $I$ and $4I$ have amplitudes $A$ and $2A$: superposing in phase gives amplitude $3A$ and intensity $9I$, in antiphase amplitude $A$ and intensity $I$. So two coherent waves of different intensities never cancel completely: the minima have intensity $(A_{2} - A_{1})^{2}$, not zero. A wave of amplitude $2A$ meeting one of amplitude $A/2$ travelling the opposite way gives a resultant that varies between $2.5A$ and $1.5A$, a stationary pattern with no true nodes. Doubling the amplitude of one of two equal waves that meet in phase takes the resultant from $2A$ to $3A$, so the intensity rises from $4A^{2}$ to $9A^{2}$: $2.25$ times.
Ketika dua atau lebih gelombang tumpang tindih di suatu titik, simpangan di sana adalah jumlah vektor dari simpangan yang akan dibuat masing-masing gelombang jika berdiri sendiri. Ini adalah prinsip superposisi.
Gelombang-gelombang itu saling menembus dan keluar tak berubah. Superposisi adalah dasar dari segala hal dalam topik ini.
Untuk pernyataan dua nilai: ketika dua atau lebih gelombang bertemu di suatu titik, simpangan resultan adalah jumlah dari simpangan-simpangan gelombang individu. Yang dijumlahkan adalah simpangannya, bukan amplitudonya atau intensitasnya, dan prinsip ini berlaku setiap kali gelombang jenis sama tumpang tindih; gelombang tidak harus koheren atau memiliki amplitudo yang sama.
Jika dua gelombang dengan amplitudo $A_{1}$ dan $A_{2}$ bertemu:
Untuk intensitas, $I \propto A^{2}$. Dua gelombang sama yang bertemu sefasa menghasilkan intensitas $(2A)^{2} = 4A^{2}$ — empat kali intensitas satu gelombang saja.
Karena $I \propto A^{2}$, kerjakan dalam amplitudo terlebih dahulu. Dua gelombang dengan intensitas $I$ dan $4I$ memiliki amplitudo $A$ dan $2A$: superposisi sefase menghasilkan amplitudo $3A$ dan intensitas $9I$, berfasa berlawanan menghasilkan amplitudo $A$ dan intensitas $I$. Jadi dua gelombang koheren dengan intensitas berbeda tidak pernah saling membatalkan sepenuhnya: minimumnya memiliki intensitas $(A_{2} - A_{1})^{2}$, bukan nol. Gelombang dengan amplitudo $2A$ yang bertemu dengan gelombang beramplitudo $A/2$ yang merambat ke arah berlawanan menghasilkan resultan yang bervariasi antara $2.5A$ dan $1.5A$, sebuah pola diam tanpa simpul sejati. Menggandakan amplitudo salah satu dari dua gelombang sama yang bertemu sefase mengubah resultan dari $2A$ menjadi $3A$, sehingga intensitas meningkat dari $4A^{2}$ menjadi $9A^{2}$: $2.25$ kali.
Two waves overlap and add. Line them up for constructive interference, or oppose them for destructive — change the phase to see both. · Dua gelombang tumpang tindih dan ditambahkan. Selaraskan untuk interferensi konstruktif, atau lawankan untuk interferensi destruktif — ubah fase untuk melihat keduanya.
| English | Bahasa Indonesia |
|---|---|
| principle of superposition/ˈprɪnsɪpl ɒv ˌsuːpəpəˈzɪʃn/ | prinsip superposisi |
| superposition/ˌsuːpəpəˈzɪʃn/ | superposisi |
| wave/weɪv/ | gelombang |
| displacement/dɪˈspleɪsmənt/ | pindahan |
| vector sum/ˈvektə sʌm/ | jumlah vektor |
| amplitude/ˈæmplɪtjuːd/ | amplitudo |
| coherent/kəʊˈhɪərənt/ | koheren |
| in phase/ɪn feɪz/ | sefasa |
| crest/krest/ | puncak |
| constructive interference/kənˈstrʌktɪv ˌɪntəˈfɪərəns/ | interferensi konstruktif |
| interference/ˌɪntəˈfɪərəns/ | intervensi |
| trough/trɒf/ | Titik terendah dari siklus bisnis adalah |
| phase difference/feɪz ˈdɪfrəns/ | selisih fase |
| destructive interference/dɪˈstrʌktɪv ˌɪntəˈfɪərəns/ | interferensi destruktif |
| minimum/ˈmɪnɪməm/ | minimum |
When two identical progressive waves 行波 travel in opposite directions and overlap, they make a stationary wave 驻波. Examples: a wave on a string reflected 反射 from a fixed end overlapping the incoming wave; sound in an air column reflected from a closed end; microwaves between an emitter and a metal sheet.
Explain how the stationary wave is formed (three or four marks, asked for a string, an air column and microwaves alike): the wave from the source travels to the far end (the wall, the closed end, the metal plate) and is reflected; the incident and reflected waves have the same frequency, wavelength and speed and travel in opposite directions; they superpose where they overlap; at the points where they always meet in phase the displacements add to give the maximum amplitude, an antinode, and at the points where they always meet in antiphase they cancel, a node. For "state the conditions": two waves of the same type, with the same frequency (and wavelength) and speed, travelling in opposite directions along the same line; equal amplitudes are needed only for the nodes to have zero amplitude. The metal plate in the microwave experiment is there to reflect the waves back along their own path.
In a stationary wave:
Particles between two nodes oscillate in phase with each other, but with different amplitudes (largest at the antinode, zero at the nodes). Particles on opposite sides of a node oscillate in antiphase 反相 (phase difference $\pi$).
So the phase difference between two points on the same loop (between adjacent nodes) is $0°$; between points on neighbouring loops it is $180°$; and between points on loops separated by one whole loop it is $0°$ again. Every point reaches its maximum displacement at the same instant and passes through zero at the same instant. A sketch of the string a quarter of a cycle after the instant of maximum displacement is a straight line along the rest position; half a cycle later it is the mirror image of the first sketch. In one period a particle at an antinode travels four amplitudes, so a particle that moves $72\ \text{mm}$ in one and a half periods has an amplitude of $12\ \text{mm}$.
Worked example. A string of length $0.80\ \text{m}$ is fixed at both ends and vibrates in its fundamental mode, where the wave speed is $240\ \text{m s}^{-1}$. Find the fundamental frequency.
One loop fits the string, so $\lambda = 2L = 1.60\ \text{m}$. Then
How a stationary wave differs from a progressive wave: it does not carry energy 能量 along its length, the pattern does not move along, and the nodes stay fixed; a progressive wave has the same amplitude everywhere and carries energy. The three differences the scheme lists: a progressive wave transfers energy and a stationary wave does not; in a progressive wave every point has the same amplitude, in a stationary wave the amplitude varies from zero at a node to a maximum at an antinode; in a progressive wave neighbouring points differ in phase, in a stationary wave all the points between two nodes are in phase.
Drive a string with a vibrator at frequency $f$ until a stationary pattern appears. Measure the distance between two well-separated nodes and divide by the number of half-wavelengths 波长 between them. Then $\lambda$ is known, and $v = f\lambda$ gives the wave speed.
For a tube closed at one end and open at the other (a resonance tube 共鸣管), the closed end is a displacement node and the open end is a displacement antinode. The fundamental 基频 has $L = \lambda/4$; the next resonance is at $L = 3\lambda/4$; and so on. For a tube open at both ends, both ends are antinodes; the fundamental is at $L = \lambda/2$.
The general rule: a closed end is a node and an open end an antinode, and a string fixed at both ends has a node at each end. So a closed pipe fits an odd number of quarter-wavelengths ($L = \lambda/4, 3\lambda/4, 5\lambda/4, \ldots$), while an open pipe and a fixed string fit whole half-wavelengths ($L = \lambda/2, \lambda, 3\lambda/2, \ldots$); a pipe open at both ends with $n$ nodes has $n + 1$ antinodes. Frequencies follow from $f = v/\lambda$: a closed pipe whose lowest note is $820\ \text{Hz}$ has $\lambda = 4L$, and a corridor $13.2\ \text{m}$ long with a reflecting door at each end resonates lowest at $f = v/2L = 330/26.4 = 12.5\ \text{Hz}$.
Worked example. A loudspeaker at the open end of a tube $0.51\ \text{m}$ long, closed at the other end, sets up a stationary wave with two nodes and two antinodes. Find the wavelength and, with $v = 340\ \text{m s}^{-1}$, the frequency.
Two nodes and two antinodes is the second mode of a closed pipe, $L = 3\lambda/4$, so $\lambda = 4 \times 0.51 / 3 = 0.68\ \text{m}$ and $f = 340 / 0.68 = 500\ \text{Hz}$. The longest wavelength that can resonate in this tube is $4L = 2.0\ \text{m}$.
The experiments the syllabus names all measure $\lambda$ from node or antinode spacing. In the resonance tube, a tube is raised out of water while a loudspeaker or tuning fork sounds at its open end; the sound is suddenly loud at the first resonance, when the air column is $\lambda/4$ long, and again at $3\lambda/4$, so the tube moves $\lambda/2$ between the two (a student needs no equipment to detect the resonance: the note becomes loud). In a dust tube, fine powder settles into heaps at the displacement nodes, $\lambda/2$ apart. With microwaves, a receiver moved along the line between the transmitter and a metal reflecting plate reads a minimum every $\lambda/2$: a receiver that starts at a minimum and passes six more minima in $1.05\ \text{m}$ has crossed $3\lambda$, so $\lambda = 0.35\ \text{m}$ and, with $v = 340\ \text{m s}^{-1}$ for the equivalent sound experiment, $f = 970\ \text{Hz}$. A stationary wave in a microwave oven melts chocolate at the antinodes, $\lambda/2$ apart, so the spot spacing and the oven's frequency ($2.45\ \text{GHz}$) give the speed of light from $c = f\lambda$.
Ketika dua gelombang merambat identik bergerak ke arah berlawanan dan tumpang tindih, mereka membentuk gelombang berdiri. Contoh: gelombang pada senar terpantul dari ujung tetap yang tumpang tindih dengan gelombang datang; bunyi dalam kolom udara terpantul dari ujung tertutup; gelombang mikro di antara pemancar dan lembaran logam.
Jelaskan bagaimana gelombang berdiri terbentuk (tiga atau empat poin, ditanyakan untuk senar, kolom udara, dan gelombang mikro serupa): gelombang dari sumber merambat ke ujung jauh (dinding, ujung tertutup, pelat logam) dan terpantul; gelombang datang dan pantul memiliki frekuensi, panjang gelombang, dan kecepatan yang sama serta merambat ke arah berlawanan; mereka superposisi di mana mereka tumpang tindih; pada titik-titik di mana mereka selalu bertemu sefase, perpindahan bertambah untuk memberikan amplitudo maksimum, antinod, dan pada titik-titik di mana mereka selalu bertemu berfasa berlawanan, mereka saling membatalkan, yaitu nod. Untuk "nyatakan kondisi": dua gelombang dari jenis yang sama, dengan frekuensi (dan panjang gelombang) dan kecepatan yang sama, merambat ke arah berlawanan sepanjang garis yang sama; amplitudo yang sama hanya diperlukan agar nod memiliki amplitudo nol. Pelat logam dalam eksperimen gelombang mikro berfungsi untuk memantulkan gelombang kembali sepanjang jalurnya sendiri.
Dalam gelombang berdiri:
Partikel di antara dua nod bergetar sefase satu sama lain, tetapi dengan amplitudo yang berbeda (terbesar di antinode, nol di nod). Partikel di sisi berlawanan dari sebuah nod bergetar berfasa berlawanan (selisih fase $\pi$).
Jadi selisih fase antara dua titik pada loop yang sama (di antara nod berdekatan) adalah $0°$; antara titik pada loop bertetangga adalah $180°$; dan antara titik pada loop yang dipisahkan oleh satu loop penuh adalah $0°$ lagi. Setiap titik mencapai perpindahan maksimumnya pada saat yang sama dan melewati nol pada saat yang sama. Sketsa senar seperempat siklus setelah saat perpindahan maksimum adalah garis lurus sepanjang posisi istirahat; setengah siklus kemudian itu adalah cerminan dari sketsa pertama. Dalam satu periode, partikel di antinode menempuh empat kali amplitudo, sehingga partikel yang bergerak $72\ \text{mm}$ dalam satu setengah periode memiliki amplitudo $12\ \text{mm}$.
Contoh kerja. Senar sepanjang $0.80\ \text{m}$ dikunci di kedua ujungnya dan bergetar dalam modus fundamentalnya, di mana kecepatan gelombangnya adalah $240\ \text{m s}^{-1}$. Temukan frekuensi fundamental.
Satu loop muat pada senar, jadi $\lambda = 2L = 1.60\ \text{m}$. Maka
Bagaimana gelombang berdiri berbeda dari gelombang merambat: ia tidak membawa energi sepanjang panahnya, polanya tidak bergerak maju, dan nod tetap; gelombang merambat memiliki amplitudo yang sama di semua tempat dan membawa energi. Tiga perbedaan yang disebutkan dalam pedoman: gelombang merambat mentransfer energi dan gelombang berdiri tidak; dalam gelombang merambat setiap titik memiliki amplitudo yang sama, dalam gelombang berdiri amplitudonya bervariasi dari nol di nod hingga maksimum di antinode; dalam gelombang merambat titik-titik tetangga berbeda fasenya, dalam gelombang berdiri semua titik di antara dua nod adalah sefase.
Kendalikan senar dengan getaran pada frekuensi $f$ hingga muncul pola diam. Ukur jarak antara dua nod yang terpisah jauh dan bagi dengan jumlah setengah-panjang-gelombang di antaranya. Kemudian $\lambda$ diketahui, dan $v = f\lambda$ memberikan kecepatan gelombang.
Untuk tabung yang tertutup di satu ujung dan terbuka di ujung lainnya (tabung resonansi), ujung tertutup adalah nod perpindahan dan ujung terbuka adalah antinode perpindahan. Fundamental memiliki $L = \lambda/4$; resonansi berikutnya ada pada $L = 3\lambda/4$; dan seterusnya. Untuk tabung yang terbuka di kedua ujungnya, kedua ujungnya adalah antinode; fundamentalnya ada pada $L = \lambda/2$.
Aturan umum: ujung tertutup adalah simpul dan ujung terbuka adalah perut, serta senar yang terikat di kedua ujungnya memiliki simpul di setiap ujungnya. Jadi pipa tertutup sesuai dengan jumlah ganjil seperempat panjang gelombang ($L = \lambda/4, 3\lambda/4, 5\lambda/4, \ldots$), sedangkan pipa terbuka dan senar terikat sesuai dengan kelipatan setengah panjang gelombang penuh ($L = \lambda/2, \lambda, 3\lambda/2, \ldots$); pipa terbuka di kedua ujungnya dengan $n$ simpul memiliki $n + 1$ perut. Frekuensi diturunkan dari $f = v/\lambda$: pipa tertutup yang nada terendahnya adalah $820\ \text{Hz}$ memiliki $\lambda = 4L$, dan lorong $13.2\ \text{m}$ panjang dengan pintu pemantul di setiap ujungnya beresonansi paling rendah pada $f = v/2L = 330/26.4 = 12.5\ \text{Hz}$.
Contoh terpadu. Loudspeaker di ujung terbuka tabung $0.51\ \text{m}$ panjang, tertutup di ujung lainnya, menghasilkan gelombang berdiri dengan dua simpul dan dua perut. Tentukan panjang gelombangnya dan, dengan $v = 340\ \text{m s}^{-1}$, frekuensinya.
Dua simpul dan dua perut adalah moda kedua pipa tertutup, $L = 3\lambda/4$, sehingga $\lambda = 4 \times 0.51 / 3 = 0.68\ \text{m}$ dan $f = 340 / 0.68 = 500\ \text{Hz}$. Panjang gelombang terpanjang yang dapat beresonansi dalam tabung ini adalah $4L = 2.0\ \text{m}$.
Eksperimen yang disebutkan dalam silabus semuanya mengukur $\lambda$ dari jarak antar node atau antinode. Pada tabung resonansi, sebuah tabung dinaikkan keluar dari air sementara speaker atau garpu tala berbunyi di ujung terbuka; suara tiba-tiba keras pada resonansi pertama, ketika kolom udara memiliki panjang $\lambda/4$, dan lagi pada $3\lambda/4$, sehingga tabung bergerak $\lambda/2$ di antara keduanya (siswa tidak memerlukan peralatan untuk mendeteksi resonansi: nada menjadi keras). Pada tabung debu, serbuk halus mengendap membentuk gundukan di node simpangan, $\lambda/2$ terpisah. Dengan gelombang mikro, penerima yang digerakkan sepanjang garis antara pemancar dan pelat pemantul logam membaca nilai minimum setiap $\lambda/2$: jika penerima dimulai dari minimum dan melewati enam minimum lainnya dalam $1.05\ \text{m}$, maka telah melintasi $3\lambda$, sehingga $\lambda = 0.35\ \text{m}$ dan, dengan $v = 340\ \text{m s}^{-1}$ untuk eksperimen bunyi setara, $f = 970\ \text{Hz}$. Gelombang diam dalam oven gelombang mikro mencairkan cokelat di antinode, $\lambda/2$ terpisah, sehingga jarak bintik dan frekuensi oven ($2.45\ \text{GHz}$) memberikan kecepatan cahaya dari $c = f\lambda$.
A string fixed at both ends only resonates at its harmonics. Drag n to see the nodes, antinodes and how the wavelength changes. · Senar yang terikat di kedua ujung hanya beresonansi pada harmoniknya. Seret n untuk melihat simpul, antinod, dan bagaimana panjang gelombang berubah.
y = y₁ + y₂
Two waves superpose: where they reinforce you get antinodes, where they cancel, nodes. · Dua gelombang superposisi: di mana mereka saling memperkuat Anda mendapatkan antinod, di mana mereka saling meniadakan, simpul.
| English | Bahasa Indonesia |
|---|---|
| node/nəʊd/ | simpul |
| progressive wave/prəˈɡresɪv weɪv/ | gelombang merambat |
| stationary wave/ˈsteɪʃənəri weɪv/ | gelombang berdiri |
| reflected/rɪˈflektɪd/ | ter pantulkan |
| wavelength/ˈweɪvleŋθ/ | panjang gelombang |
| maximum/ˈmæksɪməm/ | maksimum |
| antinode/ˌæntɪˈnəʊd/ | antinode |
| fundamental/ˌfʌndəˈmentl/ | fundamental |
| energy/ˈenədʒi/ | tenaga |
| resonance tube/ˈrezənəns tjuːb/ | pipa resonansi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Diffraction 衍射 is the spreading of a wave after it passes through a gap or around an obstacle 障碍物. All waves diffract — water, sound, light, microwaves.
For the two-mark "state what is meant by diffraction": the spreading of a wave as it passes through a gap (an aperture) or around the edge of an obstacle, into the region behind it. Diffraction changes the direction the wave travels in, but not its speed, frequency or wavelength.
The amount of spreading depends on the ratio of wavelength to gap width:
Show this with water waves in a ripple tank 水波槽: straight waves meet a barrier with a gap, and the waves curve more as the gap is made narrower. The same idea is why you can hear someone around a corner (speech has $\lambda$ near 1 m, close to the gap size) but cannot see them (visible light has $\lambda \sim 500\ \text{nm}$, far smaller than the gap).
For the strongest spreading the gap should be about one wavelength wide. So to increase the diffraction of a given wave, make the gap narrower; for a given gap, use a longer wavelength, which means a lower frequency. Radio waves of wavelength $1.5\ \text{km}$ diffract around a mountain and reach an aerial behind it; microwaves of $1.5\ \text{cm}$ do not. Sound of $0.44\ \text{kHz}$ in air has $\lambda = 330 / 440 = 0.75\ \text{m}$, so features of about $0.75\ \text{m}$ diffract it most, and of the sounds passing through a doorway $0.80\ \text{m}$ wide the low frequencies spread out most. Making the gap many wavelengths wide, or raising the frequency, reduces the spreading.
Difraksi adalah penyebaran gelombang setelah melewati celah atau mengelilingi halangan. Semua gelombang mengalami difraksi — air, suara, cahaya, gelombang mikro.
Untuk soal bernilai dua poin "nyatakan apa yang dimaksud dengan difraksi": penyebaran gelombang saat melewati celah ( aperture ) atau di sekitar tepi halangan, ke daerah di belakangnya. Difraksi mengubah arah perjalanan gelombang, tetapi tidak mengubah kecepatannya, frekuensinya, atau panjang gelombangnya.
Jumlah penyebaran bergantung pada rasio panjang gelombang terhadap lebar celah:
Tunjukkan ini dengan gelombang air di bak riak: gelombang lurus bertemu penghalang dengan celah, dan gelombang melengkung semakin besar saat celah dibuat semakin sempit. Ide yang sama adalah alasan mengapa Anda bisa mendengar seseorang di balik sudut (ucapan memiliki $\lambda$ mendekati 1 m, dekat dengan ukuran celah) tetapi tidak bisa melihat mereka (cahaya tampak memiliki $\lambda \sim 500\ \text{nm}$, jauh lebih kecil dari celah).
Untuk penyebaran terkuat, celah harus selebar satu panjang gelombang. Jadi untuk meningkatkan difraksi gelombang tertentu, buatlah celah lebih sempit; untuk celah tertentu, gunakan panjang gelombang yang lebih panjang, yang berarti frekuensi lebih rendah. Gelombang radio dengan panjang gelombang $1.5\ \text{km}$ difraksi mengelilingi gunung dan mencapai antena di belakangnya; gelombang mikro dengan $1.5\ \text{cm}$ tidak. Suara dengan $0.44\ \text{kHz}$ di udara memiliki $\lambda = 330 / 440 = 0.75\ \text{m}$, sehingga fitur sekitar $0.75\ \text{m}$ memfraksinya paling banyak, dan dari suara yang melewati pintu $0.80\ \text{m}$ lebar, frekuensi rendah menyebar paling luas. Membuat celah selingkaran panjang gelombang atau menaikkan frekuensi mengurangi penyebaran.
Two overlapping waves add where they are in phase and cancel where out of phase — change the phase to see the result. This is what makes diffraction patterns. · Dua gelombang yang tumpang tindih bertambah di tempat sefasa dan saling meniadakan di tempat berlawanan fasa — ubah fasanya untuk melihat hasilnya. Inilah yang membentuk pola difraksi.
| English | Bahasa Indonesia |
|---|---|
| diffraction/dɪˈfrækʃn/ | difraksi |
| obstacle/ˈɒbstəkl/ | halangan |
| ripple tank/ˈrɪpl tæŋk/ | bak riak |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Interference 干涉 is the superposition of two coherent 相干 waves to give a steady pattern of high-amplitude regions (constructive) and low-amplitude regions (destructive).
Two sources are coherent when they emit waves with a constant phase difference (which also needs the same frequency). Two separate lamps are not coherent — their phase changes randomly, so any pattern flickers too fast to see and you get only an average.
The one-mark definition: coherent waves have a constant phase difference, which requires the same frequency. They need not be in phase with each other: two coherent sources emitting $180°$ apart give a pattern whose central line is a minimum. Two separate lasers, or a lamp and a laser, are not coherent even when their frequencies happen to match, so no steady pattern forms.
To make coherent light from one source, pass it through two slits 狭缝 in a double-slit 双缝 setup. Both slits are lit by the same wavefront, so the two beams keep a fixed phase relationship.
To see two-source fringes you need:
In practice, for light: a single slit (or a laser) makes the two slits coherent; the slits are narrow, so each diffracts the light into the region where the two beams overlap; and the slits are close together with the screen far away, so that the fringes are wide enough to see.
For two coherent sources, what happens at a point depends on the path difference 路程差 $\Delta x$ between the two waves arriving there:
The path difference fixes the phase difference: one wavelength of path is $360°$. Waves that have travelled $100\ \text{cm}$ and $80\ \text{cm}$ from two in-phase sources of wavelength $8.0\ \text{cm}$ arrive with a path difference of $2.5\lambda$, a phase difference of $180°$, and cancel; microwaves of wavelength $4\ \text{cm}$ whose paths differ by $6\ \text{cm}$ ($1.5\lambda$) give a minimum, of zero intensity only if the two amplitudes are equal. Lowering both source frequencies equally lengthens the wavelength, so the same path difference is a smaller number of wavelengths and the point is no longer a minimum.
For two slits a distance $a$ apart, with a screen a distance $D$ away (assume $D \gg a$), light of wavelength $\lambda$ makes fringes on the screen.
The fringe spacing 条纹间距 $x$ (one fringe 条纹 to the next) is
Bright fringes (maximum 极大) are where the path difference is a whole number of $\lambda$; dark fringes (minimum 极小) where it is $(n + \tfrac{1}{2})\lambda$. The fringes are equally spaced.
To make the fringe spacing smaller: increase $a$ (slits further apart), reduce $D$ (screen closer), or use a shorter $\lambda$ (bluer light).
Explain how the pattern of bright and dark fringes is formed (three marks): the light diffracts at each slit; the two diffracted beams overlap and superpose; where the path difference from the two slits is a whole number of wavelengths the waves arrive in phase and interfere constructively, giving a bright fringe, and where it is an odd number of half-wavelengths they arrive in antiphase and interfere destructively, giving a dark fringe. A brighter source makes the bright fringes brighter but does not change their spacing; changing to blue light makes the spacing smaller, so to keep the same spacing the slits must be moved closer together; making each slit narrower increases the diffraction, so fringes appear across a wider region, with the spacing unchanged.
Worked example. In a double-slit experiment the slits are $0.50\ \text{mm}$ apart and lit by light of wavelength $600\ \text{nm}$. The screen is $2.0\ \text{m}$ away. Find the fringe spacing.
Worked example. Red light of wavelength $680\ \text{nm}$ falls on slits $0.16\ \text{mm}$ apart. The distance between the centres of the first and ninth dark fringes is $3.2\ \text{cm}$. Find the distance $D$ to the screen.
Eight fringe spacings make $3.2\ \text{cm}$, so $x = 4.0\ \text{mm}$ and $D = ax / \lambda = (0.16 \times 10^{-3})(4.0 \times 10^{-3}) / (680 \times 10^{-9}) = 0.94\ \text{m}$. On a screen $5.0\ \text{cm}$ wide centred on the pattern, with $x = 2.4\ \text{mm}$, ten bright fringes fit on each side of the central one: $21$ in all. A graph of $x$ against $a$ is a curve falling as $1/a$; a graph of $x$ against $\lambda$ (or against $D$) is a straight line through the origin with gradient $D/a$ (or $\lambda/a$), from which $a$ can be found.
Interferensi adalah superposisi dua gelombang koheren untuk menghasilkan pola stabil dari wilayah amplitudo tinggi (konstruktif) dan wilayah amplitudo rendah (destruktif).
Dua sumber koheren ketika memancarkan gelombang dengan beda fase konstan (yang juga membutuhkan frekuensi yang sama). Dua lampu terpisah tidak koheren — fase mereka berubah secara acak, sehingga pola apa pun berkedip terlalu cepat untuk dilihat dan Anda hanya mendapatkan rata-rata.
Definisi satu-mata: gelombang koheren memiliki selisih fase yang konstan, yang memerlukan frekuensi yang sama. Mereka tidak perlu sefasa satu sama lain: dua sumber koheren yang memancarkan $180°$ terpisah menghasilkan pola di mana garis tengahnya adalah minimum. Dua laser terpisah, atau sebuah lampu dan sebuah laser, tidak koheren meskipun frekuensinya kebetulan sama, sehingga tidak terbentuk pola yang stabil.
Untuk menghasilkan cahaya koheren dari satu sumber, arahkan melalui dua celah dalam susunan celah ganda. Kedua celah diterangi oleh bidang gelombang yang sama, sehingga kedua berkas mempertahankan hubungan fase yang tetap.
Untuk melihat pita dua-sumber Anda membutuhkan:
Dalam praktik, untuk cahaya: sebuah celah tunggal (atau laser) membuat kedua celah menjadi koheren; celah-celahnya sempit, sehingga masing-masing mendifraksi cahaya ke wilayah di mana kedua berkas saling tumpang tindih; dan celah-celahnya berdekatan dengan layar jauh, sehingga pita cukup lebar untuk terlihat.
Untuk dua sumber koheren, apa yang terjadi pada suatu titik bergantung pada selisih lintasan $\Delta x$ antara dua gelombang yang tiba di sana:
Selisih lintasan menentukan selisih fase: satu panjang gelombang lintasan setara dengan $360°$. Gelombang yang telah menempuh $100\ \text{cm}$ dan $80\ \text{cm}$ dari dua sumber sefasa dengan panjang gelombang $8.0\ \text{cm}$ tiba dengan selisih lintasan $2.5\lambda$, selisih fase $180°$, dan saling membatalkan; gelombang mikro dengan panjang gelombang $4\ \text{cm}$ yang lintasanya berbeda $6\ \text{cm}$ ($1.5\lambda$) menghasilkan minimum, intensitas nol hanya jika kedua amplitudonya sama. Menurunkan kedua frekuensi sumber secara sama memperpanjang panjang gelombang, sehingga selisih lintasan yang sama merupakan jumlah panjang gelombang yang lebih kecil dan titik tersebut bukan lagi minimum.
Untuk dua celah yang berjarak $a$, dengan layar berjarak $D$ (asumsikan $D \gg a$), cahaya dengan panjang gelombang $\lambda$ menghasilkan pita pada layar.
Jarak antar-pita $x$ (dari satu pita ke pita berikutnya) adalah
Pita terang (maksimum) terjadi ketika selisih lintasan merupakan bilangan bulat $\lambda$; pita gelap (minimum) terjadi ketika selisih lintasan adalah $(n + \tfrac{1}{2})\lambda$. Pita-pita ini berjarak sama.
Untuk membuat jarak antar-pita lebih kecil: tingkatkan $a$ (celah lebih berjauhan), kurangi $D$ (layar lebih dekat), atau gunakan $\lambda$ yang lebih pendek (cahaya lebih biru).
Jelaskan bagaimana pola pita terang dan gelap terbentuk (tiga mata): cahaya terdifraksi di setiap celah; dua berkas terdifraksi saling tumpang tindih dan bersuperposisi; di mana selisih lintasan dari kedua celah merupakan bilangan bulat panjang gelombang, gelombang tiba sefasa dan mengalami interferensi konstruktif, menghasilkan pita terang, dan di mana selisih lintasan merupakan bilangan ganjil setengah panjang gelombang, mereka tiba antifase dan mengalami interferensi destruktif, menghasilkan pita gelap. Sumber yang lebih terang membuat pita terang lebih cerah tetapi tidak mengubah jaraknya; beralih ke cahaya biru membuat jaraknya lebih kecil, sehingga untuk menjaga jarak yang sama, celah harus dipindahkan lebih berdekatan; membuat setiap celah lebih sempit meningkatkan difraksi, sehingga pita muncul di wilayah yang lebih luas, dengan jarak yang tak berubah.
Contoh terpecahkan. Dalam eksperimen celah ganda, celah-celah tersebut berjarak $0.50\ \text{mm}$ dan diterangi cahaya dengan panjang gelombang $600\ \text{nm}$. Layar berada pada jarak $2.0\ \text{m}$. Tentukan jarak antar-pita.
Contoh terpecahkan. Cahaya merah dengan panjang gelombang $680\ \text{nm}$ jatuh pada celah yang berjarak $0.16\ \text{mm}$. Jarak antara pusat pita gelap pertama dan kesembilan adalah $3.2\ \text{cm}$. Tentukan jarak $D$ ke layar.
Delapan jarak antar-pita membentuk $3.2\ \text{cm}$, sehingga $x = 4.0\ \text{mm}$ dan $D = ax / \lambda = (0.16 \times 10^{-3})(4.0 \times 10^{-3}) / (680 \times 10^{-9}) = 0.94\ \text{m}$. Pada layar selebar $5.0\ \text{cm}$ yang berpusat pada pola, dengan $x = 2.4\ \text{mm}$, sepuluh pita terang muat di setiap sisi pita tengah: $21$ total. Grafik $x$ terhadap $a$ adalah kurva menurun saat $1/a$; grafik $x$ terhadap $\lambda$ (atau terhadap $D$) adalah garis lurus melalui titik asal dengan kemiringan $D/a$ (atau $\lambda/a$), dari mana $a$ dapat ditemukan.
y = y₁ + y₂
In phase → constructive (bright/loud); antiphase → destructive (dark/quiet). · Sefase → konstruktif (terang/keras); antifase → destruktif (gelap/lembut).
| English | Bahasa Indonesia |
|---|---|
| slit/slɪt/ | celah |
| double-slit/ˈdʌbl slɪt/ | celah ganda |
| path difference/pæθ ˈdɪfrəns/ | selisih lintasan |
| fringe spacing/frɪndʒ ˈspeɪsɪŋ/ | jarak garis terang |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A diffraction grating 衍射光栅 has many equally spaced slits — often hundreds or thousands per millimetre. Each slit is a coherent source. A maximum is seen at angle $\theta$ from the normal 法线 to the grating when
where $d$ is the slit spacing 缝间距 (centre to centre), $n = 0, \pm 1, \pm 2, \ldots$ is the order 级次, and $\lambda$ is the wavelength.
Worked example. A diffraction grating has $500$ lines per mm. Light of wavelength $600\ \text{nm}$ is shone normally on it. Find the angle of the first-order ($n = 1$) maximum.
The slit spacing is $d = \dfrac{1}{500}\ \text{mm} = 2.0 \times 10^{-6}\ \text{m}$, so
Compared with the double slit, a grating gives much sharper maxima, because more slits add together — every other direction is cancelled by many slits.
To describe the diffraction at the grating: the light spreads out (diffracts) at every slit, and the waves from all the slits superpose; in the directions where the path difference between neighbouring slits is a whole number of wavelengths they are all in phase, so sharp maxima form there and almost nothing in between. A graph of intensity against angle is a set of narrow peaks at $\theta = 0$ and at $\pm\theta_{1}, \pm\theta_{2}, \ldots$, where $\sin\theta_{n} = n\lambda/d$: equally spaced in $\sin\theta$, so slightly further apart in $\theta$ at the higher orders.
Worked example. Light of wavelength $680\ \text{nm}$ is incident normally on a grating with $450$ lines per mm. Find the angle between the two second-order maxima.
$d = 1/450\ \text{mm} = 2.22 \times 10^{-6}\ \text{m}$, so $\sin\theta_{2} = 2 \times 680 \times 10^{-9} / (2.22 \times 10^{-6}) = 0.612$ and $\theta_{2} = 37.7°$; the two second-order beams are $2\theta_{2} = 75°$ apart.
If a grating has $N$ lines per millimetre, then $d = 1/N$ millimetres $= 10^{-3}/N$ metres. For $450$ lines per mm, $d = 1/450\ \text{mm} \approx 2.22\ \mu\text{m}$.
For a given grating and wavelength, $\sin\theta = n\lambda/d$ cannot be more than $1$, so the highest order seen is
If $d/\lambda = 3.27$, orders up to $n = 3$ exist; $n = 4$ would need $\sin\theta > 1$ and is not seen.
So the total number of maxima on a wide screen is $2n_{\text{max}} + 1$: for $700\ \text{nm}$ light and $400$ lines per mm, $d/\lambda = 3.57$, so $n_{\text{max}} = 3$ and seven beams are seen. A shorter wavelength gives more orders at smaller angles. With white light every order except the zero order is a spectrum, violet nearest the centre and red furthest out, because $\theta$ grows with $\lambda$; the zero order stays white. Two wavelengths give a maximum at the same angle when $n_{1}\lambda_{1} = n_{2}\lambda_{2}$: the third order of $400\ \text{nm}$ coincides with the second order of $600\ \text{nm}$.
Shine parallel light of unknown wavelength straight at the grating. Measure the angle $\theta_{1}$ of the first-order maximum from the centre. Then $\lambda = d \sin\theta_{1}$. Repeating for higher orders and averaging reduces error.
Measure the angle between the first-order maxima on the two sides and halve it, which cancels any error in setting the zero; higher orders give larger angles and so a smaller percentage uncertainty; and plotting $\sin\theta$ against $n$ for several orders gives a straight line through the origin of gradient $\lambda/d$, so $\lambda = Gd$ (or, for a known wavelength, $d = \lambda/G$). Two things must be right: $\theta$ is measured from the normal to the grating, not from its surface, and $d$ is the distance between adjacent lines, so $400$ lines per mm means $d = 2.5\ \mu\text{m}$, never $400$.
Sebuah kisi difraksi memiliki banyak celah yang berjarak sama — seringkali ratusan atau ribuan per milimeter. Setiap celah adalah sumber koheren. Maksimum terlihat pada sudut $\theta$ dari garis normal kisi ketika
di mana $d$ adalah jarak antar-celah (pusat ke pusat), $n = 0, \pm 1, \pm 2, \ldots$ adalah orde, dan $\lambda$ adalah panjang gelombang.
Contoh terpecahkan. Sebuah kisi difraksi memiliki $500$ garis per mm. Cahaya dengan panjang gelombang $600\ \text{nm}$ disorotkan tegak lurus padanya. Tentukan sudut maksimum orde pertama ($n = 1$).
Jarak antar-celah adalah $d = \dfrac{1}{500}\ \text{mm} = 2.0 \times 10^{-6}\ \text{m}$, sehingga
Dibandingkan dengan celah ganda, kisi memberikan maksimum yang jauh lebih tajam, karena lebih banyak celah saling menambah — setiap arah lainnya dibatalkan oleh banyak celah.
Untuk mendiagramkan difraksi pada kisi difraksi: cahaya menyebar (terdifraksi) di setiap celah, dan gelombang dari semua celah saling tumpang tindih; pada arah di mana selisih lintasan antara celah yang berdekatan adalah bilangan bulat keliling gelombang, semuanya berada dalam fase, sehingga maksimum tajam terbentuk di sana dan hampir tidak ada intensitas di antaranya. Grafik intensitas terhadap sudut menunjukkan serangkaian puncak sempit di $\theta = 0$ dan di $\pm\theta_{1}, \pm\theta_{2}, \ldots$, di mana $\sin\theta_{n} = n\lambda/d$: berjarak sama dalam $\sin\theta$, sehingga sedikit lebih jauh satu sama lain dalam $\theta$ pada orde yang lebih tinggi.
Contoh terpecahkan. Cahaya dengan panjang gelombang $680\ \text{nm}$ jatuh tegak lurus pada kisi dengan $450$ garis per mm. Tentukan sudut antara dua maksimum orde kedua.
$d = 1/450\ \text{mm} = 2.22 \times 10^{-6}\ \text{m}$, sehingga $\sin\theta_{2} = 2 \times 680 \times 10^{-9} / (2.22 \times 10^{-6}) = 0.612$ dan $\theta_{2} = 37.7°$; dua berkas orde kedua berjarak $2\theta_{2} = 75°$ satu sama lain.
Jika kisi memiliki $N$ garis per milimeter, maka $d = 1/N$ milimeter $= 10^{-3}/N$ meter. Untuk $450$ garis per mm, $d = 1/450\ \text{mm} \approx 2.22\ \mu\text{m}$.
Untuk kisi dan panjang gelombang tertentu, $\sin\theta = n\lambda/d$ tidak bisa lebih besar dari $1$, sehingga orde tertinggi yang terlihat adalah
Jika $d/\lambda = 3.27$, orde hingga $n = 3$ ada; $n = 4$ akan memerlukan $\sin\theta > 1$ dan tidak terlihat.
Jadi total jumlah maksimum pada layar lebar adalah $2n_{\text{max}} + 1$: untuk cahaya $700\ \text{nm}$ dan $400$ garis per mm, $d/\lambda = 3.57$, sehingga $n_{\text{max}} = 3$ dan tujuh berkas terlihat. Panjang gelombang yang lebih pendek menghasilkan lebih banyak orde pada sudut yang lebih kecil. Dengan cahaya putih, setiap orde kecuali orde nol adalah spektrum, violet terdekat dengan pusat dan merah terjauh keluar, karena $\theta$ bertambah seiring dengan $\lambda$; orde nol tetap putih. Dua panjang gelombang memberikan maksimum pada sudut yang sama ketika $n_{1}\lambda_{1} = n_{2}\lambda_{2}$: orde ketiga dari $400\ \text{nm}$ berimpit dengan orde kedua dari $600\ \text{nm}$.
Sinarikan cahaya paralel dengan panjang gelombang tak diketahui lurus ke kisi. Ukur sudut $\theta_{1}$ dari maksimum orde pertama dari pusat. Maka $\lambda = d \sin\theta_{1}$. Mengulang untuk orde yang lebih tinggi dan mengambil rata-rata mengurangi kesalahan.
Ukur sudut antara maksimum orde pertama di kedua sisi dan bagi dua, yang mengeliminasi kesalahan apa pun dalam pengaturan nol; orde yang lebih tinggi menghasilkan sudut yang lebih besar dan sehingga ketidakpastian persentase yang lebih kecil; dan memplot $\sin\theta$ terhadap $n$ untuk beberapa orde menghasilkan garis lurus melalui titik asal dengan gradien $\lambda/d$, sehingga $\lambda = Gd$ (atau, untuk panjang gelombang yang diketahui, $d = \lambda/G$). Dua hal harus benar: $\theta$ diukur dari normal kisi, bukan dari permukaannya, dan $d$ adalah jarak antar garis bersebelahan, jadi $400$ garis per mm berarti $d = 2.5\ \mu\text{m}$, bukan $400$.
Two waves add when in phase and cancel when out of phase — change the phase and watch the resultant. A grating's many slits make the bright fringes razor-sharp. · Dua gelombang bertambah saat sefase dan saling membatalkan saat tidak sefase — ubah fase dan saksikan hasil resultannya. Banyaknya celah pada kisi membuat garis terang sangat tajam seperti pisau cukur.
| English | Bahasa Indonesia |
|---|---|
| intensity/ɪnˈtensɪti/ | intensitas |
| antiphase/ˌæntɪˈfeɪz/ | antifase |
| diffraction grating/dɪˈfrækʃn ˈɡreɪtɪŋ/ | kisi difraksi |
| normal/ˈnɔːml/ | normal |
| slit spacing/slɪt ˈspeɪsɪŋ/ | jarak celah |
| order/ˈɔːdə/ | orde |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| principle of superposition | when two or more waves meet at a point, the resultant displacement is the sum of the displacements of the individual waves |
| stationary wave | the pattern formed when two progressive waves of the same frequency and speed travel in opposite directions and superpose, with nodes and antinodes that do not move |
| node | a point on a stationary wave where the displacement is always zero |
| antinode | a point on a stationary wave where the amplitude is a maximum |
| diffraction | the spreading of a wave as it passes through a gap or around the edge of an obstacle |
| interference | the superposition of waves from coherent sources, giving a steady pattern of maxima and minima |
| coherence | waves that have a constant phase difference (and so the same frequency) |
| path difference | the difference between the distances travelled by two waves from their sources to a point |
| fringe spacing | the distance between the centres of two adjacent bright (or dark) fringes |
| order of a maximum | the whole number $n$ in $d\sin\theta = n\lambda$, the number of wavelengths of path difference between adjacent slits |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| prinsip superposisi | ketika dua gelombang atau lebih bertemu di suatu titik, simpangan resultan adalah jumlah simpangan dari gelombang-gelombang individu |
| gelombang diam | pola yang terbentuk ketika dua gelombang merambat dengan frekuensi dan kecepatan yang sama bergerak berlawanan arah dan saling tumpang tindih, dengan nodus dan antinodus yang tidak bergerak |
| nodus | titik pada gelombang diam di mana simpangannya selalu nol |
| antinodus | titik pada gelombang diam di mana amplitudonya maksimum |
| difraksi | penyebaran gelombang saat melewati celah atau mengelilingi tepi penghalang |
| interferensi | tumpang tindih gelombang dari sumber koheren, menghasilkan pola stabil maksimum dan minimum |
| koherensi | gelombang yang memiliki beda fase konstan (dan karena itu frekuensi yang sama) |
| selisih lintasan | perbedaan antara jarak yang ditempuh oleh dua gelombang dari sumbernya ke suatu titik |
| jarak fringe | jarak antara pusat dua fringe terang (atau gelap) yang bersebelahan |
| orde maksimum | bilangan bulat $n$ dalam $d\sin\theta = n\lambda$, yaitu jumlah keliling gelombang selisih lintasan antara celah bersebelahan |
| English | Bahasa Indonesia |
|---|---|
| fringe/frɪndʒ/ | garis terang |
| polarisation/ˌpəʊləraɪˈzeɪʃn/ | polarisasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
An electric current 电流 is a flow of charge carriers 载流子. In a metal the carriers are negative conduction electrons 电子; in an electrolyte 电解质 they are positive and negative ions 离子; in a semiconductor 半导体 they may be electrons or "holes" 空穴. The conventional current 常规电流 direction is the way positive charge would flow — opposite to the real flow of electrons in a wire.
Charge is quantised 量子化: the smallest free unit of charge is the elementary charge 基本电荷
Every free charge in this syllabus is a whole-number multiple of $e$. The unit of charge is the coulomb 库仑, $\text{C}$.
So a particle can carry $4.8 \times 10^{-19}\ \text{C}$ (three electrons' worth) or $-2.4 \times 10^{-18}\ \text{C}$ (fifteen), but never $1.1 \times 10^{-19}\ \text{C}$ or $6.4 \times 10^{-20}\ \text{C}$; a set of measurements that gives a charge carrier $2.5 \times 10^{-19}\ \text{C}$ cannot be right, because that is not a multiple of $e$. For the one-mark definitions: an electric current is a flow of charge carriers; the coulomb is the charge that passes a point in one second when the current is one ampere (an ampere multiplied by a second); a charge carrier can be an electron, a proton, an ion or an $\alpha$-particle, but never a neutron.
If charge $Q$ passes a point in time $t$, the current is
Unit of current: ampere, $\text{A}$ ($= \text{C s}^{-1}$). For a changing current, the charge that has flowed in a time is the area under an $I$–$t$ graph.
Worked example. A current of $0.50\ \text{A}$ flows for $2.0$ minutes. Find the charge that passes, and how many electrons this represents. ($e = 1.60 \times 10^{-19}\ \text{C}$.)
The same two lines answer "how many electrons pass in $10$ hours at $4.0\ \text{mA}$" (convert the time to seconds: $N = It/e = 4.0 \times 10^{-3} \times 36\,000 / (1.60 \times 10^{-19}) = 9.0 \times 10^{20}$) and, the other way round, "the average current when $6.0 \times 10^{23}$ electrons pass in $24$ hours" ($I = Ne/t = 1.1\ \text{A}$). A beam of $\alpha$-particles carries a current too: each carries $2e$, so a beam current of $6.9 \times 10^{-9}\ \text{A}$ is $6.9 \times 10^{-9} / (2 \times 1.60 \times 10^{-19}) = 2.2 \times 10^{10}$ particles per second. A lightning strike that transfers $1 \times 10^{20}$ electrons in $30\ \mu\text{s}$ is a current of $Ne/t = 5.3 \times 10^{5}\ \text{A}$.
For a uniform conductor of cross-section area $A$, with $n$ charge carriers per unit volume (the number density 数密度), each carrying charge $q$, moving with average drift velocity 漂移速度 $v$:
Worked example. A copper wire of cross-sectional area $1.0 \times 10^{-6}\ \text{m}^{2}$ carries a current of $5.0\ \text{A}$. Copper has $n = 8.5 \times 10^{28}$ free electrons per $\text{m}^{3}$. Find the drift velocity. ($e = 1.60 \times 10^{-19}\ \text{C}$.)
Rearranging $I = Anvq$ gives $v = \dfrac{I}{Anq}$:
The electrons drift very slowly — less than a millimetre per second.
Two "show that" steps that often precede this calculation. The number density of a metal with one free electron per atom is $n = \dfrac{\text{density} \times N_{\text{A}}}{\text{molar mass}}$: for copper, $8.9 \times 10^{3} \times 6.02 \times 10^{23} / 0.0635 = 8.5 \times 10^{28}\ \text{m}^{-3}$. And the time for an electron to drift the length of a wire is $t = L / v$: at $3.7 \times 10^{-4}\ \text{m s}^{-1}$ a $2.0\ \text{m}$ wire takes $5\,400\ \text{s}$, an hour and a half, even though the current is established almost instantly. Between them, the equation $I = Anvq$ is used as a ratio far more often than as a calculation:
Use this to compare currents:
Current is the rate of flow of charge. Raise the voltage and current rises; raise the resistance and it falls — I = V / R. · Arus adalah laju aliran muatan. Naikkan tegangan dan arus naik; naikkan hambatan dan arus turun — I = V / R.
| English | Bahasa Indonesia |
|---|---|
| electric current/ɪˈlektrɪk ˈkʌrənt/ | arus listrik |
| charge carrier/tʃɑːdʒ ˈkærɪə/ | pembawa muatan |
| electron/ɪˈlektrɒn/ | elektron |
| electrolyte/ɪˈlektrəlaɪt/ | elektrolit |
| ion/ˈaɪɒn/ | ion |
| semiconductor/ˌsemɪkənˈdʌktə/ | semikonduktor |
| hole/həʊl/ | lubang |
| conventional current/kənˈvenʃənl ˈkʌrənt/ | arus konvensional |
| quantised/ˈkwɒntaɪzd/ | terkuantisasi |
| elementary charge/ˌelɪˈmentəri tʃɑːdʒ/ | muatan elementer |
| coulomb/ˈkuːlɒm/ | coulomb |
| drift velocity/drɪft vəˈlɒsɪti/ | kecepatan rambat |
| number density/ˈnʌmbə ˈdensɪti/ | kerapatan jumlah |
| series/ˈsɪəriːz/ | seri |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The potential difference 电势差 (p.d.) across a component is the energy 能量 transferred per unit charge as that charge passes through it:
Unit: volt 伏特, $\text{V}$ ($= \text{J C}^{-1}$).
If $1\ \text{J}$ of electrical energy changes into other forms (thermal, light, kinetic, …) when $1\ \text{C}$ of charge passes through a component, the p.d. across it is $1\ \text{V}$.
The electromotive force 电动势 (e.m.f.) of a source is the energy given per unit charge by the source. The formula is the same as for p.d.; the difference is direction: e.m.f. is energy given to the charge by the change; p.d. is energy given up by the charge to the component.
The definitions the examiner accepts: the potential difference across a component is the energy transferred (from electrical to other forms) per unit charge passing through it; the e.m.f. of a source is the energy transferred from other forms (chemical, in a cell) to electrical energy per unit charge in driving the charge round a complete circuit; one volt is one joule per coulomb. A battery "marked $9.0\ \text{V}$" therefore gives each coulomb $9.0\ \text{J}$ of electrical energy for the whole circuit, and the product of charge and p.d. is the energy transferred.
| English | Bahasa Indonesia |
|---|---|
| potential difference/pəˈtenʃl ˈdɪfrəns/ | beda potensial |
| energy/ˈenədʒi/ | tenaga |
| volt/vəʊlt/ | volt |
| electromotive force/ɪˌlektrəʊˈməʊtɪv fɔːs/ | gaya gerak listrik |
Combining $V = W/Q$ and $I = Q/t$:
Using Ohm's law $V = IR$:
Pick the form with the quantities you know. Examples:
Energy transferred in time $t$ is $E = P t$.
Worked example. Two lamps, P rated $250\ \text{V}$, $50\ \text{W}$ and Q rated $250\ \text{V}$, $200\ \text{W}$, are connected in series to a $250\ \text{V}$ supply. Which is brighter?
Their resistances at the rated p.d. are $R = V^{2}/P$: $1250\ \Omega$ for P and $313\ \Omega$ for Q. In series they carry the same current, so $P = I^{2}R$ makes the larger resistance, lamp P, dissipate more power: the "weaker" lamp is the brighter one. In parallel the p.d. is the same across both and $P = V^{2}/R$ favours the smaller resistance: Q. "State and explain which resistor dissipates more power" is answered by naming the quantity the two share (current in series, p.d. in parallel) and the form of the power equation that uses it.
Worked example. A supply delivers $2.4\ \text{kW}$ at $240\ \text{V}$ to a kettle through two cables of resistance $0.10\ \Omega$ each. Find the power lost in the cables.
The current is $I = P/V = 10\ \text{A}$, so the cables dissipate $I^{2}R = 10^{2} \times 0.20 = 20\ \text{W}$, and the kettle receives $2380\ \text{W}$. The efficiency of a circuit that exists to power one component is that component's power divided by the total power from the supply: with $6.0\ \text{V}$ across a $0.86\ \Omega$ series resistor and $4.5\ \text{V}$ across the component at the same current, the efficiency is $4.5 / (6.0 + 4.5) = 43\%$. A thermistor across a fixed p.d. dissipates $P = V^{2}/R$, so as its temperature rises and $R$ falls, the power rises. A kettle draws about $10\ \text{A}$ from a $250\ \text{V}$ supply; a mobile phone charger a fraction of an ampere.
P = VI
At a fixed voltage, power is proportional to the current it drives. · Pada tegangan tetap, daya berbanding lurus dengan arus yang dialirkannya.
| English | Bahasa Indonesia |
|---|---|
| power/ˈpaʊə/ | kuasa |
| voltage/ˈvəʊltɪdʒ/ | tegangan |
| parallel/ˈpærəlel/ | paralel |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The resistance 电阻 $R$ of a component is
Unit: ohm 欧姆, $\Omega$ ($= \text{V A}^{-1}$). Resistance depends on the conditions (such as temperature) when it is measured.
A conductor obeys Ohm's law 欧姆定律 when the current through it is proportional to the p.d. across it, as long as the conditions (especially temperature) stay constant. For such a conductor $R$ is constant and the $I$–$V$ graph is a straight line through the origin.
Ohm's law is an experimental result, not a definition. The definition $R = V/I$ works for any component; only ohmic ones have constant $R$.
For the marks: resistance is the ratio of the potential difference across a component to the current in it; the ohm is the resistance of a component in which a p.d. of one volt produces a current of one ampere (a volt per ampere); Ohm's law states that the current in a metallic conductor is directly proportional to the potential difference across it, provided that its temperature (and other physical conditions) remains constant. Of several $I$–$V$ graphs, only a straight line through the origin obeys Ohm's law; a straight line that misses the origin, or any curve, does not.
You should be able to sketch these:
Explain the shape of the filament lamp's line (three marks): as the current increases the filament's temperature rises; the lattice ions vibrate with larger amplitude, so the free electrons collide with them more often; each collision takes energy from the electrons, so the resistance increases; on the graph the ratio $V/I$ grows and the gradient falls. Run backwards for "the current decreases": the temperature falls, so the resistance falls. Two things to read off any characteristic. The resistance at a point is $V/I$ for that point, never the gradient of a curve, so a diode's resistance is very large (infinite, in practice) up to the switch-on p.d. and then falls steeply as $V$ rises further, and of four components at the same p.d. the one with the smallest current has the greatest resistance. And when a lamp (or a diode) is in series with a resistor, the two carry the same current and their p.d.s add: read the current from the component's curve at its own p.d., then use $V = IR$ for the resistor, and the supply p.d. is the sum.
For a uniform conductor of length $L$ and cross-section area $A$,
$\rho$ is the resistivity 电阻率, a property of the material, with unit $\Omega\ \text{m}$. Doubling the length doubles $R$; doubling the area halves it; halving the diameter quarters the area and so makes $R$ four times bigger.
Worked example. A copper wire of length $2.0\ \text{m}$ and cross-sectional area $1.7 \times 10^{-7}\ \text{m}^{2}$ has resistivity $1.7 \times 10^{-8}\ \Omega\ \text{m}$. Find its resistance.
Typical values: copper at room temperature $\rho \sim 1.7 \times 10^{-8}\ \Omega\ \text{m}$; an insulator 绝缘体 $\rho \sim 10^{15}\ \Omega\ \text{m}$ or more.
The resistivity of a metal rises with temperature (more lattice vibration 晶格振动 scatters 散射 the electrons), which is why the filament lamp's $I$–$V$ line curves.
Most resistivity questions are ratios. Stretching a wire keeps its volume ($A \times L$) constant, so if the length becomes $k$ times longer the area becomes $k$ times smaller and $R = \rho L / A$ becomes $k^{2}$ times larger: a wire three times as long (same mass, same metal) has nine times the resistance, and a wire whose diameter falls to $0.940$ of its value has its area multiplied by $0.884$, its length divided by $0.884$, and its resistance multiplied by $1/0.884^{2} = 1.28$.
Worked example. A copper lightning rod of resistance $9.6\ \Omega$ and length $20\ \text{m}$ has resistivity $1.7 \times 10^{-8}\ \Omega\ \text{m}$. Find its radius.
$A = \rho L / R = 1.7 \times 10^{-8} \times 20 / 9.6 = 3.5 \times 10^{-8}\ \text{m}^{2}$, and $r = \sqrt{A / \pi} = 1.1 \times 10^{-4}\ \text{m}$. Doubling the radius (same length) quarters the resistance. The other standard ratios: strands in parallel — seven identical strands share the current, so the cable's resistance is one seventh of one strand's; two wires of the same resistance where one metal has twice the resistivity need the second wire to have twice the area (a diameter $\sqrt{2}$ times larger) for the same length; a wire's resistance per unit length, $0.92\ \Omega\ \text{m}^{-1}$, multiplied by its area, $5.3 \times 10^{-7}\ \text{m}^{2}$, is its resistivity, $4.9 \times 10^{-7}\ \Omega\ \text{m}$; and a cylinder of conducting putty $60\ \text{mm}$ long and $20\ \text{mm}$ across, or a cube of side $a$ ($R = \rho a / a^{2} = \rho / a$), uses the same $R = \rho L / A$ with the shape's own length and end area. When the wire is held under tension, $R_{0} = \rho L / A$ still gives its resistance, and a stretch that lengthens it and thins it raises $R$ for both reasons. Because $\rho = RA/L = R\pi d^{2}/(4L)$, the percentage uncertainty in a measured resistivity is the sum of the percentage uncertainties in $R$ and $L$ plus twice that in $d$.
A longer wire has more resistance; a thicker one (bigger area) has less. Change the length, area and metal. · Kawat yang lebih panjang memiliki hambatan lebih besar; kawat yang lebih tebal (luas lebih besar) memiliki hambatan lebih kecil. Ubah panjang, luas, dan jenis logam.
V = R·I
Ohm's law: voltage is proportional to current — the gradient is the resistance R. · Hukum Ohm: tegangan berbanding lurus dengan arus — gradiennya adalah hambatan R.
V = aI
Drag the resistance. For an ohmic conductor voltage is proportional to current — a straight line whose gradient is the resistance. · Geser hambatan. Untuk penghantar ohmik, tegangan berbanding lurus dengan arus — garis lurus yang gradiennya adalah hambatan.
| English | Bahasa Indonesia |
|---|---|
| ohm/əʊm/ | ohm |
| Ohm's law/əʊmz lɔː/ | Hukum Ohm |
| resistance/rɪˈzɪstəns/ | penolakan |
| filament lamp/ˈfɪləmənt læmp/ | lampu filamen |
| semiconductor diode/ˌsemɪkənˈdʌktə ˈdaɪəʊd/ | dioda semikonduktor |
| diode/ˈdaɪəʊd/ | dioda |
| resistivity/ˌriːzɪˈstɪvəti/ | resistivitas |
| insulator/ˈɪnsjuːleɪtə/ | isolator |
| lattice vibration/ˈlætɪs vaɪˈbreɪʃn/ | getaran kisi |
| scatters/ˈskætəz/ | menyebarkan |
A light-dependent resistor 光敏电阻 (LDR) is a semiconductor whose resistance falls as the light intensity rises. In bright light $R$ may be a few hundred $\Omega$; in the dark it can be in the megaohms. LDRs are used in light-sensing circuits (street lamps, camera light meters). Here the light intensity 光强 controls the resistance.
| English | Bahasa Indonesia |
|---|---|
| light-dependent resistor/laɪt dɪˈpendənt rɪˈzɪstə/ | resistor peka cahaya |
| light intensity/laɪt ɪnˈtensɪti/ | intensitas cahaya |
In this syllabus a thermistor 热敏电阻 has a negative temperature coefficient 负温度系数: its resistance falls as its temperature rises. This is useful for sensing temperature — put it in a potential divider 分压器 and the output voltage changes with temperature.
This is the opposite of a metal: in a semiconductor, more thermal energy frees more charge carriers, and this matters more than the extra scattering.
A sketch of a thermistor's resistance against temperature starts at $R_{0}$ at $0\ °\text{C}$ and falls along a curve that flattens: the fall is not linear, which is the disadvantage of a thermistor as a thermometer (its scale is not uniform, so it needs calibrating). When the light on an LDR is increased, or a thermistor is warmed, its resistance falls, the current in its circuit rises, and the p.d. across it (in series with a fixed resistor) falls while the p.d. across the resistor rises; a fixed resistor and a metal wire at constant temperature keep their resistance, and a filament lamp's rises with current.
| English | Bahasa Indonesia |
|---|---|
| thermistor/ˈθɜːmɪstə/ | termistor |
| negative temperature coefficient/ˈneɡətɪv ˈtemprɪtʃə ˌkəʊɪˈfɪʃənt/ | koefisien suhu negatif |
| potential divider/pəˈtenʃl dɪˈvaɪdə/ | pembagi potensial |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| electric current | a flow of charge carriers |
| coulomb | the charge passing a point in one second when the current is one ampere |
| potential difference | the energy transferred from electrical to other forms per unit charge passing through a component |
| electromotive force (e.m.f.) | the energy transferred from other forms to electrical energy per unit charge, by a source driving charge round a complete circuit |
| volt | one joule per coulomb |
| resistance | the ratio of the potential difference across a component to the current in it |
| ohm | the resistance of a component in which a potential difference of one volt produces a current of one ampere |
| Ohm's law | the current in a metallic conductor is directly proportional to the potential difference across it, provided its temperature remains constant |
| resistivity | the constant $\rho$ in $R = \rho L / A$, a property of the material with unit $\Omega\ \text{m}$ |
| number density | the number of charge carriers per unit volume of the conductor |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The electromotive force 电动势 (e.m.f.) $\varepsilon$ of a source is the energy 能量 given to each unit of charge by the source as it drives the charge around a full circuit. Unit: volt.
The potential difference 电势差 (p.d.) across a component is the energy changed from electrical to other forms by each unit of charge as it passes through that component.
Both are in volts; they differ in energy direction:
The examiner's wording is fixed. e.m.f. is "the energy transferred per unit charge by the source in driving charge round a complete circuit"; p.d. is "the energy transferred per unit charge from electrical to other forms". Both are energy per charge, so the volt is a joule per coulomb ($1\ \text{V} = 1\ \text{J C}^{-1}$). Statements that are always true of a source: the e.m.f. is the terminal p.d. when no current flows; the total energy it gives to a charge $Q$ is $\varepsilon Q$. The name is a trap: an electromotive "force" is not a force and is not measured in newtons.
Worked example. A cell of e.m.f. $\varepsilon$ and internal resistance $r$ drives a charge $Q$ round a circuit whose terminal p.d. is $V$. Compare the energy transferred by the cell with the energy dissipated outside it.
The cell transfers $\varepsilon Q$ in total; the external circuit receives $VQ$. Since $V = \varepsilon - Ir < \varepsilon$ whenever a current flows, $VQ < \varepsilon Q$ — the difference $(\varepsilon - V)Q = IrQ$ is the energy turned to heat inside the cell.
In the lab you often build a circuit on a breadboard 面包板 (a board with rows of holes that connect components without soldering) and measure currents and p.d.s with a multimeter 万用表.
A real source has some internal resistance 内阻 $r$ — usually the resistance of the electrolyte 电解质 in a cell 电池, or the wire windings in a generator. When current $I$ flows, an internal p.d. of $Ir$ is "lost" inside the source, so the terminal p.d. 端电压 across the outside circuit is
So:
To measure $r$, change the outside resistance and plot $V_{\text{terminal}}$ against $I$: the line has $y$-intercept $\varepsilon$ and gradient $-r$.
Worked example. A cell of e.m.f. $1.5\ \text{V}$ and internal resistance $0.50\ \Omega$ is connected to a $2.5\ \Omega$ resistor. Find the current and the terminal p.d.
The e.m.f. drives the current through both resistances: $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.5 + 0.50} = 0.50\ \text{A}$. Then
The power 功率 given to the outside load is $P_{\text{ext}} = (\varepsilon - Ir) I$; the power lost inside is $P_{\text{int}} = I^{2} r$; the total power from the source is $\varepsilon I$.
The efficiency 效率 of the source is $\dfrac{P_{\text{ext}}}{\varepsilon I} = \dfrac{VI}{\varepsilon I} = \dfrac{V}{\varepsilon} = \dfrac{R}{R + r}$: a large load resistance wastes little energy inside the source.
Worked example. A cell of e.m.f. $2.0\ \text{V}$ and internal resistance $0.40\ \Omega$ drives a current of $1.5\ \text{A}$ through a wire. Show that the terminal p.d. is $1.4\ \text{V}$ and find the percentage efficiency with which the cell supplies power to the wire.
$V = \varepsilon - Ir = 2.0 - 1.5 \times 0.40 = 2.0 - 0.60 = 1.40\ \text{V}$. Power to the wire $= VI = 1.4 \times 1.5 = 2.1\ \text{W}$; total power from the cell $= \varepsilon I = 2.0 \times 1.5 = 3.0\ \text{W}$. Efficiency $= 2.1 / 3.0 = 0.70$, i.e. $70\%$. In a "show that" question, write every step and the unrounded value ($1.40\ \text{V}$) before the value you were given.
When the outside circuit changes, argue through the current. Closing a switch that adds a second resistor in parallel lowers the total external resistance, so the current in the cell rises, the internal "lost volts" $Ir$ rise, and the terminal p.d. $\varepsilon - Ir$ falls; the p.d. across the original resistor (which is the terminal p.d.) therefore falls, and so does its current, even though the cell's current rose. This chain — resistance → current in the cell → $Ir$ → terminal p.d. — is the model answer for almost every "state and explain the effect" question in this topic.
Cells joined together. In series, e.m.f.s add and internal resistances add: three cells of e.m.f. $\varepsilon$ and internal resistance $r$ give $3\varepsilon$ and $3r$. A cell connected the wrong way round subtracts its e.m.f. (ten $1.5\ \text{V}$ cells with one reversed give $8 \times 1.5 = 12\ \text{V}$). Identical cells in parallel give the same e.m.f. $\varepsilon$ with a smaller internal resistance $r/N$, so they can supply a larger current.
You must recognise and draw the standard symbols in the syllabus: cell, battery, switch, resistor, variable resistor, ammeter 电流表, voltmeter 电压表, lamp, diode (and LED 发光二极管), capacitor 电容器, inductor, thermistor, light-dependent resistor, fuse 保险丝, earth, junction. An ideal ammeter has zero resistance 电阻 and goes in series 串联. An ideal voltmeter has infinite resistance and goes in parallel 并联.
Drawing a circuit diagram. Marks are lost for symbols, not physics. Use ruler-straight lines and the standard symbols; an ammeter goes in series with the component whose current it measures and a voltmeter in parallel with (across) the component whose p.d. it measures; a cell with internal resistance is drawn as an ideal cell in series with a resistor $r$; and a variable resistor may be used in two ways, shown below. A "complete the circuit diagram" question usually wants exactly this measuring arrangement: source, ammeter and variable resistor in one series loop, voltmeter across the component under test.
V = ε − I·r
Terminal p.d. falls with current: it starts at the e.m.f. ε and drops by I·r. · Tegangan terminal turun seiring arus: dimulai dari g.g.b ε dan turun sebesar I·r.
| English | Bahasa Indonesia |
|---|---|
| electromotive force/ɪˌlektrəʊˈməʊtɪv fɔːs/ | gaya gerak listrik |
| energy/ˈenədʒi/ | tenaga |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | beda potensial |
| breadboard/ˈbredbɔːd/ | papan breadboard |
| multimeter/ˌmʌltɪˈmiːtə/ | multimeter |
| internal resistance/ɪnˈtɜːnl rɪˈzɪstəns/ | resistansi dalam |
| electrolyte/ɪˈlektrəlaɪt/ | elektrolit |
| cell/sel/ | sel |
| terminal p.d./ˈtɜːmɪnl ˌpiː ˈdiː/ | beda potensial terminal |
| open circuit/ˈəʊpən ˈsɜːkɪt/ | sirkuit terbuka |
| short circuit/ʃɔːt ˈsɜːkɪt/ | sirkuit pendek |
| power/ˈpaʊə/ | kuasa |
| efficiency/ɪˈfɪʃənsi/ | efisiensi |
| ammeter/ˈæmiːtə/ | amperemeter |
| voltmeter/ˈvəʊltmiːtə/ | voltmeter |
| LED/ˌel iː ˈdiː/ | LED |
| capacitor/kəˈpæsɪtə/ | kapasitor |
| fuse/fjuːz/ | sekat |
| resistance/rɪˈzɪstəns/ | penolakan |
| series/ˈsɪəriːz/ | seri |
| parallel/ˈpærəlel/ | paralel |
| loop/luːp/ | gelung |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
At any junction 节点, the total current flowing in equals the total current flowing out. This follows from conservation of charge 电荷守恒 — charge cannot build up at a point in a steady circuit, so charge in per second equals charge out per second.
For a junction with three wires: $I_{1} = I_{2} + I_{3}$ if currents 2 and 3 flow out and current 1 flows in.
Two ways the examiner asks it: "state the law" (one mark: the sum of the currents into a junction equals the sum of the currents out of it) and "state the conservation law behind it" (charge). Do not answer "energy" for the first law or "charge" for the second — the pairing is tested in almost every Paper 1.
Around any closed loop 回路, the total e.m.f. equals the total p.d. across the components in that loop. This follows from conservation of energy 能量守恒: as a unit of charge goes once round a loop, the energy it gains from sources equals the energy it gives up to components.
Pick a direction round the loop. Take an e.m.f. as positive when the loop direction goes from − to + of the source, and a p.d. as positive when the loop direction is the conventional current direction through the resistor.
In symbols, round any closed loop $\sum \varepsilon = \sum IR$. A source you pass from $+$ to $-$ counts as a negative e.m.f. (it is being charged, or opposes the other source), and a resistor you pass against its current counts as a negative p.d.
Worked example. Two batteries are in one loop with their e.m.f.s opposed: $12.0\ \text{V}$ with internal resistance $1.0\ \Omega$, and $8.0\ \text{V}$ with internal resistance $0.50\ \Omega$. Find the current.
Going round the loop, the net e.m.f. is $12.0 - 8.0 = 4.0\ \text{V}$ and the total resistance is $1.0 + 0.50 = 1.5\ \Omega$ (the internal resistances are in series), so $I = 4.0 / 1.5 = 2.7\ \text{A}$. The current flows in the direction driven by the larger e.m.f.
Resistors in series. Derivation using Kirchhoff's laws: there is no junction between the resistors, so by the first law the same current $I$ passes through each. By the second law the e.m.f. round the loop equals the sum of the p.d.s:
so $R_{\text{series}} = R_{1} + R_{2} + \ldots$.
Resistors in parallel. Derivation: by the second law, each resistor forms its own loop with the source, so each has the same p.d. $V$ across it. By the first law the current entering the junction equals the sum of the branch currents:
so $\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots$.
Two equal resistors $R$ in parallel give $R/2$; $N$ equal ones give $R/N$. A parallel combination is always smaller than any of its resistors; a series combination is always larger.
Worked example. A $4.0\ \Omega$ resistor and a $12\ \Omega$ resistor are connected in parallel. Find their combined resistance.
Reduce a mixed network one step at a time: replace each series chain by its sum and each parallel pair by $\dfrac{R_{1}R_{2}}{R_{1} + R_{2}}$ (the "product over sum" form, valid for two resistors only), redraw, and repeat until one resistor is left. Two rules decide which resistor dissipates the most power without any arithmetic: components carrying the same current dissipate more in the larger resistance ($P = I^{2}R$); components with the same p.d. dissipate more in the smaller resistance ($P = V^{2}/R$). In a series–parallel mix the single resistor that carries the whole current usually dissipates the most, because a parallel branch carries only a share of it.
Worked example. Three resistors, each of resistance $R$, are connected with two in series and that pair in parallel with the third. The total resistance between the ends is $8.0\ \Omega$. Find $R$.
The series pair is $2R$; in parallel with $R$: $R_{\text{T}} = \dfrac{2R \times R}{2R + R} = \dfrac{2R}{3}$. So $\dfrac{2R}{3} = 8.0$, giving $R = 12\ \Omega$. Check: the single resistor $R$ carries twice the current of the pair (same p.d., half the resistance), so it dissipates the most power.
For symmetric resistor networks, use the symmetry to spot branches with equal currents — the branch with the most current gives the most power ($P = I^{2}R$).
Worked example. In the circuit above $\varepsilon = 12\ \text{V}$ (negligible internal resistance), $R_{1} = 2.0\ \Omega$, $R_{2} = 6.0\ \Omega$ and $R_{3} = 3.0\ \Omega$. Find the three currents.
Kirchhoff's method. Junction: $I_{1} = I_{2} + I_{3}$. Loop 1 (through $\varepsilon$, $R_{1}$, $R_{2}$): $12 = 2.0 I_{1} + 6.0 I_{2}$. Loop 2 (through $R_{2}$ and $R_{3}$, no source): $0 = 6.0 I_{2} - 3.0 I_{3}$, so $I_{3} = 2 I_{2}$. Then $I_{1} = 3 I_{2}$ and $12 = 6.0 I_{2} + 6.0 I_{2}$, giving $I_{2} = 1.0\ \text{A}$, $I_{3} = 2.0\ \text{A}$ and $I_{1} = 3.0\ \text{A}$.
Reduction check. $6.0\ \Omega$ and $3.0\ \Omega$ in parallel give $2.0\ \Omega$; with $R_{1}$ the total is $4.0\ \Omega$, so $I_{1} = 12 / 4.0 = 3.0\ \text{A}$ and the p.d. across the parallel pair is $3.0 \times 2.0 = 6.0\ \text{V}$, giving $I_{2} = 6.0/6.0 = 1.0\ \text{A}$ and $I_{3} = 6.0/3.0 = 2.0\ \text{A}$. Both methods must agree; a negative answer for a current simply means you guessed its direction the wrong way.
Switch between series and parallel and add bulbs. In series they share the voltage and one break kills them all; in parallel each gets the full voltage and a break only loses its branch. · Beralih antara seri dan paralel serta tambahkan bohlam. Dalam seri mereka berbagi tegangan dan satu kerusakan mematikan semua; dalam paralel masing-masing mendapat tegangan penuh dan kerusakan hanya kehilangan cabangnya sendiri.
| English | Bahasa Indonesia |
|---|---|
| junction/ˈdʒʌŋkʃn/ | sambungan |
| conservation of charge/ˌkɒnsəˈveɪʃn ɒv tʃɑːdʒ/ | kekekalan muatan |
| closed loop/kləʊzd luːp/ | sirkuit tertutup |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | hukum kekekalan energi |
| Kirchhoff's first law/ˈkɜːkhɒfs fɜːst lɔː/ | Hukum Kirchhoff pertama |
| Kirchhoff's second law/ˈkɜːkhɒfs ˈsekənd lɔː/ | Hukum Kirchhoff kedua |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A potential divider 分压器 is two (or more) resistors in series across a source. The p.d. across each resistor is in direct proportion to its resistance:
The output (tapped between $R_{1}$ and $R_{2}$) can be set to any voltage 电压 between $0$ and $V_{\text{in}}$ by choosing the resistances. A potentiometer 电位差计 used with all three of its connections (a slider on a uniform-resistance track) gives a smoothly variable divider; the same component with only two connections is a rheostat 变阻器, which just changes the current.
Worked example. A $6.0\ \text{V}$ supply is connected across a $2.0\ \text{k}\Omega$ resistor in series with a $4.0\ \text{k}\Omega$ resistor. Find the output voltage tapped across the $4.0\ \text{k}\Omega$ resistor.
The same answer comes from the current-first route the mark scheme often lays out: $I = V_{\text{in}} / (R_{1} + R_{2}) = 6.0 / 6000 = 1.0\ \text{mA}$, then $V_{2} = I R_{2} = 1.0 \times 10^{-3} \times 4000 = 4.0\ \text{V}$. Two consequences worth remembering: the larger resistance takes the larger share of the p.d.; and the ratio formula holds only while no current is drawn from the output — a load (or a low-resistance voltmeter) in parallel with $R_{2}$ lowers the effective resistance of $R_{2}$ and so lowers $V_{2}$.
Replace one fixed resistor with a sensor 传感器 whose resistance changes with a physical quantity:
Connect the output to a transistor 晶体管 base or a comparator 比较器 to switch a load on or off when the temperature or light passes a threshold 阈值.
Which way does the output go? The output can be taken across the sensor or across the fixed resistor, and the two choices respond in opposite directions. Argue through the current, in this order, and you have the full three-mark answer:
So an output across the fixed resistor rises with light or temperature; an output across the sensor falls. If the source has internal resistance, add one more link: the larger current also increases $Ir$, so the terminal p.d. falls slightly.
Worked example. A battery of e.m.f. $9.0\ \text{V}$ and negligible internal resistance is connected in series with an LDR and a $1200\ \Omega$ resistor. In the light the LDR has a resistance of $1800\ \Omega$. Calculate the p.d. across the LDR, and state and explain what happens to it when the light intensity decreases.
$V_{\text{LDR}} = 9.0 \times \dfrac{1800}{1800 + 1200} = 5.4\ \text{V}$. Less light → the resistance of the LDR increases → the total resistance increases and the current decreases → the p.d. across the $1200\ \Omega$ resistor ($IR$) decreases → the p.d. across the LDR ($9.0 - IR$) increases.
Worked example. A thermistor in series with a $5800\ \Omega$ resistor across a $6.0\ \text{V}$ supply gives a p.d. of $2.9\ \text{V}$ across the resistor. Find the resistance of the thermistor.
Current $= 2.9 / 5800 = 5.0 \times 10^{-4}\ \text{A}$. The thermistor takes the remaining $6.0 - 2.9 = 3.1\ \text{V}$, so $R = 3.1 / (5.0 \times 10^{-4}) = 6200\ \Omega$ (or, by ratio, $R = 5800 \times 3.1/2.9$).
A potentiometer is a uniform resistance wire of length $L_{0}$ with a sliding contact (jockey 滑动触头). The resistance per unit length is uniform, so the p.d. from one end to the jockey is proportional to the length:
To compare two e.m.f.s (an unknown cell against a standard cell), connect each in turn with the jockey through a galvanometer 检流计. Slide the jockey until the galvanometer reads zero (a null — no current flows through the cell being measured, because the potentiometer's voltage there exactly opposes the cell's e.m.f.). The balance length 平衡长度 — the wire length from the end to the jockey at balance — is proportional to the e.m.f. being measured, so the two lengths are in the ratio of the e.m.f.s:
This is a null method 零点法: you find the balance (zero current) instead of measuring a current's value. Its advantage is that at balance the unknown cell gives no current, so its internal resistance does not affect the result.
Why the null method is better than a voltmeter. At balance no current is drawn from the cell being measured, so there is no $Ir$ drop inside it: the balance point measures the e.m.f., not the terminal p.d. A voltmeter always draws some current, so it reads slightly less than the e.m.f.
Reading the balance point. The p.d. per unit length of the wire is fixed by the driver cell 驱动电池 and the resistance of the wire. Anything that makes the p.d. being balanced larger moves the balance point further along the wire; anything that makes the p.d. per unit length larger (a driver cell of larger e.m.f., or a wire that takes a larger share of the driver's p.d.) makes the balance length shorter. Replacing the wire by one of the same length but greater diameter lowers its resistance ($R = \rho L / A$): if the driver cell has negligible internal resistance and nothing else is in series, the p.d. across the wire is still the full e.m.f. and the balance length does not move; if the driver has internal resistance or a series resistor, the wire's share of the e.m.f. falls, the p.d. per metre falls, and the balance length grows. Say which case you are in.
Worked example. A potentiometer wire XY of length $2.0\ \text{m}$ carries a p.d. of $1.50\ \text{V}$. A cell of e.m.f. $E$ is balanced when the jockey is $1.6\ \text{m}$ from X. Find $E$, and state what happens to the balance length if the driver cell is replaced by one of larger e.m.f.
$E = 1.50 \times \dfrac{1.6}{2.0} = 1.2\ \text{V}$. A larger driver e.m.f. gives a larger p.d. per metre, so the same $1.2\ \text{V}$ is reached at a shorter length: the jockey must move towards X.
In a series loop the same current flows everywhere and the cell's voltage splits across the components — that split is how a potential divider works. · Dalam loop seri arus yang sama mengalir di mana saja dan tegangan sel terbagi di atas komponen — pembagian inilah cara kerja pembagi potensial.
| English | Bahasa Indonesia |
|---|---|
| potential divider/pəˈtenʃl dɪˈvaɪdə/ | pembagi potensial |
| voltage/ˈvəʊltɪdʒ/ | tegangan |
| potentiometer/pəˌtentɪˈɒmɪtə/ | potensiometer |
| rheostat/ˈriːəstæt/ | reostat |
| sensor/ˈsensə/ | sensor |
| thermistor/ˈθɜːmɪstə/ | termistor |
| light-dependent resistor/laɪt dɪˈpendənt rɪˈzɪstə/ | resistor peka cahaya |
| light intensity/laɪt ɪnˈtensɪti/ | intensitas cahaya |
| transistor/trænˈzɪstə/ | transistor |
| comparator/kəmˈpæreɪtə/ | komparator |
| threshold/ˈθreʃəʊld/ | ambang batas |
| jockey/ˈdʒɒki/ | jockey |
| galvanometer/ˌɡælvəˈnɒmɪtə/ | galvanometer |
| balance length/ˈbæləns leŋθ/ | panjang keseimbangan |
| null method/nʌl ˈmeθəd/ | metode nol |
| driver cell/ˈdraɪvə sel/ | sel penggerak |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| electromotive force (e.m.f.) | the energy transferred per unit charge by a source in driving charge round a complete circuit |
| potential difference (p.d.) | the energy transferred per unit charge from electrical energy to other forms of energy |
| internal resistance | the resistance to current inside a source of e.m.f., which causes a p.d. $Ir$ across the source when a current flows |
| terminal p.d. | the p.d. across the terminals of a source, equal to e.m.f. $- Ir$ |
| Kirchhoff's first law | the sum of the currents into a junction is equal to the sum of the currents out of the junction (conservation of charge) |
| Kirchhoff's second law | the sum of the e.m.f.s round a closed loop is equal to the sum of the p.d.s round the loop (conservation of energy) |
| potential divider | two or more resistors in series across a supply, giving a p.d. across one of them that is a fraction of the supply p.d. |
| potentiometer | a uniform resistance wire with a sliding contact, used to compare p.d.s by finding the length at which a galvanometer reads zero |
| null method | a measurement made by adjusting a circuit until a meter reads zero, so no current is drawn from the component being measured |
| balance length | the length of potentiometer wire between one end and the sliding contact when the galvanometer reads zero |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Alpha particles α粒子 fired at a thin gold foil were seen to:
From this Rutherford worked out:
Order of magnitude: atom diameter $\sim 10^{-10}\ \text{m}$, nucleus diameter $\sim 10^{-15}\ \text{m}$ — the nucleus is about $10^{5}$ times smaller than the atom.
The three results and what each proves — the mark scheme pairs them exactly like this:
| Observation | Conclusion |
|---|---|
| the vast majority pass straight through with little or no deflection | most of the atom is empty space |
| a small number are deflected through small angles | there is a concentrated positive charge (the nucleus) that repels the positive α-particles |
| a very small minority (about 1 in 8000) are deflected through more than 90°, some straight back | the nucleus is very small compared with the atom, and contains almost all of the atom's mass |
Three details of the experiment are also asked: the foil is thin so that each α-particle meets at most one nucleus (and is not absorbed); the chamber is a vacuum so the α-particles are not stopped or scattered by air; gold is used because it can be beaten into a very thin sheet and has a heavy, highly charged nucleus.
Worked example. Explain why a very small minority of the α-particles are scattered through angles greater than 90°.
Only an α-particle that approaches a nucleus almost head-on is turned back. The nucleus and the α-particle are both positive, so the electrostatic 静电 repulsion is very large at small separation, and the nucleus is much more massive than the α-particle, so it is the α-particle that is turned round. Because the nucleus is tiny, very few α-particles get that close — hence the small minority. As the α-particle approaches, its kinetic energy is converted to electric potential energy; at the point of closest approach it is momentarily at rest and all of its kinetic energy has become potential energy.
The deflection depends on how close the path passes to the nucleus: the smaller the distance, the larger the angle, and the paths are symmetrical about the line through the nucleus.
An atom has:
The proton and neutron have almost the same mass ($\approx 1\ \text{u}$); the electron is about $\tfrac{1}{1836}$ of the proton's mass.
Worked example. Describe the structure of an atom of uranium-238, $^{238}_{92}\text{U}$.
A nucleus containing 92 protons and $238 - 92 = 146$ neutrons, with 92 electrons in orbit around it. "Orbital electrons" is the syllabus phrase — the electrons are outside the nucleus, and the neutral atom has as many electrons as protons.
For a nuclide 核素 written $^{A}_{Z}\text{X}$:
A neutral atom has the same number of electrons as protons.
Isotopes 同位素 are atoms of the same element (same $Z$) with different numbers of neutrons (different $A$). They behave the same chemically but differently in the nucleus. Example: $^{12}_{6}\text{C}$ and $^{14}_{6}\text{C}$ are isotopes of carbon.
The examiner's wording: isotopes are nuclei (or atoms) with the same number of protons but different numbers of neutrons — equivalently the same proton number $Z$ and different nucleon numbers $A$. Hydrogen has three:
Worked example. Tritium, $^{3}_{1}\text{H}$, is an isotope of hydrogen. State the numbers of protons, neutrons and electrons in a neutral tritium atom, and give the quark composition of its nucleus.
$Z = 1$ proton; $A - Z = 3 - 1 = 2$ neutrons; 1 electron (neutral, so electrons $=$ protons). The nucleus is one proton (uud) and two neutrons (udd, udd): 4 up quarks and 5 down quarks. Charge check: $4 \times \tfrac{2}{3} - 5 \times \tfrac{1}{3} = +1$, the proton number. A "labelled diagram" of the atom is the nucleus (1 p, 2 n) with one electron orbiting outside it.
In any nuclear process:
These two rules let you balance decay and reaction equations.
The unified atomic mass unit 统一原子质量单位, symbol $\text{u}$, is set so that an atom of $^{12}_{6}\text{C}$ has mass exactly $12\ \text{u}$. Numerically,
A proton has mass $\approx 1.007\ \text{u}$; a neutron $\approx 1.009\ \text{u}$; an electron $\approx 5.5 \times 10^{-4}\ \text{u}$.
Worked example. A nucleus X has 14 nucleons and $p$ protons. Its charge-to-mass ratio is $4.1 \times 10^{7}\ \text{C kg}^{-1}$. Find $p$ and identify X.
Charge $= pe$; mass $\approx 14\ \text{u}$. So $\dfrac{pe}{14\text{u}} = 4.1 \times 10^{7}$, giving $p = \dfrac{4.1 \times 10^{7} \times 14 \times 1.66 \times 10^{-27}}{1.60 \times 10^{-19}} = 5.96 \approx 6$. Six protons and 14 nucleons: X is carbon-14, $^{14}_{6}\text{C}$. Keep every figure until the end, then round to the nearest whole number of protons.
Partikel alfa α ditembakkan ke arah selembar emas tipis dan terlihat:
Dari hal ini Rutherford menyimpulkan:
Orde besaran: diameter atom $\sim 10^{-10}\ \text{m}$, diameter inti $\sim 10^{-15}\ \text{m}$ — inti sekitar $10^{5}$ kali lebih kecil daripada atom.
Tiga hasil dan apa yang dibuktikan masing-masing — kunci jawaban memasangkannya persis seperti ini:
| Pengamatan | Kesimpulan |
|---|---|
| mayoritas besar menembus lurus dengan sedikit atau tanpa pembelokan | sebagian besar atom adalah ruang kosong |
| sejumlah kecil mengalami pembelokan melalui sudut kecil | terdapat muatan positif terkonsentrasi (inti) yang menolak partikel α positif |
| sebagian sangat kecil (sekitar 1 dari 8000) membelok lebih dari 90°, beberapa memantul lurus kembali | inti sangat kecil dibandingkan atom, dan mengandung hampir seluruh massa atom |
Tiga detail percobaan juga ditanyakan: daun emas tipis sehingga setiap partikel α bertemu paling banyak satu inti (dan tidak diserap); ruang vakum agar partikel α tidak dihentikan atau disebar oleh udara; emas digunakan karena dapat dipalu menjadi lembaran sangat tipis dan memiliki inti berat serta bermuatan tinggi.
Contoh terpecahkan. Jelaskan mengapa sebagian sangat kecil partikel α disebar melalui sudut lebih besar dari 90°.
Hanya partikel α yang mendekati inti hampir head-on yang dipantulkan kembali. Inti dan partikel α keduanya positif, sehingga tolakan elektrostatik sangat besar pada jarak pemisahan yang kecil, dan inti jauh lebih masif daripada partikel α, sehingga partikel αlah yang berbelok balik. Karena inti sangat kecil, sangat sedikit partikel α yang mendekat sedekat itu — karenanya sebagian kecil. Saat partikel α mendekat, energi kinetiknya diubah menjadi energi potensial listrik; pada titik pendekatan terdekat ia berhenti sesaat dan seluruh energi kinetiknya telah menjadi energi potensial.
Pembelokan bergantung pada seberapa dekat lintasan melewati inti: semakin kecil jaraknya, semakin besar sudutnya, dan lintasan simetris terhadap garis yang melalui inti.
Atom memiliki:
Proton dan neutron memiliki massa hampir sama ($\approx 1\ \text{u}$); elektron sekitar $\tfrac{1}{1836}$ dari massa proton.
Contoh terpecahkan. Deskripsikan struktur atom uranium-238, $^{238}_{92}\text{U}$.
Inti berisi 92 proton dan $238 - 92 = 146$ neutron, dengan 92 elektron mengorbitnya. "Elektron orbital" adalah frasa kurikulum — elektron berada di luar inti, dan atom netral memiliki jumlah elektron sebanyak proton.
Untuk nuklida yang ditulis $^{A}_{Z}\text{X}$:
Atom netral memiliki jumlah elektron yang sama dengan proton.
Isotop adalah atom dari unsur yang sama (sama $Z$) dengan jumlah neutron berbeda (berbeda $A$). Mereka berperilaku sama secara kimia tetapi berbeda di dalam inti. Contoh: $^{12}_{6}\text{C}$ dan $^{14}_{6}\text{C}$ adalah isotop karbon.
Kalimat penguji: isotop adalah inti (atau atom) dengan jumlah proton yang sama tetapi jumlah neutron berbeda — ekuivalen dengan nomor proton yang sama $Z$ dan nomor nukleon berbeda $A$. Hidrogen memiliki tiga:
Contoh terpecahkan. Tritium, $^{3}_{1}\text{H}$, adalah isotop hidrogen. Sebutkan jumlah proton, neutron, dan elektron dalam atom tritium netral, dan berikan komposisi kuark dari intinya.
$Z = 1$ proton; $A - Z = 3 - 1 = 2$ neutron; 1 elektron (netral, sehingga elektron $=$ proton). Inti atom terdiri dari satu proton (uud) dan dua neutron (udd, udd): 4 kuark naik dan 5 kuark turun. Pengecekan muatan: $4 \times \tfrac{2}{3} - 5 \times \tfrac{1}{3} = +1$, nomor proton. "Diagram berlabel" atom adalah inti (1 p, 2 n) dengan satu elektron yang mengorbit di luarnya.
Dalam setiap proses nuklir:
Dua aturan ini memungkinkan Anda menyetarakan persamaan peluruhan dan reaksi.
Satuan massa atom terpadu, simbol $\text{u}$, ditetapkan sedemikian rupa sehingga sebuah atom $^{12}_{6}\text{C}$ memiliki massa tepat $12\ \text{u}$. Secara numeris,
Sebuah proton memiliki massa $\approx 1.007\ \text{u}$; sebuah neutron $\approx 1.009\ \text{u}$; sebuah elektron $\approx 5.5 \times 10^{-4}\ \text{u}$.
Contoh soal terpecahkan. Sebuah inti X memiliki 14 nukleon dan $p$ proton. Rasio muatan-terhadap-massanya adalah $4.1 \times 10^{7}\ \text{C kg}^{-1}$. Temukan $p$ dan identifikasi X.
Muatan $= pe$; massa $\approx 14\ \text{u}$. Maka $\dfrac{pe}{14\text{u}} = 4.1 \times 10^{7}$, menghasilkan $p = \dfrac{4.1 \times 10^{7} \times 14 \times 1.66 \times 10^{-27}}{1.60 \times 10^{-19}} = 5.96 \approx 6$. Enam proton dan 14 nukleon: X adalah karbon-14, $^{14}_{6}\text{C}$. Pertahankan setiap angka hingga akhir, lalu bulatkan ke bilangan bulat terdekat untuk jumlah proton.
Connect observations to the nuclear model of the atom. · Hubungkan observasi dengan model inti atom.
A = A₀·bᵗ
Activity decays exponentially — set the base b below 1. · Aktivitas menurun secara eksponensial — atur basis b di bawah 1.
| English | Bahasa Indonesia |
|---|---|
| alpha particle/ˈælfə ˈpɑːtɪkl/ | partikel alfa |
| deflection/dɪˈflekʃn/ | deviasi |
| nucleus/ˈnjuːklɪəs/ | inti sel |
| electrostatic/ɪˌlektrəʊˈstætɪk/ | elektrostatik |
| energy/ˈenədʒi/ | tenaga |
| proton/ˈprəʊtɒn/ | proton |
| neutron/ˈnjuːtrɒn/ | neutron |
| electron/ɪˈlektrɒn/ | elektron |
| nuclide/ˈnjuːklaɪd/ | nuklida |
| nucleon number/ˈnjuːklɪən ˈnʌmbə/ | nomor nukleon |
| nucleon/ˈnjuːklɪən/ | nukleus |
| proton number/ˈprəʊtɒn ˈnʌmbə/ | nombor proton |
| isotope/ˈaɪsətəʊp/ | isotop |
An unstable nucleus rearranges itself and gives out one of three kinds of radiation 辐射. This is radioactive 放射性 decay. Each kind has its own properties.
A cloud chamber 云室 makes the tracks visible: each α-particle leaves a short, straight, thick trail of tiny droplets as it ionises the air. The short equal lengths show the α-particles all carry about the same energy.
Two types of beta particle β粒子:
Properties (both types):
A nucleus often gives out a γ-photon as a "tidy-up" step after an α or β decay leaves the daughter nucleus 子核 in an excited state 激发态.
Mass and charge, side by side — a table question asks for exactly these, in units of $\text{u}$ and $e$:
| Particle | Mass | Charge |
|---|---|---|
| proton | $1\ \text{u}$ (1.007 u) | $+e$ |
| neutron | $1\ \text{u}$ (1.009 u) | $0$ |
| electron, $\beta^{-}$ | $\tfrac{1}{1836}\ \text{u}$ ($5.5 \times 10^{-4}\ \text{u}$) | $-e$ |
| positron, $\beta^{+}$ | $\tfrac{1}{1836}\ \text{u}$ | $+e$ |
| $\alpha$-particle | $4\ \text{u}$ | $+2e$ |
| $\gamma$-photon | $0$ | $0$ |
| neutrino, antineutrino | almost zero | $0$ |
Worked example. Compare an α-particle with a $\beta^{+}$ particle in terms of their masses and charges (3 marks).
Both are positively charged, but the α-particle's charge ($+2e$) is twice that of the $\beta^{+}$ ($+e$). The α-particle's mass ($4\ \text{u}$) is about $7300$ times the mass of the $\beta^{+}$ ($5.5 \times 10^{-4}\ \text{u}$). Give a ratio, not just "heavier": twice the charge and about 7000 times the mass are the marking points.
Because they carry charge, α- and β-particles are deflected by electric and magnetic fields — in opposite directions for opposite signs, the light β far more than the heavy α — while γ-rays, being uncharged, pass straight through.
Every particle has an antiparticle 反粒子 with the same mass but opposite charge. The positron is the antiparticle of the electron.
In β-decay, a third particle is always given out as well:
Neutrinos and antineutrinos have zero charge, very small mass, and barely interact — they are very hard to detect, but they must be there to balance energy, momentum 动量 and other conserved quantities in β-decay.
In α-decay the energy 能量 released is shared between just two particles (the daughter nucleus and the α). Conservation of momentum and energy then fixes the α's energy to one value (discrete).
In β-decay the energy is shared between three particles (the daughter nucleus, the β, and the (anti)neutrino). The β can take any share from zero up to a maximum, so its energy spectrum is continuous.
Worked example. The energy spectrum of the $\beta^{-}$ particles from a source is continuous, from zero up to a maximum. Explain why (3 marks).
Each decay releases a fixed amount of energy. That energy is shared between the $\beta^{-}$ particle, the antineutrino and the recoiling daughter nucleus. The antineutrino can take any share, so the $\beta^{-}$ particle is left with any energy from zero up to the maximum — the maximum being when the antineutrino carries away almost none. (For an α-particle there are only two bodies, so momentum conservation fixes the share and the energy is discrete.)
A general α-decay:
(check: $A = (A-4) + 4$, $Z = (Z-2) + 2$.)
A general $\beta^{-}$ decay:
A general $\beta^{+}$ decay:
Worked example. Uranium-238, $^{238}_{92}\text{U}$, decays by α-emission; carbon-14, $^{14}_{6}\text{C}$, decays by $\beta^{-}$-emission. Find each daughter nuclide.
α-decay lowers $A$ by 4 and $Z$ by 2; $\beta^{-}$-decay leaves $A$ unchanged and raises $Z$ by 1:
Determine quantitatively the changes. α-emission: $A$ decreases by 4, $Z$ decreases by 2. $\beta^{-}$-emission: $A$ unchanged, $Z$ increases by 1. $\beta^{+}$-emission: $A$ unchanged, $Z$ decreases by 1. γ-emission: no change in either. A decay chain that ends at a known nuclide is solved from these: the number of α-decays is $(A_{\text{start}} - A_{\text{end}})/4$, and then the number of $\beta^{-}$-decays makes $Z$ come out right.
Worked example. Thorium-230, $^{230}_{90}\text{Th}$, decays in stages, by α- and $\beta^{-}$-emission, to lead-206, $^{206}_{82}\text{Pb}$. Find the number of each kind of decay.
α-decays: $(230 - 206)/4 = 6$. Six α-decays alone would lower $Z$ by 12, to 78; the final $Z$ is 82, so there are $82 - 78 = 4$ $\beta^{-}$-decays.
A nucleus at rest recoils. When a stationary nucleus emits an α-particle, momentum is conserved: the daughter nucleus and the α-particle move off in opposite directions with momenta of equal magnitude. Since $p = mv$, the lighter α-particle moves much faster, and since $E_{\text{k}} = p^{2}/2m$ it also takes most of the kinetic energy — the shares are in inverse proportion to the masses.
Worked example. A stationary nucleus P of mass $243\ \text{u}$ emits an α-particle of mass $4\ \text{u}$ at $1.5 \times 10^{7}\ \text{m s}^{-1}$. Find the speed of the daughter nucleus Q and the ratio of the kinetic energy of the α-particle to that of Q.
Q has mass $243 - 4 = 239\ \text{u}$. Momentum: $239 v_{\text{Q}} = 4 \times 1.5 \times 10^{7}$, so $v_{\text{Q}} = 2.5 \times 10^{5}\ \text{m s}^{-1}$, in the opposite direction to the α-particle. Kinetic energies: $\dfrac{E_{\alpha}}{E_{\text{Q}}} = \dfrac{p^{2}/2m_{\alpha}}{p^{2}/2m_{\text{Q}}} = \dfrac{m_{\text{Q}}}{m_{\alpha}} = \dfrac{239}{4} \approx 60$. The α-particle takes about 98% of the energy released.
Inti atom yang tidak stabil mengatur ulang dirinya sendiri dan memancarkan salah satu dari tiga jenis radiasi. Ini adalah peluruhan radioaktif. Setiap jenis memiliki sifatnya masing-masing.
Awan ruang membuat jejak terlihat: setiap partikel α meninggalkan jejak pendek, lurus, dan tebal berupa tetesan-tetesan kecil saat mengionisasi udara. Panjang kesamaan yang pendek menunjukkan bahwa semua partikel α membawa energi yang hampir sama.
Dua jenis partikel beta β:
Sifat (kedua jenis):
Inti sering memancarkan foton γ sebagai langkah "perapi" setelah peluruhan α atau β meninggalkan inti anak dalam keadaan tereksitasi.
Massa dan muatan, berdampingan — pertanyaan tabel meminta keduanya secara eksplisit, dalam satuan $\text{u}$ dan $e$:
| Partikel | Massa | Muatan |
|---|---|---|
| proton | $1\ \text{u}$ (1.007 u) | $+e$ |
| neutron | $1\ \text{u}$ (1.009 u) | $0$ |
| elektron, $\beta^{-}$ | $\tfrac{1}{1836}\ \text{u}$ ($5.5 \times 10^{-4}\ \text{u}$) | $-e$ |
| positron, $\beta^{+}$ | $\tfrac{1}{1836}\ \text{u}$ | $+e$ |
| $\alpha$-partikel | $4\ \text{u}$ | $+2e$ |
| $\gamma$-foton | $0$ | $0$ |
| neutrino, antineutrino | hampir nol | $0$ |
Contoh soal terpecahkan. Bandingkan partikel α dengan partikel $\beta^{+}$ berdasarkan massa dan muatan mereka (3 poin).
Keduanya bermuatan positif, tetapi muatan partikel α ($+2e$) dua kali lipat dari muatan $\beta^{+}$ ($+e$). Massa partikel α ($4\ \text{u}$) sekitar $7300$ kali massa partikel $\beta^{+}$ ($5.5 \times 10^{-4}\ \text{u}$). Berikan rasio, bukan hanya "lebih berat": muatan dua kali lipat dan massa sekitar 7000 kali lipat adalah poin penilaian.
Karena membawa muatan, partikel α dan β dibelokkan oleh medan listrik dan magnetik — dalam arah berlawanan untuk tanda berlawanan, partikel β yang ringan jauh lebih banyak dibelokkan daripada partikel α yang berat — sementara sinar γ, karena tidak bermuatan, menembus lurus.
Setiap partikel memiliki antipartikel dengan massa yang sama tetapi muatan berlawanan. Positron adalah antipartikel dari elektron.
Dalam peluruhan β, partikel ketiga selalu dipancarkan juga:
Neutrino dan antineutrino memiliki muatan nol, massa sangat kecil, dan hampir tidak berinteraksi — mereka sangat sulit dideteksi, tetapi mereka harus ada untuk menyeimbangkan energi, momentum dan besaran kekal lainnya dalam peluruhan β.
Dalam peluruhan α, energi yang dilepaskan dibagi hanya antara dua partikel (inti anak dan α). Hukum kekekalan momentum dan energi kemudian menetapkan energi α pada satu nilai tertentu (diskrit).
Dalam peluruhan β, energi dibagi antara tiga partikel (inti anak, β, dan (anti)neutrino). Partikel β dapat mengambil sebagian dari nol hingga maksimum, sehingga spektrum energinya bersifat kontinu.
Contoh terpecahkan. Spektrum energi dari partikel $\beta^{-}$ dari suatu sumber bersifat kontinu, mulai dari nol hingga maksimum. Jelaskan mengapa (3 poin).
Setiap peluruhan melepaskan sejumlah energi tetap. Energi tersebut dibagi antara partikel $\beta^{-}$, antineutrino, dan inti anak yang memantul balik. Antineutrino dapat mengambil sebagian berapa pun, sehingga partikel $\beta^{-}$ tersisa dengan energi berkisar dari nol hingga maksimum — di mana maksimum terjadi ketika antineutrino membawa hampir tidak ada energi. (Untuk partikel α hanya ada dua badan, sehingga kekekalan momentum menetapkan bagiannya dan energinya bersifat diskrit.)
Peluruhan α umum:
(cek: $A = (A-4) + 4$, $Z = (Z-2) + 2$.)
Peluruhan $\beta^{-}$ umum:
Peluruhan $\beta^{+}$ umum:
Contoh terpecahkan. Uranium-238, $^{238}_{92}\text{U}$, meluruh melalui pemancaran α; karbon-14, $^{14}_{6}\text{C}$, meluruh melalui pemancaran $\beta^{-}$. Tentukan setiap nuclid anak.
Peluruhan α menurunkan $A$ sebesar 4 dan $Z$ sebesar 2; peluruhan $\beta^{-}$ meninggalkan $A$ tak berubah dan menaikkan $Z$ sebesar 1:
Tentukan perubahan secara kuantitatif. Pemancaran α: $A$ berkurang 4, $Z$ berkurang 2. Pemancaran $\beta^{-}$: $A$ tak berubah, $Z$ bertambah 1. Pemancaran $\beta^{+}$: $A$ tak berubah, $Z$ berkurang 1. Pemancaran γ: tidak ada perubahan pada keduanya. Rantai peluruhan yang berakhir pada nuclid yang diketahui diselesaikan dari hal ini: jumlah peluruhan α adalah $(A_{\text{start}} - A_{\text{end}})/4$, dan kemudian jumlah peluruhan $\beta^{-}$ membuat $Z$ menjadi tepat.
Contoh terpecahkan. Thorium-230, $^{230}_{90}\text{Th}$, meluruh bertahap melalui pemancaran α dan $\beta^{-}$, menuju timbal-206, $^{206}_{82}\text{Pb}$. Tentukan jumlah masing-masing jenis peluruhan.
Peluruhan α: $(230 - 206)/4 = 6$. Enam peluruhan α saja akan menurunkan $Z$ sebesar 12, menjadi 78; hasil akhir $Z$ adalah 82, jadi terdapat $82 - 78 = 4$ peluruhan $\beta^{-}$.
Nukleus yang diam mengalami pemantulan balik. Ketika nukleus yang diam memancarkan partikel α, momentum kekal: inti anak dan partikel α bergerak menjauh dalam arah berlawanan dengan besar momentum yang sama. Karena $p = mv$, partikel α yang lebih ringan bergerak jauh lebih cepat, dan karena $E_{\text{k}} = p^{2}/2m$ ia juga mengambil sebagian besar energi kinetik — bagian-bagiannya berbanding terbalik dengan massa.
Contoh terpecahkan. Nukleus diam P bermassa $243\ \text{u}$ memancarkan partikel α bermassa $4\ \text{u}$ dengan kecepatan $1.5 \times 10^{7}\ \text{m s}^{-1}$. Tentukan kecepatan inti anak Q dan rasio energi kinetik partikel α terhadap energi kinetik Q.
Q memiliki massa $243 - 4 = 239\ \text{u}$. Momentum: $239 v_{\text{Q}} = 4 \times 1.5 \times 10^{7}$, sehingga $v_{\text{Q}} = 2.5 \times 10^{5}\ \text{m s}^{-1}$, ke arah berlawanan dengan partikel α. Energi kinetik: $\dfrac{E_{\alpha}}{E_{\text{Q}}} = \dfrac{p^{2}/2m_{\alpha}}{p^{2}/2m_{\text{Q}}} = \dfrac{m_{\text{Q}}}{m_{\alpha}} = \dfrac{239}{4} \approx 60$. Partikel α mengambil sekitar 98% dari energi yang dilepaskan.
| English | Bahasa Indonesia |
|---|---|
| cloud chamber/klaʊd ˈtʃeɪmbə/ | ruang awan |
| beta particle/ˈbiːtə ˈpɑːtɪkl/ | partikel beta |
| positron/ˈpɒzɪtrɒn/ | positron |
| antiparticle/ˌæntɪˈpɑːtɪkl/ | antipartikel |
| continuous/kənˈtɪnjuːəs/ | kontinu |
| photon/ˈfəʊtɒn/ | foton |
| electromagnetic spectrum/ɪˌlektrəʊməɡˈnetɪk ˈspektrəm/ | spektrum elektromagnetik |
| attenuated/əˈtenjuːeɪtɪd/ | melemah |
| gamma ray/ˈɡæmə reɪ/ | sinar gamma |
| daughter nucleus/ˈdɔːtə ˈnjuːklɪəs/ | inti anak |
| excited state/ekˈsaɪtɪd steɪt/ | keadaan tereksitasi |
| neutrino/njuːˈtriːnəʊ/ | neutrino |
| antineutrino/ˌæntɪnjuːˈtriːnəʊ/ | antineutrino |
| momentum/məʊˈmentəm/ | momentum |
| fundamental particles/ˌfʌndəˈmentl ˈpɑːtɪklz/ | partikel fundamental |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Some particles are fundamental particles 基本粒子 (point-like, with no smaller parts as far as we know); others are built from fundamental ones.
The one-mark definition: a fundamental particle is one that cannot be broken down into smaller particles (it has no internal structure). Electrons, neutrinos and quarks are fundamental; protons and neutrons are not.
A quark 夸克 is a fundamental particle. There are six flavours 味:
Each quark has an antiquark 反夸克 with the same size of charge but the opposite sign: $\bar{u}$ (charge $-\tfrac{2}{3}e$), $\bar{d}$ (charge $+\tfrac{1}{3}e$). No other quark property is tested.
Written as the table a question asks you to complete:
| Quark | Charge | Antiquark | Charge |
|---|---|---|---|
| up (u), charm (c), top (t) | $+\tfrac{2}{3}e$ | $\bar{\text{u}}$, $\bar{\text{c}}$, $\bar{\text{t}}$ | $-\tfrac{2}{3}e$ |
| down (d), strange (s), bottom (b) | $-\tfrac{1}{3}e$ | $\bar{\text{d}}$, $\bar{\text{s}}$, $\bar{\text{b}}$ | $+\tfrac{1}{3}e$ |
Worked example. By reference to quark composition, show that the charge of a proton is $+1.6 \times 10^{-19}\ \text{C}$.
A proton is uud: charge $= \tfrac{2}{3}e + \tfrac{2}{3}e - \tfrac{1}{3}e = +e = +1.6 \times 10^{-19}\ \text{C}$. Write the three fractions and their sum — the mark is for the arithmetic, not the answer.
Particles built from quarks are hadrons 强子. Two types:
Protons and neutrons are not fundamental — they are baryons made of quarks.
"Compare baryons and mesons in terms of their constituent particles" (2 marks): both are hadrons made of quarks; a baryon is three quarks (or three antiquarks, for an antibaryon), a meson is one quark and one antiquark. Any charge you are given must come out of the quark charges:
Worked example. (a) A meson has charge $-1e$. Give a possible quark composition. (b) A meson Q has charge 0. Give a possible composition. (c) A baryon is made of three quarks of different flavours, u, d and s. Find its charge. (d) Give the quark composition of an antineutron and its charge.
(a) d$\bar{\text{u}}$: $-\tfrac{1}{3} - \tfrac{2}{3} = -1$ (s$\bar{\text{u}}$ also works). (b) u$\bar{\text{u}}$: $+\tfrac{2}{3} - \tfrac{2}{3} = 0$ (or d$\bar{\text{d}}$). (c) uds: $\tfrac{2}{3} - \tfrac{1}{3} - \tfrac{1}{3} = 0$. (d) The antineutron is the antiparticle of udd, so it is $\bar{\text{u}}\bar{\text{d}}\bar{\text{d}}$: $-\tfrac{2}{3} + \tfrac{1}{3} + \tfrac{1}{3} = 0$ — the same mass and charge as the neutron, since both have charge zero.
In $\beta^{-}$ decay a neutron turns into a proton; in quark terms, one down quark turns into an up quark:
In $\beta^{+}$ decay a proton turns into a neutron; one up quark turns into a down quark:
"Describe $\beta^{+}$ decay in terms of the fundamental particles involved" (2 marks): an up quark in a proton changes into a down quark, so the proton becomes a neutron, and a positron and an electron neutrino are emitted — $\text{u} \to \text{d} + \beta^{+} + \nu_{\text{e}}$. For $\beta^{-}$ decay swap the roles: a down quark becomes an up quark, emitting an electron and an electron antineutrino. Name the (anti)neutrino: it is the "other lepton" the question asks for, and it is what makes the β energy continuous.
Leptons 轻子 are fundamental particles that are not made of quarks. The leptons you need are the electron and the electron neutrino, with their antiparticles the positron and the electron antineutrino. (Heavier leptons — the muon and tau — exist but are not tested.) Asked for "two different leptons", give electron and neutrino.
To answer "which are fundamental?": quarks and leptons (electrons, positrons, neutrinos, antineutrinos) are fundamental; protons, neutrons, baryons, mesons and hadrons are not — they are built from quarks.
Worked example. In the list — antineutrino, $\beta^{+}$ particle, neutron, positron, proton — underline the hadrons.
Neutron and proton (three quarks each). The antineutrino and the positron are leptons; the $\beta^{+}$ particle is a positron. The same list asked as "which are not fundamental?" has the same answer: neutron and proton.
Antimatter. Every particle has an antiparticle of the same mass and opposite charge; a positron and an electron are alike in mass (and in the size of their charge) and differ in the sign of their charge. An antihydrogen atom is an antiproton (charge $-e$, made of $\bar{\text{u}}\bar{\text{u}}\bar{\text{d}}$) with a positron in orbit.
Beberapa partikel adalah partikel fundamental (titik-titik, tanpa komponen lebih kecil sejauh yang kita ketahui); yang lain dibangun dari partikel fundamental.
Definisi bernilai satu poin: partikel fundamental adalah partikel yang tidak dapat diuraikan menjadi partikel yang lebih kecil (tidak memiliki struktur internal). Elektron, neutrino, dan kuark adalah fundamental; proton dan neutron bukan.
Kuark adalah partikel fundamental. Ada enam rasa:
Setiap kuark memiliki antikuark dengan besaran muatan yang sama tetapi tanda berlawanan: $\bar{u}$ (muatan $-\tfrac{2}{3}e$), $\bar{d}$ (muatan $+\tfrac{1}{3}e$). Tidak ada properti kuark lainnya yang diujikan.
Ditulis sebagai tabel yang meminta Anda untuk melengkapi:
| Kuark | Muatan | Antikuark | Muatan |
|---|---|---|---|
| up (u), charm (c), top (t) | $+\tfrac{2}{3}e$ | $\bar{\text{u}}$, $\bar{\text{c}}$, $\bar{\text{t}}$ | $-\tfrac{2}{3}e$ |
| down (d), strange (s), bottom (b) | $-\tfrac{1}{3}e$ | $\bar{\text{d}}$, $\bar{\text{s}}$, $\bar{\text{b}}$ | $+\tfrac{1}{3}e$ |
Contoh terpecahkan. Dengan merujuk kepada komposisi kuark, tunjukkan bahawa cas proton ialah $+1.6 \times 10^{-19}\ \text{C}$.
Proton ialah uud: cas $= \tfrac{2}{3}e + \tfrac{2}{3}e - \tfrac{1}{3}e = +e = +1.6 \times 10^{-19}\ \text{C}$. Tulis tiga pecahan dan hasil tambah mereka — markah diberikan untuk pengiraan, bukan jawapan.
Zarah yang dibina daripada kuark dipanggil hadron. Dua jenis:
Proton dan neutron adalah bukan zarah asas — mereka adalah baryon yang terdiri daripada kuark.
"Bandingkan baryon dan meson dari segi zarah penyusunnya" (2 markah): kedua-duanya adalah hadron yang dibuat daripada kuark; baryon terdiri daripada tiga kuark (atau tiga antikuarak, bagi antibaryon), manakala meson terdiri daripada satu kuark dan satu antikuarak. Sebarang cas yang diberikan mestilah bersumber daripada cas kuark tersebut:
Contoh terpecahkan. (a) Satu meson mempunyai cas $-1e$. Berikan satu kemungkinan komposisi kuark. (b) Satu meson Q mempunyai cas 0. Berikan satu kemungkinan komposisi. (c) Satu baryon terdiri daripada tiga kuark dengan perisa berbeza, iaitu u, d dan s. Cari casnya. (d) Berikan komposisi kuark bagi antineutron dan casnya.
(a) d$\bar{\text{u}}$: $-\tfrac{1}{3} - \tfrac{2}{3} = -1$ (s$\bar{\text{u}}$ juga bisa). (b) u$\bar{\text{u}}$: $+\tfrac{2}{3} - \tfrac{2}{3} = 0$ (atau d$\bar{\text{d}}$). (c) uds: $\tfrac{2}{3} - \tfrac{1}{3} - \tfrac{1}{3} = 0$. (d) Antineutron adalah antipartikel dari udd, jadi ia adalah $\bar{\text{u}}\bar{\text{d}}\bar{\text{d}}$: $-\tfrac{2}{3} + \tfrac{1}{3} + \tfrac{1}{3} = 0$ — massa dan muatan yang sama dengan neutron, karena keduanya memiliki muatan nol.
Dalam penjejakan $\beta^{-}$, neutron bertukar menjadi proton; dalam istilah kuark, satu kuark bawah bertukar menjadi kuark atas:
Dalam penjejakan $\beta^{+}$, proton bertukar menjadi neutron; satu kuark atas bertukar menjadi kuark bawah:
"Huraikan penjejakan $\beta^{+}$ dari segi zarah-zarah asas yang terlibat" (2 markah): satu kuark atas dalam proton berubah menjadi kuark bawah, maka proton menjadi neutron, dan positron serta neutrino elektron dipancarkan — $\text{u} \to \text{d} + \beta^{+} + \nu_{\text{e}}$. Untuk penjejakan $\beta^{-}$, tukar peranan: satu kuark bawah menjadi kuark atas, memancarkan elektron dan antineutrino elektron. Nyatakan nama (anti)neutrino: ia adalah "lepton lain" yang diminta oleh soalan, dan ia menyebabkan tenaga β adalah berterusan.
Lepton adalah zarah asas yang tidak terdiri daripada kuark. Lepton yang perlu anda tahu ialah elektron dan neutrino elektron, dengan antimaterinya分别是 positron dan antineutrino elektron. (Lepton yang lebih berat — muon dan tau — wujud tetapi tidak diuji.) Jika diminta "dua lepton berbeza", nyatakan elektron dan neutrino.
Untuk menjawab "yang mana adalah zarah asas": kuark dan lepton (elektron, positron, neutrino, antineutrino) adalah asas; proton, neutron, baryon, meson dan hadron adalah bukan asas — mereka dibina daripada kuark.
Contoh terpecahkan. Dalam senarai berikut — antineutrino, zarah $\beta^{+}$, neutron, positron, proton — gariskan hadron.
Neutron dan proton (masing-masing tiga kuark). Antineutrino dan positron adalah lepton; zarah $\beta^{+}$ merupakan positron. Senarai yang sama jika ditanya "yang mana bukan zarah asas?" mempunyai jawapan yang sama: neutron dan proton.
Antimateri. Setiap zarah mempunyai antimateri dengan jisim yang sama dan cas yang bertentangan; positron dan elektron sepadan dalam jisim (dan saiz cas) dan berbeza dalam tanda cas. Atom antihidrogen ialah antiproton (cas $-e$, terdiri daripada $\bar{\text{u}}\bar{\text{u}}\bar{\text{d}}$) dengan sebuah positron dalam orbit.
Sort particles by the family or interaction that defines them. · Urutkan partikel berdasarkan keluarga atau interaksi yang mendefinisikannya.
| English | Bahasa Indonesia |
|---|---|
| quark/kwɑːk/ | kuark |
| conservation of charge/ˌkɒnsəˈveɪʃn ɒv tʃɑːdʒ/ | kekekalan muatan |
| unified atomic mass unit/ˈjuːnɪfaɪd əˈtɒmɪk mæs ˈjuːnɪt/ | satuan massa atom terpadu |
| radioactive/ˌreɪdɪəʊˈæktɪv/ | radioaktif |
| radiation/ˌreɪdɪˈeɪʃn/ | radiasi |
| ionising/ˈaɪənaɪzɪŋ/ | ionisasi |
| discrete/dɪˈskriːt/ | diskrit |
| fundamental/ˌfʌndəˈmentl/ | fundamental |
| flavours/ˈfleɪvəz/ | rasa |
| antiquark/ˈæntɪkwɑːk/ | antikuark |
| hadrons/ˈhædrɒnz/ | hadron |
| baryons/ˈbæriɒnz/ | barion |
| mesons/ˈmiːzɒnz/ | meson |
| leptons/ˈleptɒnz/ | lepton |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| proton number $Z$ | the number of protons in a nucleus |
| nucleon number $A$ | the total number of protons and neutrons in a nucleus |
| isotopes | nuclei of the same element (same number of protons) with different numbers of neutrons |
| nuclide | a nucleus with a particular number of protons and a particular number of neutrons |
| unified atomic mass unit | one twelfth of the mass of an atom of carbon-12 |
| antiparticle | a particle with the same mass as the corresponding particle but the opposite charge |
| fundamental particle | a particle that cannot be broken down into smaller particles |
| hadron | a particle made of quarks: a baryon (three quarks) or a meson (a quark and an antiquark) |
| lepton | a fundamental particle that is not made of quarks, such as the electron or the neutrino |
| α-particle | a helium-4 nucleus: two protons and two neutrons, charge $+2e$, mass $4\ \text{u}$ |
| $\beta^{-}$ / $\beta^{+}$ particle | an electron / a positron emitted from a nucleus when a neutron becomes a proton / a proton becomes a neutron |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| nombor proton $Z$ | bilangan proton dalam nukleus |
| nombor nukleon $A$ | jumlah bilangan proton dan neutron dalam nukleus |
| isotop | nukleus unsur yang sama (bilangan proton yang sama) dengan bilangan neutron yang berbeza |
| nuklid | nukleus dengan bilangan proton tertentu dan bilangan neutron tertentu |
| unit jisim atom seragam | satu belasan jisim atom karbon-12 |
| antimateri | zarah dengan jisim yang sama seperti zarah berkaitan tetapi cas yang bertentangan |
| zarah asas | zarah yang tidak boleh dipecahkan kepada zarah yang lebih kecil |
| hadron | zarah yang terdiri daripada kuark: baryon (tiga kuark) atau meson (satu kuark dan satu antikuarak) |
| lepton | zarah asas yang tidak terdiri daripada kuark, seperti elektron atau neutrino |
| zarah α | nukleus helium-4: dua proton dan dua neutron, cas $+2e$, jisim $4\ \text{u}$ |
| zarah $\beta^{-}$ / $\beta^{+}$ | elektron / positron yang dipancarkan dari nukleus apabila neutron menjadi proton / proton menjadi neutron |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The radian 弧度 is the angle made at the centre of a circle by an arc 弧 whose length equals the radius. For an arc of length $s$ on a circle of radius $r$, the angle in radians is
Radians have no unit (a ratio of lengths). A full circle has $s = 2\pi r$, so $\theta = 2\pi\ \text{rad}$. A half-circle is $\pi\ \text{rad}$; a quarter is $\pi/2\ \text{rad}$.
To convert: $1\ \text{rad} = 180°/\pi \approx 57.3°$. Set your calculator to radians for this topic; "degree" mode will give wrong answers.
The one-mark definition. The radian is the angle subtended at the centre of a circle by an arc equal in length to the radius of the circle. Give it in words — $\theta = s/r$ on its own is not a definition — and keep $s = r\theta$ ready for turning a distance along the arc into an angle, or back.
Radian adalah sudut yang terbentuk di pusat lingkaran oleh busur whose panjangnya sama dengan jari-jari. Untuk busur sepanjang $s$ pada lingkaran berjari-jari $r$, sudut dalam radian adalah
Radian tidak memiliki satuan (rasio panjang). Satu lingkaran penuh memiliki $s = 2\pi r$, jadi $\theta = 2\pi\ \text{rad}$. Setengah lingkaran adalah $\pi\ \text{rad}$; seperempat adalah $\pi/2\ \text{rad}$.
Untuk konversi: $1\ \text{rad} = 180°/\pi \approx 57.3°$. Atur kalkulator Anda ke mode radian untuk topik ini; mode "derajat" akan menghasilkan jawaban yang salah.
Definisi bernilai satu poin. Radian adalah sudut yang dibentuk di pusat lingkaran oleh busur yang panjangnya sama dengan jari-jari lingkaran. Berikan dalam kata-kata — $\theta = s/r$ sendiri bukan definisi — dan siap gunakan $s = r\theta$ untuk mengubah jarak sepanjang busur menjadi sudut, atau sebaliknya.
| English | Bahasa Indonesia |
|---|---|
| radian/ˈreɪdɪən/ | radian |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
| acceleration/əkˌseləˈreɪʃn/ | percepatan |
| arc/ɑːk/ | busur |
An object moves in a circle of radius $r$ at constant speed $v$. Define:
For uniform motion $\omega$ is constant and
Unit: $\text{rad s}^{-1}$.
What stays constant and what varies. In uniform circular motion the speed, the angular speed, the period and the magnitude of the acceleration are constant; the velocity, the displacement from the centre and the acceleration all change continuously, because their direction changes. Asked to "state two quantities that vary", give two of velocity, acceleration, displacement and momentum — and say it is the direction that varies.
If the object goes once round ($2\pi\ \text{rad}$, one revolution 圈) in time $T$ (the period 周期), then
where $f = 1/T$ is the frequency 频率 of turning (Hz).
In one period $T$ the object travels a distance $2\pi r$ (the circumference 周长) at constant speed, so
This links the linear (tangential 切向) speed $v$ with the angular speed $\omega$. At a larger radius (for the same angular speed) the linear speed is larger — a child on the edge of a merry-go-round moves faster than one near the centre, even though both go round once in the same time.
Worked example. A fairground ride of radius $4.0\ \text{m}$ completes one turn every $8.0\ \text{s}$. Find its angular speed and the linear speed of a rider on the edge.
Worked example. The minute hand of a clock turns once every hour. A piece of modelling clay on the hand moves a total distance of $0.44\ \text{m}$ in $1400\ \text{s}$. Find the angular speed of the hand, the angle the clay turns through in that time, and its distance from the centre of the clock.
$\omega = 2\pi/T = 2\pi/3600 = 1.75 \times 10^{-3}\ \text{rad s}^{-1}$. Angular displacement $\theta = \omega t = 1.75 \times 10^{-3} \times 1400 = 2.44\ \text{rad}$. Distance along the arc $s = r\theta$, so $r = s/\theta = 0.44/2.44 = 0.18\ \text{m}$. Check: $v = s/t = 3.1 \times 10^{-4}\ \text{m s}^{-1}$ and $r\omega = 0.18 \times 1.75 \times 10^{-3} = 3.1 \times 10^{-4}\ \text{m s}^{-1}$ — the same. A common trap in these questions is a radius measured from the rim: a lump $1.2\ \text{cm}$ in from the edge of a $9.3\ \text{cm}$ disc moves in a circle of radius $8.1\ \text{cm}$.
Worked example. A bicycle chain passes round a pedal cog of radius $0.095\ \text{m}$ and a rear-wheel cog of radius $0.038\ \text{m}$. The pedals turn at $1.2$ revolutions per second. Find the speed of the chain and the angular speed of the rear cog. The chain is then moved to a smaller rear cog while the bicycle's speed stays the same; explain, without calculation, what happens to the angular speed of the pedals.
Pedal angular speed $\omega_{1} = 2\pi \times 1.2 = 7.5\ \text{rad s}^{-1}$. Every link of the chain moves at the speed of the rim of the pedal cog: $v = r_{1}\omega_{1} = 0.095 \times 7.5 = 0.72\ \text{m s}^{-1}$. The rear cog's rim moves at the same speed, so $\omega_{2} = v/r_{2} = 0.72/0.038 = 19\ \text{rad s}^{-1}$. With the bicycle's speed unchanged, the rear wheel keeps the same angular speed; the smaller cog's rim therefore moves more slowly ($v = r\omega$ with a smaller $r$), so the chain moves more slowly, and the pedal cog, driven by the same chain, turns at a lower angular speed.
*Kuda-kuda berputar: setiap penunggang berputar melalui sudut yang sama setiap detik.
Sebuah benda bergerak dalam lingkaran berjari-jari $r$ dengan kecepatan tetap $v$. Definisikan:
Untuk gerak seragam, $\omega$ konstan dan
Satuan: $\text{rad s}^{-1}$.
Apa yang tetap konstan dan apa yang berubah. Dalam gerak melingkar seragam, kecepatan, kecepatan sudut, periode, dan besar percepatan adalah konstan; vektornya, perpindahan dari pusat, dan percepatan semuanya berubah terus-menerus karena arah-nya berubah. Jika diminta "nyatakan dua besaran yang berubah", sebutkan dua dari vektor kecepatan, percepatan, perpindahan, dan momentum — dan jelaskan bahwa yang berubah adalah arahnya.
Jika benda tersebut menempuh satu putaran penuh ($2\pi\ \text{rad}$, satu putaran) dalam waktu $T$ (periode), maka
di mana $f = 1/T$ adalah frekuensi perputaran (Hz).
Dalam satu periode $T$, benda menempuh jarak $2\pi r$ (keliling) dengan kecepatan konstan, sehingga
Ini menghubungkan kecepatan linear (tangensial) $v$ dengan kecepatan sudut $\omega$. Pada jari-jari yang lebih besar (untuk kecepatan sudut yang sama), kecepatan linear lebih besar — seorang anak di tepi kincir berputar bergerak lebih cepat daripada yang dekat pusat, meskipun keduanya menyelesaikan satu putaran dalam waktu yang sama.
Contoh dikerjakan. Kuda-kuda berjari-jari $4.0\ \text{m}$ menyelesaikan satu putaran setiap $8.0\ \text{s}$. Temukan kecepatan sudutnya dan kecepatan linear penunggang di tepinya.
Contoh dikerjakan. Jarum menit jam berputar sekali setiap jam. Selembar tanah liat pada jarum tersebut bergerak sejauh total $0.44\ \text{m}$ dalam waktu $1400\ \text{s}$. Temukan kecepatan sudut jarum, sudut yang ditempuh tanah liat dalam waktu itu, dan jaraknya dari pusat jam.
$\omega = 2\pi/T = 2\pi/3600 = 1.75 \times 10^{-3}\ \text{rad s}^{-1}$. Pergeseran sudut $\theta = \omega t = 1.75 \times 10^{-3} \times 1400 = 2.44\ \text{rad}$. Jarak sepanjang busur $s = r\theta$, sehingga $r = s/\theta = 0.44/2.44 = 0.18\ \text{m}$. Cek: $v = s/t = 3.1 \times 10^{-4}\ \text{m s}^{-1}$ dan $r\omega = 0.18 \times 1.75 \times 10^{-3} = 3.1 \times 10^{-4}\ \text{m s}^{-1}$ — sama. jebakan umum dalam pertanyaan ini adalah jari-jari yang diukur dari tepi: gumpalan $1.2\ \text{cm}$ masuk dari tepi cakram $9.3\ \text{cm}$ bergerak dalam lingkaran berjari-jari $8.1\ \text{cm}$.
Contoh terpecahkan. Rantai sepeda melingkari gigi pedal berjari-jari $0.095\ \text{m}$ dan gigi roda belakang berjari-jari $0.038\ \text{m}$. Pedal berputar dengan kecepatan $1.2$ revolusi per detik. Tentukan kecepatan rantai dan kecepatan sudut gigi roda belakang. Kemudian, rantai dipindahkan ke gigi roda belakang yang lebih kecil sementara kecepatan sepeda tetap sama; jelaskan, tanpa perhitungan, apa yang terjadi pada kecepatan sudut pedal.
Kecepatan sudut pedal $\omega_{1} = 2\pi \times 1.2 = 7.5\ \text{rad s}^{-1}$. Setiap mata rantai bergerak dengan kecepatan tepi gigi pedal: $v = r_{1}\omega_{1} = 0.095 \times 7.5 = 0.72\ \text{m s}^{-1}$. Tepi gigi roda belakang bergerak dengan kecepatan yang sama, sehingga $\omega_{2} = v/r_{2} = 0.72/0.038 = 19\ \text{rad s}^{-1}$. Dengan kecepatan sepeda yang tidak berubah, roda belakang mempertahankan kecepatan sudut yang sama; tepi gigi yang lebih kecil tersebut maka bergerak lebih lambat ($v = r\omega$ dengan $r$ yang lebih kecil), sehingga rantai bergerak lebih lambat, dan gigi pedal, yang digerakkan oleh rantai yang sama, berputar dengan kecepatan sudut yang lebih rendah.
s = rθ
Angular speed turns angle per time; arc length s = rθ. · Kecepatan sudut mengubah sudut per waktu; panjang busur s = rθ.
| English | Bahasa Indonesia |
|---|---|
| angular speed/ˈæŋɡjʊlə spiːd/ | kelajuan sudut |
| angular displacement/ˈæŋɡjʊlə dɪˈspleɪsmənt/ | pencucupan sudut |
| period/ˈpɪərɪəd/ | tempoh |
| frequency/ˈfriːkwənsi/ | frekuensi |
| revolution/ˌrevəˈluːʃn/ | revolusi |
| circumference/sɜːˈkʌmfrəns/ | keliling |
| tangential/tænˈdʒenʃl/ | tangensial |
| centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ | percepatan sentripetal |
| perpendicular/ˌpɜːpənˈdɪkjʊlə/ | tegak lurus |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
An object moving in a circle at constant speed still has a changing velocity 速度 — its direction keeps changing, even though its size stays the same. A changing velocity needs an acceleration 加速度. This acceleration points towards the centre and is the centripetal acceleration 向心加速度.
The two-mark description. In uniform circular motion the speed is constant but the velocity is continuously changing, because its direction changes; the acceleration is constant in magnitude and is always directed towards the centre of the circle, perpendicular to the velocity. "State what is meant by centripetal acceleration": the acceleration of an object moving along a circular path, directed towards the centre of the circle.
The two forms are equal because $v = r\omega$. Pick the one with the quantities you have.
The centripetal acceleration is perpendicular 垂直 to the velocity at every instant — never along the direction of motion. (If part of it were along the motion, the speed would change.) Unit: $\text{m s}^{-2}$.
Why the speed does not change. The centripetal force is perpendicular to the displacement at every instant, so it does no work on the object; its kinetic energy, and so its speed, stays constant. Only the direction of the velocity changes — which is exactly what an acceleration perpendicular to the velocity does.
Worked example. The Earth is a sphere of radius $6.37 \times 10^{6}\ \text{m}$ rotating once in $24$ hours. Cambridge is at latitude 纬度 $52.2°$ north. Find the radius of the circle in which Cambridge moves, its speed, and its centripetal acceleration. A student of mass $58.6\ \text{kg}$ stands on bathroom scales there; state and explain the effect of the rotation on the reading.
The circle is about the axis, so $r = R\cos\lambda = 6.37 \times 10^{6} \times \cos 52.2° = 3.90 \times 10^{6}\ \text{m}$. $\omega = 2\pi/(24 \times 3600) = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$, so $v = r\omega = 284\ \text{m s}^{-1}$ and $a = r\omega^{2} = 3.90 \times 10^{6} \times (7.27 \times 10^{-5})^{2} = 2.1 \times 10^{-2}\ \text{m s}^{-2}$. The student needs a resultant force $ma = 1.2\ \text{N}$ towards the axis, which can only come from the weight exceeding the upward contact force; so the scales read less than the weight $mg = 575\ \text{N}$ — by about one newton (the acceleration has a component $a\cos\lambda = 0.013\ \text{m s}^{-2}$ along the vertical, i.e. $0.7\ \text{N}$ of the $1.2\ \text{N}$). Small, but the sign and the reason are the marks.
Sebuah benda yang bergerak dalam lingkaran dengan kecepatan konstan tetap memiliki kecepatan yang berubah — arahnya terus berubah, meskipun besarnya tetap sama. Perubahan kecepatan memerlukan percepatan. Percepatan ini mengarah menuju pusat dan disebut percepatan sentripetal.
Deskripsi dua nilai. Dalam gerak melingkar beraturan, kecepatannya konstan tetapi kecepatannya terus-menerus berubah karena arahnya berubah; percepatannya konstan dalam besar dan selalu diarahkan menuju pusat lingkaran, tegak lurus terhadap kecepatan. "Nyatakan arti dari percepatan sentripetal": percepatan sebuah benda yang bergerak sepanjang lintasan melingkar, diarahkan menuju pusat lingkaran.
Kedua bentuk tersebut setara karena $v = r\omega$. Pilihlah salah satu yang memuat besaran yang Anda miliki.
Percepatan sentripetal adalah tegak lurus terhadap kecepatan pada setiap saat — tidak pernah searah dengan arah gerak. (Jika sebagian darinya berada sejajar dengan gerak, kecepatannya akan berubah.) Satuan: $\text{m s}^{-2}$.
Mengapa kecepatan tidak berubah. Gaya sentripetal tegak lurus terhadap perpindahan pada setiap saat, sehingga gaya ini tidak melakukan usaha pada benda; energi kinetiknya, dan karenanya kecepatannya, tetap konstan. Hanya arah kecepatan yang berubah — yang merupakan persis apa yang dilakukan oleh percepatan yang tegak lurus terhadap kecepatan.
Contoh terpecahkan. Bumi adalah bola berjari-jari $6.37 \times 10^{6}\ \text{m}$ yang berotasi sekali dalam $24$ jam. Cambridge terletak pada lintang $52.2°$ utara. Tentukan jari-jari lingkaran tempat Cambridge bergerak, kecepatannya, dan percepatan sentripetalnya. Seorang siswa bermassa $58.6\ \text{kg}$ berdiri di atas timbangan mandi di sana; nyatakan dan jelaskan efek rotasi terhadap pembacaan timbangan.
Lingkaran tersebut berada tentang sumbu, sehingga $r = R\cos\lambda = 6.37 \times 10^{6} \times \cos 52.2° = 3.90 \times 10^{6}\ \text{m}$. $\omega = 2\pi/(24 \times 3600) = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$, sehingga $v = r\omega = 284\ \text{m s}^{-1}$ dan $a = r\omega^{2} = 3.90 \times 10^{6} \times (7.27 \times 10^{-5})^{2} = 2.1 \times 10^{-2}\ \text{m s}^{-2}$. Siswa membutuhkan gaya resultan $ma = 1.2\ \text{N}$ menuju sumbu, yang hanya dapat berasal dari berat yang melebihi gaya kontak ke atas; sehingga timbangan menunjukkan lebih sedikit daripada berat $mg = 575\ \text{N}$ — sekitar satu newton (percepatan memiliki komponen $a\cos\lambda = 0.013\ \text{m s}^{-2}$ sepanjang vertikal, yaitu $0.7\ \text{N}$ dari $1.2\ \text{N}$). Kecil, tetapi tanda dan alasannya adalah poin penilaiannya.
| English | Bahasa Indonesia |
|---|---|
| latitude/ˈlætɪtjuːd/ | lintang |
| weight/weɪt/ | berat |
| tension/ˈtenʃn/ | tegangan |
| friction/ˈfrɪkʃn/ | gesekan |
| normal contact force/ˈnɔːml ˈkɒntækt fɔːs/ | gaya kontak normal |
By Newton's second law, the resultant force 力 on a body in circular motion at constant speed is
This is the centripetal force 向心力. It always points towards the centre — perpendicular to the velocity.
The centripetal force is not a new kind of force — it is the net result of the real forces acting (tension, gravity, friction, electric attraction, normal contact force, …). In a problem, work out which real force(s) provide it.
Worked example. A $0.20\ \text{kg}$ ball on a string is whirled in a horizontal circle of radius $0.50\ \text{m}$ at $3.0\ \text{m s}^{-1}$. Find the centripetal force (the tension in the string).
where $k = 1/(4\pi\varepsilon_{0})$ and $Z$ is the nuclear charge. Solve for $v$ to get the orbital speed; then $T = 2\pi r/v$.
A ball on a string swung in a horizontal circle (a conical pendulum 圆锥摆), a ball rolling round the inside of a smooth cone, a chair on a fairground swing-ride and a car on a banked track are all the same problem: one force (the tension or the normal contact force) is tilted at an angle $\theta$ to the vertical. Resolve it. Its vertical component balances the weight; its horizontal component is the whole centripetal force:
Never add a separate "centripetal force" to the diagram: the two real forces are the weight and the tilted force, and their resultant is horizontal, towards the centre.
Worked example. A steel ball moves in a horizontal circle of radius $0.12\ \text{m}$ on the smooth inside surface of a cone whose surface makes $52°$ with the horizontal. Name the two forces on the ball, state the direction of their resultant, and find the speed of the ball and the period of its motion.
The forces are the weight (vertically down) and the normal contact force from the cone's surface (perpendicular to the surface, so at $52°$ to the vertical); their resultant is horizontal, towards the centre of the circle. Vertically $N\cos 52° = mg$; horizontally $N\sin 52° = mv^{2}/r$. Dividing: $v^{2} = rg\tan 52° = 0.12 \times 9.81 \times 1.28 = 1.51$, so $v = 1.2\ \text{m s}^{-1}$, and $T = 2\pi r/v = 2\pi \times 0.12/1.23 = 0.61\ \text{s}$. The mass cancels — the speed does not depend on it.
Worked example. A sphere of mass $0.29\ \text{kg}$ hangs from a spring of spring constant $40\ \text{N m}^{-1}$ and unstretched length $6.0\ \text{cm}$. It is set moving in a horizontal circle so that the spring makes $30°$ with the vertical. Find the tension, the radius of the circle and the speed.
Vertically $T\cos 30° = mg$, so $T = 0.29 \times 9.81/0.866 = 3.3\ \text{N}$. The extension is $x = T/k = 3.28/40 = 0.082\ \text{m}$, so the spring's length is $0.060 + 0.082 = 0.142\ \text{m}$ and $r = 0.142\sin 30° = 0.071\ \text{m}$. Horizontally $T\sin 30° = mv^{2}/r$: $v^{2} = 3.28 \times 0.5 \times 0.071/0.29 = 0.40$, $v = 0.63\ \text{m s}^{-1}$. Hooke's law supplies the length; the circle supplies the rest.
When the circle is upright, the speed is not constant (gravity does work) — but at each instant the net force towards the centre still equals $m v^{2}/r$:
Worked example. A car goes round a vertical loop of radius $2.0\ \text{m}$. Find the minimum speed at the top for the car to keep contact with the track (take $g = 9.81\ \text{m s}^{-2}$).
At the slowest speed the track force is zero, so gravity alone provides the centripetal force: $mg = m v_{\text{min}}^{2}/r$, giving $v_{\text{min}} = \sqrt{gr}$:
The constant-speed result ($v = r\omega$, $\omega$ constant) holds for horizontal circles, or where the force only bends the path (orbits in gravity, charges in a magnetic field 磁场).
Worked example. A helium atom is modelled as a nucleus of charge $+2e$ with two electrons in the same circular orbit of radius $170\ \text{pm}$, always on opposite sides of the nucleus. Find the resultant electric force on one electron and its speed.
Each electron is attracted by the nucleus, distance $r$ away, and repelled by the other electron, distance $2r$ away, along the same line: $F = \dfrac{1}{4\pi\varepsilon_{0}}\left(\dfrac{2e^{2}}{r^{2}} - \dfrac{e^{2}}{(2r)^{2}}\right) = \dfrac{e^{2}}{4\pi\varepsilon_{0}r^{2}} \times 1.75 = \dfrac{2.31 \times 10^{-28}}{(1.7 \times 10^{-10})^{2}} \times 1.75 = 1.4 \times 10^{-8}\ \text{N}$. This is the centripetal force: $v = \sqrt{Fr/m} = \sqrt{1.40 \times 10^{-8} \times 1.7 \times 10^{-10}/(9.11 \times 10^{-31})} = 1.6 \times 10^{6}\ \text{m s}^{-1}$.
Worked example. A proton (mass $1.67 \times 10^{-27}\ \text{kg}$) moving at $2.0 \times 10^{6}\ \text{m s}^{-1}$ enters a uniform magnetic field of flux density $0.50\ \text{T}$ at right angles to the field. Find the radius of its path and the time for one revolution.
The magnetic force $Bqv$ is always perpendicular to the velocity, so it is a centripetal force and the speed is constant: $Bqv = mv^{2}/r$ gives $r = \dfrac{mv}{Bq} = \dfrac{1.67 \times 10^{-27} \times 2.0 \times 10^{6}}{0.50 \times 1.60 \times 10^{-19}} = 4.2 \times 10^{-2}\ \text{m}$, and $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{Bq} = 1.3 \times 10^{-7}\ \text{s}$ — independent of the speed, which is why a faster proton makes a larger circle in the same time. For a satellite the gravitational force plays the same part: $\dfrac{GMm}{r^{2}} = mr\omega^{2}$, and the orbital speed and period follow from $r$ alone.
Menurut hukum kedua Newton, gaya resultan pada benda dalam gerak melingkar dengan kecepatan konstan adalah
Ini adalah gaya sentripetal. Ia selalu mengarah menuju pusat — tegak lurus terhadap kecepatan.
Gaya sentripetal bukanlah jenis gaya baru — ia adalah hasil gabungan dari gaya-gaya nyata yang bekerja (tegangan, gravitasi, gesekan, tarikan listrik, gaya kontak normal, ...). Dalam suatu soal, tentukan gaya nyata mana yang menyediakannya.
Contoh terpecahkan. Bola $0.20\ \text{kg}$ pada tali diputar dalam lingkaran horizontal berjari-jari $0.50\ \text{m}$ pada $3.0\ \text{m s}^{-1}$. Tentukan gaya sentripetal (tegangan pada tali).
di mana $k = 1/(4\pi\varepsilon_{0})$ dan $Z$ adalah muatan inti. Selesaikan untuk $v$ untuk mendapatkan kecepatan orbit; lalu $T = 2\pi r/v$.
Bola pada tali yang diayunkan dalam lingkaran horizontal (a pendulum kerucut), bola yang berguling di bagian dalam kerucut halus, kursi di wahana ayunan taman hiburan dan mobil di lintasan miring semuanya merupakan masalah yang sama: satu gaya (tegangan atau gaya kontak normal) miring pada sudut $\theta$ terhadap vertikal. Resolusikan. Komponen vertikal-nya menyeimbangkan berat; komponen horizontal-nya adalah seluruh gaya sentripetal:
Jangan pernah menambahkan "gaya sentripetal" terpisah ke dalam diagram: dua gaya nyata adalah berat dan gaya miring, dan resultannya horizontal, menuju pusat.
Contoh terpecahkan. Sebuah bola baja bergerak dalam lingkaran horizontal berjari-jari $0.12\ \text{m}$ pada permukaan bagian dalam kerucut yang halus yang permukaannya membuat sudut $52°$ terhadap horizontal. Sebutkan dua gaya pada bola, nyatakan arah resultannya, dan temukan kecepatan bola serta periode geraknya.
Gaya-gaya tersebut adalah berat (ke bawah secara vertikal) dan gaya kontak normal dari permukaan kerucut (tegak lurus terhadap permukaan, sehingga pada $52°$ terhadap vertikal); resultannya horizontal, menuju pusat lingkaran. Vertikal $N\cos 52° = mg$; horizontal $N\sin 52° = mv^{2}/r$. Membagi: $v^{2} = rg\tan 52° = 0.12 \times 9.81 \times 1.28 = 1.51$, sehingga $v = 1.2\ \text{m s}^{-1}$, dan $T = 2\pi r/v = 2\pi \times 0.12/1.23 = 0.61\ \text{s}$. Massa saling menghilangkan — kecepatan tidak bergantung padanya.
Contoh terpecahkan. Sebuah bola bermassa $0.29\ \text{kg}$ tergantung pada pegas dengan konstanta pegas $40\ \text{N m}^{-1}$ dan panjang tak meregang $6.0\ \text{cm}$. Bola ini digerakkan dalam lingkaran horizontal sehingga pegas membuat sudut $30°$ dengan vertikal. Temukan tegangan, jari-jari lingkaran dan kecepatannya.
Vertikal $T\cos 30° = mg$, sehingga $T = 0.29 \times 9.81/0.866 = 3.3\ \text{N}$. Perpanjangan adalah $x = T/k = 3.28/40 = 0.082\ \text{m}$, sehingga panjang pegas adalah $0.060 + 0.082 = 0.142\ \text{m}$ dan $r = 0.142\sin 30° = 0.071\ \text{m}$. Horizontal $T\sin 30° = mv^{2}/r$: $v^{2} = 3.28 \times 0.5 \times 0.071/0.29 = 0.40$, $v = 0.63\ \text{m s}^{-1}$. Hukum Hooke menyediakan panjang; lingkaran menyediakan sisanya.
Ketika lingkaran tegak, kecepatannya tidak konstan (gravitasi melakukan kerja) — tetapi pada setiap saat gaya total menuju pusat tetap sama dengan $m v^{2}/r$:
Contoh terpecahkan. Sebuah mobil melintas di lintasan vertikal berjari-jari $2.0\ \text{m}$. Temukan kecepatan minimum di bagian atas agar mobil tetap bersentuhan dengan lintasan (ambil $g = 9.81\ \text{m s}^{-2}$).
Pada kecepatan terendah gaya lintasan adalah nol, sehingga gravitasi saja menyediakan gaya sentripetal: $mg = m v_{\text{min}}^{2}/r$, memberikan $v_{\text{min}} = \sqrt{gr}$:
Hasil kecepatan konstan ($v = r\omega$, $\omega$ konstan) berlaku untuk lingkaran horizontal, atau di mana gaya hanya membengkokkan lintasan (orbit dalam gravitasi, muatan dalam medan magnet).
Contoh terpecahkan. Atom helium dimodelkan sebagai inti bermuatan $+2e$ dengan dua elektron dalam orbit lingkaran yang sama berjari-jari $170\ \text{pm}$, selalu berada di sisi berlawanan dari inti. Temukan resultan gaya listrik pada satu elektron dan kecepatannya.
Setiap elektron ditarik oleh inti, jarak $r$ menjauh, dan ditolak oleh elektron lain, jarak $2r$ menjauh, sepanjang garis yang sama: $F = \dfrac{1}{4\pi\varepsilon_{0}}\left(\dfrac{2e^{2}}{r^{2}} - \dfrac{e^{2}}{(2r)^{2}}\right) = \dfrac{e^{2}}{4\pi\varepsilon_{0}r^{2}} \times 1.75 = \dfrac{2.31 \times 10^{-28}}{(1.7 \times 10^{-10})^{2}} \times 1.75 = 1.4 \times 10^{-8}\ \text{N}$. Ini adalah gaya sentripetal: $v = \sqrt{Fr/m} = \sqrt{1.40 \times 10^{-8} \times 1.7 \times 10^{-10}/(9.11 \times 10^{-31})} = 1.6 \times 10^{6}\ \text{m s}^{-1}$.
Contoh terpecahkan. Sebuah proton (massa $1.67 \times 10^{-27}\ \text{kg}$) bergerak pada $2.0 \times 10^{6}\ \text{m s}^{-1}$ memasuki medan magnet seragam dengan densitas fluks $0.50\ \text{T}$ secara tegak lurus terhadap medan. Temukan jari-jari lintasannya dan waktu untuk satu revolusi.
Gaya magnet $Bqv$ selalu tegak lurus terhadap kecepatan, sehingga itu adalah gaya sentripetal dan kecepatannya konstan: $Bqv = mv^{2}/r$ memberikan $r = \dfrac{mv}{Bq} = \dfrac{1.67 \times 10^{-27} \times 2.0 \times 10^{6}}{0.50 \times 1.60 \times 10^{-19}} = 4.2 \times 10^{-2}\ \text{m}$, dan $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{Bq} = 1.3 \times 10^{-7}\ \text{s}$ — independen dari kecepatan, itulah sebabnya proton yang lebih cepat membuat lingkaran yang lebih besar dalam waktu yang sama. Untuk satelit, gaya gravitasi memainkan peran yang sama: $\dfrac{GMm}{r^{2}} = mr\omega^{2}$, dan kecepatan orbit dan periode mengikuti dari $r$ saja.
F = mv²/r
For circular motion the force needed grows with the square of the speed — double v, four times the force. · Untuk gerak melingkar, gaya yang dibutuhkan bertambah seiring dengan kuadrat kecepatan — ganda v, empat kali lipat gayanya.
| English | Bahasa Indonesia |
|---|---|
| force/fɔːs/ | gaya |
| centripetal force/senˈtrɪpɪtl fɔːs/ | kekuatan sentripetal |
| Banked corner/bæŋkt ˈkɔːnə/ | Tikungan miring |
| banked/bæŋkt/ | miring (banked) |
| satellite/ˈsætəlaɪt/ | satelit |
| orbit/ˈɔːbɪt/ | orbit |
| gravitational attraction/ˌɡrævɪˈteɪʃənl əˈtrækʃn/ | tarikan gravitasi |
| electron/ɪˈlektrɒn/ | elektron |
| electrostatic/ɪˌlektrəʊˈstætɪk/ | elektrostatik |
| nucleus/ˈnjuːklɪəs/ | inti sel |
| conical pendulum/ˈkɒnɪkl ˈpendjʊləm/ | pendulum kerucut |
| magnetic field/mæɡˈnetɪk fiːld/ | medan magnet |
| English | Bahasa Indonesia |
|---|---|
| radial/ˈreɪdɪəl/ | radial |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| radian | the angle subtended at the centre of a circle by an arc equal in length to the radius |
| angular displacement | the angle, in radians, through which the radius has turned from its starting position |
| angular speed | the angle swept out by the radius per unit time (the rate of change of angular displacement) |
| period | the time taken for one complete revolution |
| uniform circular motion | motion in a circle at constant speed, with a velocity that changes continuously in direction and an acceleration of constant magnitude directed towards the centre |
| centripetal acceleration | the acceleration of an object moving in a circular path, directed towards the centre of the circle and perpendicular to the velocity |
| centripetal force | the resultant force on an object in circular motion, directed towards the centre, of magnitude $mv^{2}/r$ |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| radian | sudut yang dibentuk di pusat lingkaran oleh busur yang panjangnya sama dengan jari-jari |
| perpindahan sudut | sudut, dalam radian, yang telah diputar oleh jari-jari dari posisi awalnya |
| kecepatan sudut | sudut yang disapu oleh jari-jari per satuan waktu (laju perubahan perpindahan sudut) |
| periode | waktu yang diperlukan untuk satu putaran lengkap |
| gerak melingkar seragam | gerak dalam lingkaran dengan kecepatan konstan, dengan kecepatan yang berubah terus-menerus dalam arah dan percepatan dengan besaran konstan yang mengarah ke pusat |
| percepatan sentripetal | percepatan suatu benda yang bergerak dalam lintasan melingkar, berarah menuju pusat lingkaran dan tegak lurus terhadap kecepatan |
| gaya sentripetal | resultan gaya pada benda dalam gerak melingkar, berarah menuju pusat, dengan besaran $mv^{2}/r$ |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A gravitational field 重力场 is a region where a mass 质量 feels a force 力 from other masses. The gravitational field strength 重力场强度 $g$ at a point is the gravitational force per unit mass on a small test mass 检验质量 placed there:
Unit: $\text{N kg}^{-1}$ (the same as $\text{m s}^{-2}$ — the acceleration of free fall in the field). $g$ is a vector 矢量, pointing the way the force acts — towards the source mass.
The examiner's wording. A gravitational field is a region of space in which a mass experiences a force. Gravitational field strength at a point is the gravitational force per unit mass acting on a small test mass placed at the point. The direction of a field line at a point is the direction of the force on a (small test) mass placed there. All three are one-mark definitions; "force per unit mass" is the phrase that scores, "the force on 1 kg" is not.
A gravitational field is drawn with field lines 场线 that point the way the force acts on a test mass:
Closer lines mean a stronger field.
Why $g$ is constant near the surface, in terms of field lines (a two-mark explanation): over a region whose size is small compared with the Earth's radius, the radial field lines are almost parallel and their spacing hardly changes with height, so the field strength — which the spacing represents — is almost the same at the top of the region as at the bottom. Over the whole planet the lines spread out with distance and the field falls.
Worked example. A point P represents a point mass. Draw lines to represent the gravitational field around P, and state what the direction of a line shows.
Straight radial lines, evenly spaced around P, with arrows pointing inwards towards P. The direction of a line is the direction of the force on a small test mass placed there — always towards the mass that produces the field, because gravity only attracts.
Medan gravitasi adalah daerah di mana massa merasakan gaya dari massa lainnya. Kekuatan medan gravitasi $g$ pada suatu titik adalah gaya gravitasi per satuan massa pada massa uji kecil yang ditempatkan di sana:
Satuan: $\text{N kg}^{-1}$ (sama dengan $\text{m s}^{-2}$ — percepatan jatuh bebas di dalam medan tersebut). $g$ adalah vektor, menunjuk ke arah gaya bekerja — menuju massa sumber.
Istilah penguji. Sebuah medan gravitasi adalah daerah ruang di mana suatu massa mengalami gaya. Kekuatan medan gravitasi di suatu titik adalah gaya gravitasi per satuan massa yang bekerja pada massa uji kecil yang diletakkan di titik tersebut. Arah garis medan di suatu titik adalah arah gaya pada (massa uji) kecil yang diletakkan di sana. Ketiganya adalah definisi bernilai satu poin; "gaya per satuan massa" adalah frasa yang mendapat nilai, "gaya pada 1 kg" tidak mendapatkan nilai penuh.
Medan gravitasi digambar dengan garis medan yang menunjuk ke arah gaya bekerja pada massa uji:
Garis yang lebih rapat menunjukkan medan yang lebih kuat.
Mengapa $g$ konstan dekat permukaan, dalam istilah garis medan (penjelasan bernilai dua poin): di atas wilayah yang ukurannya kecil dibandingkan jari-jari Bumi, garis medan radial hampir paralel dan jaraknya hampir tidak berubah seiring ketinggian, sehingga kekuatan medan — yang diwakili oleh jarak antar garis — hampir sama di bagian atas wilayah tersebut dibandingkan dengan bagian bawahnya. Di seluruh planet, garis-garis tersebut menyebar menjauh seiring bertambahnya jarak dan medan melemah.
Contoh terpecahkan. Titik P mewakili massa titik. Gambar garis untuk merepresentasikan medan gravitasi di sekitar P, dan nyatakan apa arti arah garis tersebut.
Garis radial lurus, berjarak sama di sekitar P, dengan anak panah menunjuk ke dalam menuju P. Arah garis adalah arah gaya pada massa uji kecil yang diletakkan di sana — selalu menuju massa yang menghasilkan medan, karena gravitasi hanya menarik.
Change the mass. The field lines point inward and get denser close in, where the field is stronger — a radial field around a point mass. · Ubah massanya. Garis-garis medan mengarah ke dalam dan menjadi lebih rapat di dekat pusat, di mana medan lebih kuat — medan radial di sekitar massa titik.
| English | Bahasa Indonesia |
|---|---|
| gravitational field/ˌɡrævɪˈteɪʃənl fiːld/ | medan gravitasi |
| mass/mæs/ | massa |
| force/fɔːs/ | gaya |
| gravitational field strength/ˌɡrævɪˈteɪʃənl fiːld streŋθ/ | kekuatan medan gravitasi |
| test mass/test mæs/ | massa uji |
| vector/ˈvektə/ | vektor |
| field line/fiːld laɪn/ | garis medan |
| point mass/pɔɪnt mæs/ | massa titik |
| radial/ˈreɪdɪəl/ | radial |
| uniform field/ˈjuːnɪfɔːm fiːld/ | medan seragam |
| Newton's law of gravitation/ˈnjuːtnz lɔː ɒv ˌɡrævɪˈteɪʃn/ | Hukum gravitasi Newton |
| universal gravitational constant/ˌjuːnɪˈvɜːsl ˌɡrævɪˈteɪʃənl ˈkɒnstənt/ | konstanta gravitasi universal |
| scalar/ˈskeɪlə/ | skalar |
| gravitational potential/ˌɡrævɪˈteɪʃənl pəˈtenʃl/ | potensial gravitasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
For two point masses $m_{1}, m_{2}$ a distance $r$ apart, the force on each is
pulling them together along the line joining them. This is Newton's law of gravitation 万有引力定律. The constant $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$ is the universal gravitational constant 万有引力常量.
In words, as the mark scheme wants it: the gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation — both proportionalities, and "point masses". Asked to "state the equation and the meaning of any other symbols", give $F = Gm_{1}m_{2}/r^{2}$ with $G$ the gravitational constant and $r$ the distance between the centres.
Worked example. Two isolated uniform spheres, each of mass $1.0 \times 10^{3}\ \text{kg}$, have their centres $2.0\ \text{m}$ apart. Find the gravitational force between them and comment on its size.
$F = \dfrac{6.67 \times 10^{-11} \times 1.0 \times 10^{3} \times 1.0 \times 10^{3}}{2.0^{2}} = 1.7 \times 10^{-5}\ \text{N}$. This is about $10^{-9}$ of each sphere's weight ($9.8 \times 10^{3}\ \text{N}$): gravity between everyday objects is negligible, and only a planet-sized mass produces a noticeable field. "Isolated" in a question means that no other masses need be considered.
For a uniform sphere (such as a planet or star), the field at any point outside is the same as that of a point mass equal to the total mass at the centre. So from above the surface, you can treat the Earth as a point mass at its centre. (Points inside a sphere are different, and are not in the syllabus.)
Between two masses the fields subtract and the potentials add. Field strength is a vector: on the line joining two spheres the two fields point in opposite directions, so there is a point where they cancel and the resultant field is zero. Potential is a scalar: the two potentials simply add, and both are negative, so the potential is least negative (a maximum) at the point where the field is zero.
Worked example. Two identical isolated uniform spheres X and Y, each of mass $M$ and radius $R$, have their centres a distance $L$ apart. Point P lies on the line joining the centres. State where on the line the resultant field strength is zero, and find the gravitational potential there.
By symmetry the fields of X and Y are equal and opposite at the midpoint, $L/2$ from each centre, so the resultant field is zero there. The potential at P is the sum $\phi = -\dfrac{GM}{L/2} - \dfrac{GM}{L/2} = -\dfrac{4GM}{L}$. A mass released at P would stay there; a mass released anywhere else on the line falls towards the nearer sphere.
Put the gravitational force on a test mass $m$ at distance $r$ from a point mass $M$ into $g = F/m$:
So $g$ falls off as $1/r^{2}$ as you move away from the source.
The two-mark derivation. The force on a test mass $m$ at distance $r$ from a point mass $M$ is $F = GMm/r^{2}$ (Newton's law). Field strength is force per unit mass, $g = F/m$, so $g = GM/r^{2}$. Write both lines: the law and the definition are the two marks. Because $g \propto 1/r^{2}$, halving the distance makes the field four times stronger: at distance $x/2$ the field is $4g$, in the same direction, towards $M$.
Worked example. Find the gravitational field strength at the Earth's surface. (Earth's mass $M = 6.0 \times 10^{24}\ \text{kg}$, radius $R = 6.4 \times 10^{6}\ \text{m}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)
The Earth's radius is $R \approx 6.4 \times 10^{6}\ \text{m}$. Rising to height $h$ changes the distance from the centre from $R$ to $R + h$. For $h \ll R$ (any building or mountain), $(R + h)/R \approx 1$, so $g$ barely changes — going from $5\ \text{m}$ to $10\ \text{m}$ high changes $r$ by about one part in a million. In the laboratory, $g$ is effectively constant.
Untuk dua massa titik $m_{1}, m_{2}$ yang terpisah sejauh $r$, gaya pada masing-masing adalah
menarik mereka bersama sepanjang garis penghubung. Ini adalah hukum gravitasi Newton. Konstanta $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$ adalah konstanta gravitasi universal.
Dalam kata-kata, sesuai dengan kunci jawaban: gaya gravitasi antara dua massa titik sebanding dengan hasil kali massa mereka dan berbanding terbalik dengan kuadrat jarak pemisahan — kedua proporsionalitas tersebut, serta "massa titik". Jika diminta untuk "nyatakan persamaan dan makna simbol lainnya", berikan $F = Gm_{1}m_{2}/r^{2}$ dengan $G$ sebagai konstanta gravitasi dan $r$ sebagai jarak antara pusat-pusatnya.
Contoh terpecahkan. Dua bola seragam terisolasi, masing-masing bermassa $1.0 \times 10^{3}\ \text{kg}$, memiliki jarak pusat $2.0\ \text{m}$. Hitunglah gaya gravitasi di antara keduanya dan komentari besarnya.
$F = \dfrac{6.67 \times 10^{-11} \times 1.0 \times 10^{3} \times 1.0 \times 10^{3}}{2.0^{2}} = 1.7 \times 10^{-5}\ \text{N}$. Ini sekitar $10^{-9}$ dari berat setiap bola ($9.8 \times 10^{3}\ \text{N}$): gravitasi antara benda sehari-hari dapat diabaikan, dan hanya massa sebesar planet yang menghasilkan medan yang terasa. "Terisolasi" dalam soal berarti tidak perlu mempertimbangkan massa lain.
Untuk bola seragam (seperti planet atau bintang), medan di setiap titik di luar sama dengan yang dari massa titik yang besarnya sama dengan massa total di pusat. Jadi dari atas permukaan, Anda dapat memperlakukan Bumi sebagai massa titik di pusatnya. (Titik di dalam bola berbeda, dan tidak termasuk dalam silabus.)
Antara dua massa medan saling mengurangi dan potensial saling menjumlahkan. Kekuatan medan adalah vektor: pada garis yang menghubungkan dua bola, kedua medan mengarah berlawanan arah, sehingga ada titik di mana mereka saling meniadakan dan medan resultan menjadi nol. Potensial adalah skalar: kedua potensial hanya dijumlahkan, dan keduanya negatif, sehingga potensial paling kurang negatif (maksimum) terjadi di titik di mana medan nol.
Contoh terpecahkan. Dua bola seragam terisolasi identik X dan Y, masing-masing bermassa $M$ dan berjari-jari $R$, memiliki jarak antarpusel $L$. Titik P terletak pada garis yang menghubungkan pusat-pusat tersebut. Nyatakan di mana pada garis tersebut kekuatan medan resultan bernilai nol, dan temukan potensial gravitasi di sana.
Berdasarkan simetri, medan X dan Y sama besar dan berlawanan arah di titik tengah, $L/2$ dari setiap pusat, sehingga medan resultan nol di sana. Potensial di P adalah jumlah $\phi = -\dfrac{GM}{L/2} - \dfrac{GM}{L/2} = -\dfrac{4GM}{L}$. Massa yang dilepaskan di P akan tetap di sana; massa yang dilepaskan di tempat lain pada garis tersebut jatuh menuju bola yang lebih dekat.
Masukkan gaya gravitasi pada massa uji $m$ pada jarak $r$ dari massa titik $M$ ke dalam $g = F/m$:
Sehingga $g$ menurun sebanding dengan $1/r^{2}$ saat Anda menjauh dari sumber.
Penurunan dua poin. Gaya pada massa uji $m$ pada jarak $r$ dari massa titik $M$ adalah $F = GMm/r^{2}$ (hukum Newton). Kekuatan medan adalah gaya per satuan massa, $g = F/m$, sehingga $g = GM/r^{2}$. Tulis kedua baris: hukum dan definisi adalah dua poin tersebut. Karena $g \propto 1/r^{2}$, menggandakan jarak membuat medan empat kali lebih kuat: pada jarak $x/2$ medan adalah $4g$, dalam arah yang sama, menuju $M$.
Contoh terpecahkan. Temukan kekuatan medan gravitasi di permukaan Bumi. (Massa Bumi $M = 6.0 \times 10^{24}\ \text{kg}$, jari-jari $R = 6.4 \times 10^{6}\ \text{m}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)
Jari-jari Bumi adalah $R \approx 6.4 \times 10^{6}\ \text{m}$. Naik ke ketinggian $h$ mengubah jarak dari pusat dari $R$ menjadi $R + h$. Untuk $h \ll R$ (bangunan atau gunung apa pun), $(R + h)/R \approx 1$, sehingga $g$ hampir tidak berubah — naik dari $5\ \text{m}$ ke $10\ \text{m}$ tinggi mengubah $r$ sekitar satu bagian dalam satu juta. Di laboratorium, $g$ efektif konstan.
F ∝ Mm / r²
Gravity pulls inward and weakens with the square of the distance. · Gravitasi menarik ke dalam dan melemah dengan kuadrat jarak.
For a satellite 卫星 of mass $m$ in a circular orbit 轨道 of radius $r$ around a body of mass $M$, gravity provides the centripetal force 向心力:
Cancel $m$ (the orbital speed does not depend on the satellite's mass):
Worked example. A satellite orbits the Earth in a circular orbit of radius $r = 7.0 \times 10^{6}\ \text{m}$. Find its orbital speed. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$.)
The period 周期 follows from $T = 2\pi r / v$:
This is Kepler's third law 开普勒第三定律 for circular orbits: $T^{2} \propto r^{3}$. A plot of $T^{2}$ against $r^{3}$ is a straight line through the origin with gradient $4\pi^{2}/(GM)$, so orbital data gives the central mass.
The two-mark derivation, as the exam sets it. For a satellite of mass $m$ in a circular orbit of radius $R$ and period $T$ around a planet of mass $M$: the gravitational force provides the centripetal force, $\dfrac{GMm}{R^{2}} = mR\omega^{2}$, and $\omega = 2\pi/T$; so $\dfrac{GM}{R^{2}} = \dfrac{4\pi^{2}R}{T^{2}}$, giving $T^{2} = \dfrac{4\pi^{2}R^{3}}{GM}$. For an orbit at height $h$ above a planet of radius $R_{\text{p}}$ the orbital radius is $R_{\text{p}} + h$, so $T^{2} = 4\pi^{2}(R_{\text{p}} + h)^{3}/GM$: a graph of $T^{2}$ against $(R_{\text{p}} + h)^{3}$ is a straight line through the origin whose gradient gives $M$.
Worked example. Satellite X, of mass $M_{\text{s}}$, orbits a planet at a distance $4R$ from its centre; satellite Y, of mass $2M_{\text{s}}$, orbits at $3R$. Compare their speeds, periods and kinetic energies.
From $v = \sqrt{GM/r}$, $\dfrac{v_{\text{Y}}}{v_{\text{X}}} = \sqrt{\dfrac{4R}{3R}} = 1.15$: Y moves faster, and the satellite masses do not enter. From $T \propto r^{3/2}$, $\dfrac{T_{\text{Y}}}{T_{\text{X}}} = \left(\dfrac{3}{4}\right)^{3/2} = 0.65$. Kinetic energy $= \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, so $\dfrac{E_{\text{Y}}}{E_{\text{X}}} = \dfrac{2M_{\text{s}}/3R}{M_{\text{s}}/4R} = \dfrac{8}{3}$. In ratio questions cancel $G$, $M$ and the satellite mass first; only the radii and masses that differ survive.
Binary stars. Two stars of masses $M_{\text{A}}$ and $M_{\text{B}}$ a distance $d$ apart orbit their common centre of mass 质心, always on opposite sides of it, with the same period. The centre of mass divides $d$ in the inverse ratio of the masses, $M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ with $r_{\text{A}} + r_{\text{B}} = d$, and the same gravitational force $GM_{\text{A}}M_{\text{B}}/d^{2}$ is the centripetal force on each star.
Worked example. A binary star consists of star A, mass $4.0 \times 10^{30}\ \text{kg}$, and star B, mass $2.0 \times 10^{30}\ \text{kg}$, with centres $3.3 \times 10^{11}\ \text{m}$ apart. Find the distance of A from the centre of mass and the period of the orbit.
$M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ with $r_{\text{A}} + r_{\text{B}} = 3.3 \times 10^{11}$ gives $r_{\text{A}} = \dfrac{2.0}{6.0} \times 3.3 \times 10^{11} = 1.1 \times 10^{11}\ \text{m}$ (and $r_{\text{B}} = 2.2 \times 10^{11}\ \text{m}$). For star A: $\dfrac{GM_{\text{A}}M_{\text{B}}}{d^{2}} = M_{\text{A}}r_{\text{A}}\omega^{2}$, so $\omega^{2} = \dfrac{GM_{\text{B}}}{d^{2}r_{\text{A}}} = \dfrac{6.67 \times 10^{-11} \times 2.0 \times 10^{30}}{(3.3 \times 10^{11})^{2} \times 1.1 \times 10^{11}} = 1.1 \times 10^{-14}\ \text{s}^{-2}$, $\omega = 1.06 \times 10^{-7}\ \text{rad s}^{-1}$ and $T = 2\pi/\omega = 6.0 \times 10^{7}\ \text{s}$, about $1.9$ years. Using $d$ in the force but $r_{\text{A}}$ in the centripetal term is the whole point of the question.
A geostationary 地球同步 satellite:
It must have the same angular speed 角速度 as the Earth, in the same direction, in the equatorial plane (or it would drift north–south during the day). From $T = 24\ \text{h}$ and $T^{2} = 4\pi^{2} r^{3}/(GM)$, the radius is $r \approx 4.2 \times 10^{7}\ \text{m}$ (about $3.6 \times 10^{7}\ \text{m}$ above the surface).
Worked example. Calculate the radius of a geostationary orbit around the Earth ($M = 5.98 \times 10^{24}\ \text{kg}$), and explain why a satellite with the same orbital radius and period may still not be geostationary.
$T = 24 \times 3600 = 8.64 \times 10^{4}\ \text{s}$; $r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (8.64 \times 10^{4})^{2}}{4\pi^{2}} = 7.5 \times 10^{22}\ \text{m}^{3}$, so $r = 4.2 \times 10^{7}\ \text{m}$ — a height of about $3.6 \times 10^{7}\ \text{m}$ above the surface. A satellite with this period is geostationary only if its orbit is in the plane of the equator and it travels from west to east; with the same period in a tilted orbit (over the poles, say) it returns to the same point each day but is not always above it. Mars turns once in about $25$ hours, so a satellite that stays above one point on Mars has a $25$-hour period and the same two other features.
Untuk satelit bermassa $m$ dalam orbit melingkar berjari-jari $r$ mengelilingi benda bermassa $M$, gravitasi menyediakan gaya sentripetal:
Coret $m$ (kecepatan orbit tidak bergantung pada massa satelit):
Contoh terpecahkan. Sebuah satelit mengorbit Bumi dalam orbit melingkar berjari-jari $r = 7.0 \times 10^{6}\ \text{m}$. Temukan kecepatan orbitnya. (Untuk Bumi, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$.)
Periode diturunkan dari $T = 2\pi r / v$:
Ini adalah hukum Kepler ketiga untuk orbit melingkar: $T^{2} \propto r^{3}$. Plot $T^{2}$ terhadap $r^{3}$ adalah garis lurus melalui titik asal dengan gradien $4\pi^{2}/(GM)$, sehingga data orbit memberikan massa pusat.
Penurunan nilai dua poin, sesuai format soal ujian. Untuk satelit bermassa $m$ dalam orbit melingkar berjari-jari $R$ dan periode $T$ mengelilingi planet bermassa $M$: gaya gravitasi menyediakan gaya sentripetal, $\dfrac{GMm}{R^{2}} = mR\omega^{2}$, dan $\omega = 2\pi/T$; sehingga $\dfrac{GM}{R^{2}} = \dfrac{4\pi^{2}R}{T^{2}}$, menghasilkan $T^{2} = \dfrac{4\pi^{2}R^{3}}{GM}$. Untuk orbit pada ketinggian $h$ di atas planet berjari-jari $R_{\text{p}}$, jari-jari orbit adalah $R_{\text{p}} + h$, sehingga $T^{2} = 4\pi^{2}(R_{\text{p}} + h)^{3}/GM$: grafik $T^{2}$ terhadap $(R_{\text{p}} + h)^{3}$ merupakan garis lurus melalui titik asal yang gradiennya memberikan $M$.
Contoh terpecahkan. Satelit X, bermassa $M_{\text{s}}$, mengorbit sebuah planet pada jarak $4R$ dari pusatnya; satelit Y, bermassa $2M_{\text{s}}$, mengorbit pada $3R$. Bandingkan kecepatan, periode, dan energi kinetik keduanya.
Dari $v = \sqrt{GM/r}$, $\dfrac{v_{\text{Y}}}{v_{\text{X}}} = \sqrt{\dfrac{4R}{3R}} = 1.15$: Y bergerak lebih cepat, dan massa satelit tidak masuk dalam persamaan. Dari $T \propto r^{3/2}$, $\dfrac{T_{\text{Y}}}{T_{\text{X}}} = \left(\dfrac{3}{4}\right)^{3/2} = 0.65$. Energi kinetik $= \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, sehingga $\dfrac{E_{\text{Y}}}{E_{\text{X}}} = \dfrac{2M_{\text{s}}/3R}{M_{\text{s}}/4R} = \dfrac{8}{3}$. Pada soal perbandingan, coret $G$, $M$ dan massa satelit terlebih dahulu; hanya jari-jari dan massa yang berbeda yang tersisa.
Bintang ganda. Dua bintang dengan massa $M_{\text{A}}$ dan $M_{\text{B}}$ berjarak $d$ satu sama lain mengorbit pusat massa bersama mereka, selalu berada di sisi yang berlawanan dari pusat tersebut, dengan periode yang sama. Pusat massa membagi $d$ dalam rasio terbalik dari massa-massa tersebut, $M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ dengan $r_{\text{A}} + r_{\text{B}} = d$, dan gaya gravitasi yang sama $GM_{\text{A}}M_{\text{B}}/d^{2}$ adalah gaya sentripetal pada masing-masing bintang.
Contoh terpecahkan. Sebuah bintang ganda terdiri dari bintang A, bermassa $4.0 \times 10^{30}\ \text{kg}$, dan bintang B, bermassa $2.0 \times 10^{30}\ \text{kg}$, dengan jarak antar pusat $3.3 \times 10^{11}\ \text{m}$. Tentukan jarak A dari pusat massa dan periode orbitnya.
$M_{\text{A}}r_{\text{A}} = M_{\text{B}}r_{\text{B}}$ dengan $r_{\text{A}} + r_{\text{B}} = 3.3 \times 10^{11}$ menghasilkan $r_{\text{A}} = \dfrac{2.0}{6.0} \times 3.3 \times 10^{11} = 1.1 \times 10^{11}\ \text{m}$ (dan $r_{\text{B}} = 2.2 \times 10^{11}\ \text{m}$). Untuk bintang A: $\dfrac{GM_{\text{A}}M_{\text{B}}}{d^{2}} = M_{\text{A}}r_{\text{A}}\omega^{2}$, sehingga $\omega^{2} = \dfrac{GM_{\text{B}}}{d^{2}r_{\text{A}}} = \dfrac{6.67 \times 10^{-11} \times 2.0 \times 10^{30}}{(3.3 \times 10^{11})^{2} \times 1.1 \times 10^{11}} = 1.1 \times 10^{-14}\ \text{s}^{-2}$, $\omega = 1.06 \times 10^{-7}\ \text{rad s}^{-1}$ dan $T = 2\pi/\omega = 6.0 \times 10^{7}\ \text{s}$, sekitar $1.9$ tahun. Menggunakan $d$ dalam suku gaya tetapi $r_{\text{A}}$ dalam suku sentripetal merupakan inti utama dari pertanyaan ini.
Sebuah satelit geostasioner:
Satelit tersebut harus memiliki kecepatan sudut yang sama dengan Bumi, dalam arah yang sama, di bidang khatulistiwa (jika tidak, ia akan bergeser ke utara-selatan selama hari). Dari $T = 24\ \text{h}$ dan $T^{2} = 4\pi^{2} r^{3}/(GM)$, jari-jarinya adalah $r \approx 4.2 \times 10^{7}\ \text{m}$ (sekitar $3.6 \times 10^{7}\ \text{m}$ di atas permukaan).
Contoh terpecahkan. Hitung jari-jari orbit geostasioner di sekitar Bumi ($M = 5.98 \times 10^{24}\ \text{kg}$), dan jelaskan mengapa satelit dengan jari-jari dan periode orbit yang sama mungkin saja belum tentu geostasioner.
$T = 24 \times 3600 = 8.64 \times 10^{4}\ \text{s}$; $r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (8.64 \times 10^{4})^{2}}{4\pi^{2}} = 7.5 \times 10^{22}\ \text{m}^{3}$, sehingga $r = 4.2 \times 10^{7}\ \text{m}$ — ketinggian sekitar $3.6 \times 10^{7}\ \text{m}$ di atas permukaan. Satelit dengan periode ini bersifat geostasioner hanya jika orbitnya berada di bidang khatulistiwa dan bergerak dari barat ke timur; dengan periode yang sama pada orbit miring (misalnya melewati kutub), ia kembali ke titik yang sama setiap hari tetapi tidak selalu berada tepat di atasnya. Mars berputar sekali dalam waktu sekitar $25$ jam, sehingga satelit yang tetap di atas satu titik di Mars memiliki periode $25$ jam dan dua karakteristik lainnya yang sama.
| English | Bahasa Indonesia |
|---|---|
| orbit/ˈɔːbɪt/ | orbit |
| centripetal force/senˈtrɪpɪtl fɔːs/ | kekuatan sentripetal |
| satellite/ˈsætəlaɪt/ | satelit |
| period/ˈpɪərɪəd/ | tempoh |
| Kepler's third law/ˈkepləz θɜːd lɔː/ | Hukum ketiga Kepler |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| centre of mass/ˈsentə ɒv mæs/ | pusat massa |
| geostationary/ˌdʒiːəʊˈsteɪʃənəri/ | geostasioner |
| equator/ɪˈkweɪtə/ | ekuator |
| angular speed/ˈæŋɡjʊlə spiːd/ | kelajuan sudut |
| infinity/ɪnˈfɪnɪti/ | tak hingga |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | energi potensial gravitasi |
| escape velocity/eˈskeɪp vəˈlɒsɪti/ | kecepatan lepas |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Gravitational potential 引力势 $\phi$ at a point is the work done per unit mass in bringing a small test mass from infinity 无穷远 to that point:
Unit: $\text{J kg}^{-1}$.
The potential is taken as zero at infinity. As the test mass falls in towards the source, gravity does the work for you, so $\phi$ is negative everywhere except at infinity. For a point mass $M$ at distance $r$:
$\phi$ is a scalar 标量. For several masses, add the potentials.
The two-mark definition, and why it is negative. Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to the point. Potential is defined as zero at infinity; gravity is attractive, so as the test mass comes in from infinity the field does work on it and the mass gives out energy — the work that an external agent must do is negative. Hence the potential at every finite distance is below zero. (Electric potential near a positive charge is positive for the opposite reason: that field repels, so work must be done to bring a positive test charge in.)
Similarities and differences with electric potential. Both are defined as work done per unit mass or per unit positive charge from infinity, both are scalars, both vary as $1/r$ from a point source and both are zero at infinity. The difference: gravitational potential is always negative, because gravity only attracts, whereas electric potential can be positive or negative depending on the sign of the charge.
Reading a potential graph. Two graphs appear in questions. On $\phi$ against $r$ the curve is $-GM/r$: the surface value $\phi_{\text{s}} = -GM/R$ gives the mass of the planet, and the gradient at any point gives the field strength there, since $g = -\dfrac{\Delta\phi}{\Delta r}$ — field strength is the negative of the potential gradient (this is the "relationship between potential and field strength" a question may ask for). On $\phi$ against $1/r$ the graph is a straight line through the origin with gradient $-GM$, which is the neater way to extract $M$.
Worked example. The Moon is an isolated uniform sphere of mass $7.3 \times 10^{22}\ \text{kg}$ and radius $1.7 \times 10^{6}\ \text{m}$. Calculate the gravitational potential at its surface, and the minimum speed with which a particle must leave the surface to escape.
$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{22}}{1.7 \times 10^{6}} = -2.9 \times 10^{6}\ \text{J kg}^{-1}$. To escape, the particle's kinetic energy per unit mass must equal the depth of the potential well: $\tfrac{1}{2}v^{2} = 2.9 \times 10^{6}$, so $v = \sqrt{2 \times 2.9 \times 10^{6}} = 2.4 \times 10^{3}\ \text{m s}^{-1}$. The minus sign carries the meaning: energy of $2.9\ \text{MJ}$ per kilogram must be supplied to lift the particle out.
If a test mass $m$ sits where the potential is $\phi$, the gravitational potential energy 重力势能 of the pair is
Like the potential, $E_{\text{P}}$ is negative and reaches zero only at infinite separation. Closer masses have more negative potential energy (more tightly bound).
For small height changes near the surface, $r$ barely changes, so $\Delta E_{\text{P}} \approx mg\Delta h$. For large changes (a satellite moving to a higher orbit) use $-GMm/r$ at each radius and take the difference:
which is positive (energy must be supplied to raise the satellite).
Worked example. A satellite of mass $1200\ \text{kg}$ is moved from a circular orbit of radius $7.0 \times 10^{6}\ \text{m}$ to one of radius $8.0 \times 10^{6}\ \text{m}$ around the Earth ($GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$). Find the changes in its gravitational potential energy, its kinetic energy and its total energy.
$\Delta E_{\text{P}} = GMm\left(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}}\right) = 4.0 \times 10^{14} \times 1200 \times \left(\dfrac{1}{7.0 \times 10^{6}} - \dfrac{1}{8.0 \times 10^{6}}\right) = +8.6 \times 10^{9}\ \text{J}$. In a circular orbit $E_{\text{K}} = \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, so $\Delta E_{\text{K}} = \dfrac{GMm}{2}\left(\dfrac{1}{r_{2}} - \dfrac{1}{r_{1}}\right) = -4.3 \times 10^{9}\ \text{J}$: the higher satellite moves more slowly. The total energy $E_{\text{K}} + E_{\text{P}} = -\dfrac{GMm}{2r}$ rises by $+4.3 \times 10^{9}\ \text{J}$, which is the energy the rocket motor must supply. A satellite's total energy is negative — it is bound — and it becomes less negative as the orbit widens.
To escape from radius $r$ to infinity, an object's kinetic energy 动能 must equal the size of its gravitational potential energy:
At the Earth's surface, the escape velocity 逃逸速度 is $\approx 11\ \text{km s}^{-1}$. It does not depend on the object's mass.
Worked example. Find the escape velocity from the Earth's surface. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$, $R = 6.4 \times 10^{6}\ \text{m}$.)
Worked example. A particle is projected vertically upwards from the surface of an isolated planet of radius $R$, where the gravitational potential is $\phi_{\text{s}}$. Show that it just reaches a distance $r$ from the centre if its launch speed $v$ satisfies $\tfrac{1}{2}v^{2} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, and deduce the escape speed.
Potential varies as $1/r$, so at distance $r$ it is $\phi_{\text{s}}R/r$. When the particle stops, its kinetic energy per unit mass has all become potential energy per unit mass: $\tfrac{1}{2}v^{2} = \phi(r) - \phi_{\text{s}} = \phi_{\text{s}}\dfrac{R}{r} - \phi_{\text{s}} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, which is positive because $\phi_{\text{s}}$ is negative. Letting $r \to \infty$ gives $\tfrac{1}{2}v_{\text{esc}}^{2} = -\phi_{\text{s}} = GM/R$ — the escape-speed result again, now read straight off the potential.
Potensial gravitasi $\phi$ di suatu titik adalah usaha per satuan massa untuk memindahkan massa uji kecil dari tak hingga ke titik tersebut:
Satuan: $\text{J kg}^{-1}$.
Potensial dianggap nol di tak hingga. Saat massa uji jatuh mendekati sumber, gravitasi melakukan usaha untuk Anda, sehingga $\phi$ bernilai negatif di mana pun kecuali di tak hingga. Untuk massa titik $M$ pada jarak $r$:
$\phi$ adalah skalar. Untuk beberapa massa, jumlahkan potensial-potensialnya.
Definisi bernilai dua poin, dan mengapa nilainya negatif. Potensial gravitasi di suatu titik adalah usaha per satuan massa untuk memindahkan massa uji kecil dari tak hingga ke titik tersebut. Potensial didefinisikan nol di tak hingga; gravitasi bersifat tarik-menarik, sehingga saat massa uji datang dari tak hingga, medan melakukan usaha padanya dan massa melepaskan energi — usaha yang harus dilakukan oleh agen eksternal adalah negatif. Oleh karena itu, potensial pada setiap jarak berhingga berada di bawah nol. (Potensial listrik di dekat muatan positif bernilai positif karena alasan sebaliknya: medan tersebut menolak, sehingga usaha harus dilakukan untuk mendekatkan muatan uji positif.)
Kesamaan dan perbedaan dengan potensial listrik. Keduanya didefinisikan sebagai usaha per satuan massa atau per satuan muatan positif dari tak hingga, keduanya adalah skalar, keduanya berubah sebanding dengan $1/r$ dari sumber titik, dan keduanya bernilai nol di tak hingga. Perbedaannya: potensial gravitasi selalu negatif, karena gravitasi hanya menarik, sedangkan potensial listrik bisa positif atau negatif tergantung tanda muatannya.
Membaca grafik potensial. Dua grafik muncul dalam soal. Pada $\phi$ terhadap $r$, kurva berbentuk $-GM/r$: nilai permukaan $\phi_{\text{s}} = -GM/R$ memberikan massa planet, dan gradien pada titik manapun memberikan kekuatan medan di sana, karena $g = -\dfrac{\Delta\phi}{\Delta r}$ — kekuatan medan adalah negatif dari gradien potensial (ini adalah "hubungan antara potensial dan kekuatan medan" yang mungkin ditanyakan oleh soal). Pada $\phi$ terhadap $1/r$, grafiknya adalah garis lurus melalui titik asal dengan gradien $-GM$, yang merupakan cara lebih rapi untuk mengekstrak $M$.
Contoh terpecahkan. Bulan adalah bola seragam terisolasi bermassa $7.3 \times 10^{22}\ \text{kg}$ dan jari-jari $1.7 \times 10^{6}\ \text{m}$. Hitung potensial gravitasi di permukaannya, dan kecepatan minimum dengan mana partikel harus meninggalkan permukaan untuk lepas.
$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.3 \times 10^{22}}{1.7 \times 10^{6}} = -2.9 \times 10^{6}\ \text{J kg}^{-1}$. Untuk keluar, energi kinetik partikel per satuan massa harus sama dengan kedalaman sumur potensial: $\tfrac{1}{2}v^{2} = 2.9 \times 10^{6}$, sehingga $v = \sqrt{2 \times 2.9 \times 10^{6}} = 2.4 \times 10^{3}\ \text{m s}^{-1}$. Tanda minus membawa makna: energi $2.9\ \text{MJ}$ per kilogram harus disuplai untuk mengangkat partikel keluar.
Jika sebuah massa uji $m$ berada di tempat yang memiliki potensial $\phi$, maka energi potensial gravitasi dari pasangan tersebut adalah
Seperti potensialnya, $E_{\text{P}}$ bernilai negatif dan hanya mencapai nol pada pemisahan tak hingga. Massa yang lebih dekat memiliki energi potensial yang lebih negatif (terikat lebih kuat).
Untuk perubahan ketinggian kecil di dekat permukaan, $r$ hampir tidak berubah, sehingga $\Delta E_{\text{P}} \approx mg\Delta h$. Untuk perubahan besar (satelit berpindah ke orbit lebih tinggi) gunakan $-GMm/r$ pada setiap jari-jari dan ambil selisihnya:
yang merupakan nilai positif (energi harus disuplai untuk menaikkan satelit).
Contoh terpecahkan. Sebuah satelit bermassa $1200\ \text{kg}$ dipindahkan dari orbit melingkar berjari-jari $7.0 \times 10^{6}\ \text{m}$ ke orbit berjari-jari $8.0 \times 10^{6}\ \text{m}$ mengelilingi Bumi ($GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$). Tentukan perubahan energi potensial gravitasinya, energi kinetiknya, dan energinya total.
$\Delta E_{\text{P}} = GMm\left(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}}\right) = 4.0 \times 10^{14} \times 1200 \times \left(\dfrac{1}{7.0 \times 10^{6}} - \dfrac{1}{8.0 \times 10^{6}}\right) = +8.6 \times 10^{9}\ \text{J}$. Dalam orbit melingkar $E_{\text{K}} = \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$, sehingga $\Delta E_{\text{K}} = \dfrac{GMm}{2}\left(\dfrac{1}{r_{2}} - \dfrac{1}{r_{1}}\right) = -4.3 \times 10^{9}\ \text{J}$: satelit yang lebih tinggi bergerak lebih lambat. Energi total $E_{\text{K}} + E_{\text{P}} = -\dfrac{GMm}{2r}$ meningkat sebesar $+4.3 \times 10^{9}\ \text{J}$, yang merupakan energi yang harus disuplai oleh mesin roket. Energi total satelit bernilai negatif — ia terikat — dan menjadi kurang negatif seiring orbit yang semakin melebar.
Untuk lepas dari jari-jari $r$ menuju tak hingga, energi kinetik suatu benda harus sama dengan besaran energi potensial gravitasinya:
Di permukaan Bumi, kecepatan lepas adalah $\approx 11\ \text{km s}^{-1}$. Nilai ini tidak bergantung pada massa benda.
Contoh terpecahkan. Tentukan kecepatan lepas dari permukaan Bumi. (Untuk Bumi, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$, $R = 6.4 \times 10^{6}\ \text{m}$.)
Contoh terpecahkan. Sebuah partikel diluncurkan vertikal ke atas dari permukaan planet terisolasi berjari-jari $R$, di mana potensial gravitasinya adalah $\phi_{\text{s}}$. Tunjukkan bahwa partikel itu tepat sampai pada jarak $r$ dari pusat jika kecepatan peluncurannya $v$ memenuhi $\tfrac{1}{2}v^{2} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, dan turunkan kecepatan lepasnya.
Potensial bervariasi sebanding dengan $1/r$, sehingga pada jarak $r$ nilainya adalah $\phi_{\text{s}}R/r$. Ketika partikel berhenti, energi kinetik per satuan massanya seluruhnya telah menjadi energi potensial per satuan massa: $\tfrac{1}{2}v^{2} = \phi(r) - \phi_{\text{s}} = \phi_{\text{s}}\dfrac{R}{r} - \phi_{\text{s}} = -\phi_{\text{s}}\left(1 - \dfrac{R}{r}\right)$, yang bernilai positif karena $\phi_{\text{s}}$ bernilai negatif. Memisalkan $r \to \infty$ menghasilkan $\tfrac{1}{2}v_{\text{esc}}^{2} = -\phi_{\text{s}} = GM/R$ — hasil kecepatan lepas lagi, kali ini langsung dibaca dari grafik potensial.
V = −GM / r
Potential ∝ −1/r — deep near the mass, flattening with distance. · Potensial ∝ −1/r — mendalam dekat massa, mendatar seiring jarak.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| gravitational field | a region of space in which a mass experiences a force |
| gravitational field strength | the gravitational force per unit mass acting on a small test mass placed at the point |
| field line (direction) | the direction of the force on a small test mass placed at that point |
| Newton's law of gravitation | the gravitational force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation |
| gravitational potential | the work done per unit mass in bringing a small test mass from infinity to the point |
| gravitational potential energy (two point masses) | the work done in bringing the two masses from infinity to their separation, $E_{\text{P}} = -GMm/r$ |
| geostationary orbit | an orbit with a period of 24 hours, from west to east, directly above the equator, so the satellite stays above the same point on the surface |
| escape speed | the minimum speed at which an object must leave the surface to reach infinity with zero kinetic energy |
| centre of mass (binary star) | the point about which both stars orbit, dividing their separation in the inverse ratio of their masses |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| medan gravitasi | daerah ruang di mana massa mengalami gaya |
| kekuatan medan gravitasi | gaya gravitasi per satuan massa yang bekerja pada massa uji kecil yang ditempatkan di titik tersebut |
| garis medan (arah) | arah gaya pada massa uji kecil yang ditempatkan di titik tersebut |
| hukum gravitasi Newton | gaya gravitasi antara dua massa titik berbanding lurus dengan hasil kali massa mereka dan berbanding terbalik dengan kuadrat jarak pisahnya |
| potensial gravitasi | usaha per satuan massa dalam memindahkan massa uji kecil dari tak hingga ke titik tersebut |
| energi potensial gravitasi (dua massa titik) | usaha yang dilakukan untuk membawa kedua massa dari tak hingga ke jarak pisahnya, $E_{\text{P}} = -GMm/r$ |
| orbit geostasioner | orbit dengan periode 24 jam, dari barat ke timur, tepat di atas ekuator, sehingga satelit tetap berada di atas titik yang sama di permukaan |
| kecepatan lepas | kecepatan minimum yang harus dimiliki benda untuk meninggalkan permukaan dan mencapai tak hingga dengan energi kinetik nol |
| pusat massa (bintang ganda) | titik di sekitar kedua bintang mengorbit, membagi jarak pisah mereka dalam perbandingan terbalik dari massa masing-masing |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Heat 热量 (thermal energy) flows from a higher temperature to a lower temperature. When two bodies touch, energy moves until their temperatures are equal — they reach thermal equilibrium 热平衡. At equilibrium there is no net flow of energy.
Two regions at the same temperature 温度 are in thermal equilibrium with each other — no net energy flows, even though particles still exchange energy.
Temperature decides the direction of heat flow. It is not a measure of how much thermal energy 热能 a body holds. A small cup of boiling water (100 °C) holds far less energy than a swimming pool at 25 °C, but a piece of metal put in the cup gains energy while one put in the pool loses it.
The examiner's wording. Two objects are in thermal equilibrium when there is no net transfer of thermal energy between them; that happens when they are at the same temperature. Asked for "the reason why two objects at the same temperature are in thermal equilibrium", the one mark is: there is no net flow of thermal energy between them — energy still passes both ways, but equally. Thermal energy is the energy transferred because of a temperature difference; temperature is what decides the direction.
Worked example. Two metal cuboids P and Q are in thermal contact and in thermal equilibrium. State what this means, and describe what happens when a hotter cuboid is placed against them.
No net thermal energy passes between P and Q, because they are at the same temperature. A hotter cuboid transfers thermal energy to them (from higher to lower temperature) until all three reach one common temperature; then, and only then, is the whole group in thermal equilibrium.
Watch energy transfer until two objects reach the same temperature. · Perhatikan perpindahan energi hingga dua benda mencapai suhu yang sama.
| English | Bahasa Indonesia |
|---|---|
| thermal equilibrium/ˈθɜːml ˌiːkwɪˈlɪbrɪəm/ | kesetimbangan termal |
| heat/hiːt/ | panas |
| thermal energy/ˈθɜːml ˈenədʒi/ | energi termal |
| temperature/ˈtemprɪtʃə/ | suhu |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Any physical property that changes in a repeatable way with temperature can make a thermometer 温度计. Examples:
Different thermometers can read slightly differently if the property does not change in a straight line; they agree only at the calibration 校准 points.
Fixed points and calibration. A thermometer is calibrated at two fixed points 固定点 that are easy to reproduce — the ice point ($0\ ^{\circ}\text{C}$, pure melting ice) and the steam point ($100\ ^{\circ}\text{C}$, steam above boiling water at standard pressure). The value of the property is measured at each, and the scale between them is drawn assuming the property changes linearly with temperature. That assumption is the weakness: mercury's expansion, a metal's resistance and a thermocouple's e.m.f. all vary slightly differently between the fixed points, so two thermometers that agree at $0$ and $100\ ^{\circ}\text{C}$ can disagree at $50\ ^{\circ}\text{C}$.
Why a liquid-in-glass thermometer does not measure thermodynamic temperature (the one-mark reason): its reading depends on the property of a particular substance — the expansion of mercury or alcohol — and on the assumption that this expansion is linear between the fixed points. The thermodynamic scale depends on no substance at all.
Why water is a poor thermometric liquid. Its density does not change steadily with temperature: it is greatest at $4\ ^{\circ}\text{C}$, so between $0$ and $8\ ^{\circ}\text{C}$ two different temperatures give the same density and the reading is ambiguous, and the change per degree is small, so the thermometer is insensitive. Mercury's density falls steadily and almost linearly over the whole range, which is why it was chosen.
Worked example. A platinum resistance thermometer is a coil of platinum wire in a glass tube, connected to a circuit that measures its resistance. Explain how it measures temperature, and give one disadvantage compared with a thermocouple.
The resistance of the platinum increases with temperature in a known, almost linear way; the resistance is measured at the ice point and the steam point, and any other temperature is read from where the measured resistance falls on the linear scale between them (or from a calibration graph). Disadvantage: the coil, tube and the fluid around them have a large thermal capacity, so the thermometer responds slowly and cannot follow a rapidly changing temperature; it is also bulky and needs a circuit to read it. (Its advantages: accurate, stable and usable over a very wide range.)
Worked example. In a constant-volume gas thermometer the pressure of the gas is $1.05 \times 10^{5}\ \text{Pa}$ when the bulb is in melting ice and $1.19 \times 10^{5}\ \text{Pa}$ when it is in a warm room. Find the thermodynamic temperature of the room, and explain why this thermometer, unlike a liquid-in-glass one, gives thermodynamic temperature directly.
For a fixed mass of gas at constant volume the pressure is proportional to the thermodynamic temperature, so $\dfrac{T}{273.15} = \dfrac{1.19 \times 10^{5}}{1.05 \times 10^{5}}$, giving $T = 310\ \text{K}$ ($36\ ^{\circ}\text{C}$). The pressure of a (nearly) ideal gas depends only on its temperature, not on which gas it is, so the reading does not rely on the property of a particular substance — the defining feature of the thermodynamic scale.
Kelvin index = Celsius index + 2.73 · Indeks Kelvin = Indeks Celcius + 2.73
Slide Celsius temperature and see the Kelvin scale shift by 273. · Geserkan suhu Celcius dan lihat skala Kelvin bergeser sebesar 273.
| English | Bahasa Indonesia |
|---|---|
| thermometer/θɜːˈmɒmɪtə/ | termometer |
| capillary/kəˈpɪləri/ | kapiler |
| resistance/rɪˈzɪstəns/ | penolakan |
| thermocouple/ˈθɜːməkʌpl/ | termokopel |
| electromotive force/ɪˌlektrəʊˈməʊtɪv fɔːs/ | gaya gerak listrik |
| calibration/ˌkælɪˈbreɪʃn/ | kalibrasi |
| fixed points/fɪkst pɔɪnts/ | titik tetap |
The thermodynamic temperature 热力学温度 (or absolute temperature 绝对温度) scale does not depend on any one substance — only on the laws of thermodynamics. Its unit is the kelvin 开尔文 (K).
The lowest possible temperature is zero kelvin ($0\ \text{K}$), called absolute zero 绝对零度. There a system has its least possible internal energy 内能 — particles have no random motion to speak of. Nothing can be cooled below this.
As the exam asks it. "State the magnitude and unit of absolute zero on the thermodynamic scale": $0\ \text{K}$ (the unit is the kelvin). "State the temperature of absolute zero on the Celsius scale": $-273.15\ ^{\circ}\text{C}$ (accept $-273\ ^{\circ}\text{C}$). "What is absolute zero?": the temperature at which a system has its minimum internal energy — the particles have the least kinetic and potential energy they can have — and below which it is impossible to go. The thermodynamic scale is fixed by absolute zero and by the triple point 三相点 of water, defined as $273.16\ \text{K}$; it does not depend on the property of any particular substance.
The Celsius 摄氏度 scale $\theta$ is shifted from the thermodynamic scale by a fixed amount:
So $0\ ^{\circ}\text{C} = 273.15\ \text{K}$ and $100\ ^{\circ}\text{C} = 373.15\ \text{K}$. A kelvin and a degree Celsius are the same size, so a temperature difference of $1\ \text{K}$ equals $1\ ^{\circ}\text{C}$ — but the absolute values differ by $273.15$.
In gas-law calculations you must always use absolute temperatures in kelvin. Using °C gives wrong answers.
| English | Bahasa Indonesia |
|---|---|
| absolute temperature/ˈæbsəluːt ˈtemprɪtʃə/ | suhu absolut |
| thermodynamic temperature/ˌθɜːməʊdaɪˈnæmɪk ˈtemprɪtʃə/ | suhu termodinamika |
| kelvin/ˈkelvɪn/ | kelvin |
| absolute zero/ˈæbsəluːt ˈzɪərəʊ/ | nol mutlak |
| internal energy/ɪnˈtɜːnl ˈenədʒi/ | energi dalam |
| Celsius/ˈselsɪəs/ | Celsius |
| triple point/ˈtrɪpl pɔɪnt/ | titik tiga |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The specific heat capacity 比热容 $c$ of a substance is the energy 能量 needed to raise the temperature of unit mass by one kelvin:
Unit: $\text{J kg}^{-1}\ \text{K}^{-1}$.
The two-mark definition: specific heat capacity is the energy required per unit mass of the substance to raise its temperature by one kelvin (or by one degree). Both "per unit mass" and "per unit temperature rise" are needed; "the energy to heat 1 kg by 1 K" scores, "the energy to heat the substance" does not.
Examples:
To find an unknown $c$ by experiment: supply known energy $Q$ electrically ($Q = VIt$, from the power 功率), then measure the temperature rise $\Delta T$ of a known mass 质量 $m$. Then $c = Q/(m\Delta T)$. Reduce heat loss with insulation 隔热 and use a rise of about 10 K (big enough to measure well, small enough to limit losses).
Worked example. How much energy is needed to heat $0.50\ \text{kg}$ of water from $20\ ^{\circ}\text{C}$ to $100\ ^{\circ}\text{C}$? (Specific heat capacity of water $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$.)
A temperature difference is the same in K and °C, so $\Delta T = 80$:
When two bodies reach thermal equilibrium with no heat lost to the surroundings, the energy gained by the colder one equals the energy lost by the hotter one:
Worked example. $0.20\ \text{kg}$ of water at $80\ ^{\circ}\text{C}$ is mixed with $0.30\ \text{kg}$ of water at $20\ ^{\circ}\text{C}$, with no heat lost. Find the final temperature.
The heat lost by the hot water equals the heat gained by the cold water (the $c$ of water cancels):
Worked example. A beaker of mass $42\ \text{g}$ and specific heat capacity $840\ \text{J kg}^{-1}\ \text{K}^{-1}$ contains $120\ \text{g}$ of a liquid. A heater supplies energy at $810\ \text{W}$ and the temperature of the beaker and liquid rises from $19\ ^{\circ}\text{C}$ to $47\ ^{\circ}\text{C}$ in $15\ \text{s}$. Find the specific heat capacity of the liquid. The experiment is then repeated with water in place of the liquid; state and explain how the temperature rise differs.
Energy supplied $Q = Pt = 810 \times 15 = 1.22 \times 10^{4}\ \text{J}$, $\Delta T = 28\ \text{K}$. It heats both the beaker and the liquid: $Q = m_{\text{b}}c_{\text{b}}\Delta T + m_{\text{l}}c_{\text{l}}\Delta T$, so $1.215 \times 10^{4} = 0.042 \times 840 \times 28 + 0.120 \times c_{\text{l}} \times 28$, i.e. $1.215 \times 10^{4} = 988 + 3.36c_{\text{l}}$, giving $c_{\text{l}} = 3.3 \times 10^{3}\ \text{J kg}^{-1}\ \text{K}^{-1}$. Forgetting the beaker is the usual lost mark. With water ($c = 4200$, higher) the same energy in the same time produces a smaller temperature rise, since $\Delta T = Q/(mc)$ and $mc$ is larger. If energy is lost to the surroundings the true $Q$ absorbed is less than $Pt$, so a value of $c$ found this way is an overestimate.
Worked example. Two metal blocks X and Y are placed in contact and insulated from the surroundings. X (mass $0.50\ \text{kg}$, initially $80\ ^{\circ}\text{C}$, $c = 390\ \text{J kg}^{-1}\ \text{K}^{-1}$) and Y (mass $0.50\ \text{kg}$, initially $20\ ^{\circ}\text{C}$, $c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$) reach a common final temperature. Explain, in terms of energy, why the final temperature is nearer to Y's starting temperature, and find it.
Thermal energy flows from X (hotter) to Y until they are in thermal equilibrium; the energy lost by X equals the energy gained by Y. Because Y has the larger specific heat capacity, a given amount of energy changes its temperature less than it changes X's, so Y warms by fewer degrees than X cools and the final temperature lies nearer to $20\ ^{\circ}\text{C}$. Numerically: $0.50 \times 390 \times (80 - T) = 0.50 \times 900 \times (T - 20)$, so $390(80 - T) = 900(T - 20)$, $31200 + 18000 = 1290T$, $T = 38\ ^{\circ}\text{C}$.
Worked example. An aluminium block has volume $3.612 \times 10^{-3}\ \text{m}^{3}$ and density $2.70 \times 10^{3}\ \text{kg m}^{-3}$. Find the energy needed to raise its temperature by $40\ \text{K}$ ($c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$).
Mass $= \rho V = 2.70 \times 10^{3} \times 3.612 \times 10^{-3} = 9.75\ \text{kg}$, so $Q = mc\Delta T = 9.75 \times 900 \times 40 = 3.5 \times 10^{5}\ \text{J}$. A mass hidden behind a density and a volume is a common first step.
Pick a material, set the mass and the temperature rise, and read the energy. Water needs far more energy than the metals. · Pilih material, tentukan massa dan kenaikan suhu, lalu baca energinya. Air membutuhkan jauh lebih banyak energi daripada logam.
Q = mcΔT
The heat needed is proportional to the temperature rise — the gradient depends on mass and the material's specific heat capacity. · Pemanasan yang dibutuhkan berbanding lurus dengan kenaikan suhu — gradien tergantung pada massa dan kapasitas panas spesifik material.
| English | Bahasa Indonesia |
|---|---|
| energy/ˈenədʒi/ | tenaga |
| mass/mæs/ | massa |
| specific heat capacity/spəˈsɪfɪk hiːt kəˈpæsɪti/ | kapasitas panas spesifik |
| power/ˈpaʊə/ | kuasa |
| insulation/ˌɪnsjuːˈleɪʃn/ | isolasi |
When a substance changes state (solid ↔ liquid, or liquid ↔ gas) at constant temperature, energy must be supplied (or removed) with no temperature change. This energy is the latent heat 潜热.
The specific latent heat 比潜热 $L$ is the energy to change the state of unit mass at constant temperature:
Unit: $\text{J kg}^{-1}$.
The two-mark definition: specific latent heat is the energy required per unit mass to change the state of a substance without a change of temperature. "Without a change in temperature" (or "at constant temperature") is the second mark and the one most often missed. Add "from solid to liquid" for fusion or "from liquid to gas" for vaporisation when a particular one is asked for.
Two kinds:
For water at atmospheric pressure: $L_{\text{f}} \approx 3.34 \times 10^{5}\ \text{J kg}^{-1}$ (at $0\ ^{\circ}\text{C}$); $L_{\text{v}} \approx 2.26 \times 10^{6}\ \text{J kg}^{-1}$ (at $100\ ^{\circ}\text{C}$). So $L_{\text{v}}$ is about 7 times $L_{\text{f}}$.
Worked example. A $2.0\ \text{kW}$ heater boils water already at $100\ ^{\circ}\text{C}$. How long does it take to turn $0.10\ \text{kg}$ of this water into steam? ($L_{\text{v}} = 2.26 \times 10^{6}\ \text{J kg}^{-1}$, no heat lost.)
The energy needed is $Q = mL_{\text{v}} = 0.10 \times 2.26 \times 10^{6} = 2.26 \times 10^{5}\ \text{J}$. From $Q = Pt$,
Worked example. Water in a kettle stays at $100\ ^{\circ}\text{C}$ while it boils, even though the element keeps heating it. Explain this with reference to molecular energies (3 marks).
Temperature is a measure of the mean kinetic energy of the molecules. During boiling the energy supplied is used to separate the molecules — to do work against the attractive forces between them and against the atmosphere as the vapour expands — so it increases the potential energy of the molecules, not their kinetic energy. With the mean kinetic energy unchanged, the temperature stays constant until all the water has become steam.
Two reasons, both from the particle picture of matter:
Write both reasons and name the energies: the marks are for bonds broken (potential energy increased) — some in melting, all in boiling, and work done against the atmosphere in the large expansion. "Boiling needs more energy" restates the question.
Worked example. A dish holds $7.2 \times 10^{-5}\ \text{m}^{3}$ of a liquid of density $710\ \text{kg m}^{-3}$ and specific latent heat of vaporisation $3.6 \times 10^{5}\ \text{J kg}^{-1}$. It evaporates completely. Find the energy absorbed, and suggest, with a reason, whether the substance's specific latent heat of fusion is likely to be smaller or larger than this.
Mass $= \rho V = 710 \times 7.2 \times 10^{-5} = 5.1 \times 10^{-2}\ \text{kg}$, so the energy absorbed is $Q = mL_{\text{v}} = 5.1 \times 10^{-2} \times 3.6 \times 10^{5} = 1.8 \times 10^{4}\ \text{J}$. Its $L_{\text{f}}$ is likely to be smaller: melting breaks only some of the intermolecular bonds and the volume barely changes, whereas vaporising breaks them all and does work against the atmosphere — for water $L_{\text{f}}$ is about a seventh of $L_{\text{v}}$.
If a problem mixes temperature change and a phase change 相变 (e.g. ice at $-5\ ^{\circ}\text{C}$ warming to water at $30\ ^{\circ}\text{C}$):
Total: $Q_{1} + Q_{2} + Q_{3}$. A phase change is at constant temperature, so use $mL$ there, not $mc\Delta T$.
When a question gives heater power $P$ and asks for the time, use $Q = Pt$ (assuming no heat loss). Insulating (lagging) the container and using a small mass are common ways to improve the experiment.
Worked example. An ice cube of mass $37.0\ \text{g}$ at $0.0\ ^{\circ}\text{C}$ is dropped into $208\ \text{g}$ of water at $26.4\ ^{\circ}\text{C}$ in an insulated beaker. Find the final temperature when all the ice has melted. ($c_{\text{water}} = 4.20\ \text{kJ kg}^{-1}\ \text{K}^{-1}$, $L_{\text{f}} = 334\ \text{kJ kg}^{-1}$.)
The ice gains energy twice — to melt, then to warm from $0$ to $T$ — and the water loses energy cooling from $26.4$ to $T$. Working in grams and kilojoules: $37.0 \times 0.334 + 37.0 \times 4.20 \times 10^{-3}\,T = 208 \times 4.20 \times 10^{-3}\,(26.4 - T)$, i.e. $12.4 + 0.155T = 23.1 - 0.874T$, so $1.03T = 10.7$ and $T = 10\ ^{\circ}\text{C}$. Three energy terms, one equation; the commonest error is to forget that the melted ice must also be warmed.
| English | Bahasa Indonesia |
|---|---|
| volume/ˈvɒljuːm/ | volume |
| specific latent heat/spəˈsɪfɪk ˈleɪtənt hiːt/ | kalor laten spesifik |
| latent heat/ˈleɪtənt hiːt/ | kalor laten |
| fusion/ˈfjuːʒn/ | fusi |
| vaporisation/ˌveɪpəraɪˈzeɪʃn/ | penguapan |
| specific latent heat of fusion/spəˈsɪfɪk ˈleɪtənt hiːt ɒv ˈfjuːʒn/ | kalor laten spesifik peleburan |
| specific latent heat of vaporisation/spəˈsɪfɪk ˈleɪtənt hiːt ɒv ˌveɪpəraɪˈzeɪʃn/ | kalor laten spesifik penguapan |
| atmospheric pressure/ˌætməsˈferɪk ˈpreʃə/ | tekanan atmosfer |
| phase change/feɪz tʃeɪndʒ/ | perubahan fase |
| intermolecular/ˌɪntəməˈlekjʊlə/ | antarmolekul |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| thermal equilibrium | the state of two objects between which there is no net transfer of thermal energy; they are at the same temperature |
| thermal energy | energy transferred from one object to another because of a temperature difference |
| thermodynamic temperature scale | a temperature scale that does not depend on the property of any particular substance, fixed by absolute zero and the triple point of water |
| absolute zero | the lowest possible temperature, $0\ \text{K}$ ($-273.15\ ^{\circ}\text{C}$), at which a system has its minimum internal energy |
| kelvin and Celsius | $T/\text{K} = \theta/^{\circ}\text{C} + 273.15$; a temperature difference is the same number in both |
| specific heat capacity | the energy required per unit mass of a substance to raise its temperature by one kelvin |
| specific latent heat | the energy required per unit mass to change the state of a substance without a change of temperature |
| specific latent heat of fusion | the specific latent heat for the change from solid to liquid |
| specific latent heat of vaporisation | the specific latent heat for the change from liquid to gas |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Amount of substance 物质的量 is an SI base quantity. Its unit is the mole 摩尔 (mol) — one of the seven SI base units, with the kilogram, metre, second, ampere and kelvin 开尔文 from Topic 1.
One mole of any substance has a number of particles equal to the Avogadro constant 阿伏伽德罗常量:
A "particle" means whatever you are counting — atoms 原子 for a monatomic 单原子 element like helium, molecules 分子 for $\text{O}_{2}$ or $\text{H}_{2}\text{O}$. Always say what you are counting.
For $n$ moles, the number of particles is $N = n N_{\text{A}}$.
The molar mass 摩尔质量 $M_{\text{m}}$ is the mass of one mole ($\text{kg mol}^{-1}$ or $\text{g mol}^{-1}$). Mass of $n$ moles is $M = n M_{\text{m}}$. Mass of one particle is $m_{0} = M_{\text{m}} / N_{\text{A}}$.
The examiner's wording. The Avogadro constant is the number of atoms (or molecules) in one mole of a substance — one mark; adding "in $0.012\ \text{kg}$ of carbon-12" is accepted but not needed. Asked for "the relationship between $N_{\text{A}}$, $R$ and $k$", write $k = R/N_{\text{A}}$ (or $R = N_{\text{A}}k$).
Worked example. Oxygen has a molar mass of $32\ \text{g mol}^{-1}$. Find the mass of one oxygen molecule, and the number of molecules in $8.0\ \text{g}$ of oxygen.
$m_{0} = \dfrac{M_{\text{m}}}{N_{\text{A}}} = \dfrac{0.032}{6.02 \times 10^{23}} = 5.3 \times 10^{-26}\ \text{kg}$. $8.0\ \text{g}$ is $n = 8.0/32 = 0.25\ \text{mol}$, so $N = nN_{\text{A}} = 1.5 \times 10^{23}$ molecules. Keep molar masses in $\text{kg mol}^{-1}$ when the answer is in kilograms — the factor of $1000$ is the usual slip.
particles = n x Avogadro constant · partikel = n x konstanta Avogadro
Change amount of substance and see particle number scale directly. · Ubah jumlah zat dan lihat nomor partikel berskala langsung.
| English | Bahasa Indonesia |
|---|---|
| mole/məʊl/ | mol |
| amount of substance/əˈmaʊnt ɒv ˈsʌbstəns/ | kuantiti bahan |
| kelvin/ˈkelvɪn/ | kelvin |
| Avogadro constant/ˌævəˈɡædrəʊ ˈkɒnstənt/ | Pemalar Avogadro |
| atom/ˈætəm/ | atom |
| monatomic/ˌmɒnəˈtɒmɪk/ | monatomik |
| molecule/ˈmɒlɪkjuːl/ | molekul |
| molar mass/ˈməʊlə mæs/ | massa molar |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
An ideal gas 理想气体 obeys $pV \propto T$ exactly, where $T$ is the thermodynamic temperature 热力学温度.
The equation of state 状态方程 can be written two equal ways:
Here:
Since $N = n N_{\text{A}}$, we get $k = R/N_{\text{A}}$: $k$ is the gas constant per molecule, as $R$ is per mole.
The two-mark definition. An ideal gas is one that obeys $pV = nRT$ (or $pV \propto T$, with $T$ the thermodynamic temperature) at all values of pressure, volume and temperature. Both marks need the equation (or the proportionality with $T$ named as thermodynamic) and "at all values" or "for all $p$, $V$ and $T$". Asked to "state the meaning of each symbol in $pV = NkT$": $p$ is the pressure, $V$ the volume, $N$ the number of molecules, $k$ the Boltzmann constant and $T$ the thermodynamic temperature — a real gas is nearly ideal at low pressure and high temperature, where its molecules are far apart.
List the variables you have, find the unknown, and choose the form that matches your "amount" (moles → $nRT$; molecules → $NkT$). Always use SI units: Pa, m³, K.
Worked example. A cylinder of volume $0.020\ \text{m}^{3}$ holds gas at $27\ ^{\circ}\text{C}$ and a pressure of $2.0 \times 10^{5}\ \text{Pa}$. How many moles of gas are there? ($R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$.)
Convert to kelvin: $T = 27 + 273 = 300\ \text{K}$. Then from $pV = nRT$,
If a fixed amount of gas changes from state 1 to state 2:
Worked example. A fixed mass of gas at $300\ \text{K}$ occupies $0.50\ \text{m}^{3}$. It is heated to $450\ \text{K}$ at constant pressure. Find the new volume.
At constant pressure $V/T$ is constant, so
Worked example. A sealed vessel of volume $0.0500\ \text{m}^{3}$ contains $0.0424\ \text{kg}$ of an ideal gas at $227\ ^{\circ}\text{C}$ and $1.37 \times 10^{5}\ \text{Pa}$. Find the amount of gas, the mass of one molecule, and the mean-square speed of its molecules.
$T = 227 + 273 = 500\ \text{K}$. $n = \dfrac{pV}{RT} = \dfrac{1.37 \times 10^{5} \times 0.0500}{8.31 \times 500} = 1.65\ \text{mol}$. The molar mass is $0.0424/1.65 = 0.0257\ \text{kg mol}^{-1}$, so one molecule has mass $0.0257/(6.02 \times 10^{23}) = 4.3 \times 10^{-26}\ \text{kg}$. From $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$, $\langle c^{2}\rangle = \dfrac{3kT}{m} = \dfrac{3 \times 1.38 \times 10^{-23} \times 500}{4.27 \times 10^{-26}} = 4.8 \times 10^{5}\ \text{m}^{2}\ \text{s}^{-2}$ (an r.m.s. speed of about $700\ \text{m s}^{-1}$).
Worked example. Cylinder X (volume $0.0260\ \text{m}^{3}$, $0.740\ \text{mol}$) and cylinder Y (volume $0.0180\ \text{m}^{3}$, $0.320\ \text{mol}$) contain ideal gas and are in thermal equilibrium with each other at $290\ \text{K}$. Find the pressure in each, and explain what happens to the number of molecules in each cylinder if a tap joining them is opened.
$p_{\text{X}} = \dfrac{nRT}{V} = \dfrac{0.740 \times 8.31 \times 290}{0.0260} = 6.86 \times 10^{4}\ \text{Pa}$ and $p_{\text{Y}} = \dfrac{0.320 \times 8.31 \times 290}{0.0180} = 4.28 \times 10^{4}\ \text{Pa}$. Gas flows from the higher pressure (X) to the lower (Y) until the pressures are equal; the temperature is unchanged, so the final pressure is $p = \dfrac{(n_{\text{X}} + n_{\text{Y}})RT}{V_{\text{X}} + V_{\text{Y}}} = 5.8 \times 10^{4}\ \text{Pa}$, and X ends with $n_{\text{X}} = pV_{\text{X}}/RT = 0.63\ \text{mol}$ — it loses about $0.11\ \text{mol}$ to Y.
Worked example. A fixed amount of ideal gas at temperature $T$ is in state X, with pressure $2p$ and volume $V$. It is cooled at constant volume to state Y, where its pressure is $p$; then heated at constant pressure to state Z, where its volume is $2V$; then returned to X. Find the temperatures at Y and Z, and describe how the internal energy changes round the cycle.
X to Y is at constant volume, so $p/T$ is constant: halving the pressure halves the temperature, $T_{\text{Y}} = T/2$. Y to Z is at constant pressure, so $V/T$ is constant: doubling the volume doubles the temperature, $T_{\text{Z}} = T$. The internal energy of an ideal gas depends only on temperature, so it falls from X to Y (by $\tfrac{3}{2}nR \cdot T/2$), rises by the same amount from Y to Z, and is unchanged from Z back to X. Read each leg off the diagram before writing an equation: vertical means constant $V$, horizontal means constant $p$.
Special cases:
A common mistake is using °C instead of K — $pV \propto T$ only holds with $T$ in kelvin.
p = k / V
At constant temperature p ∝ 1/V — squeeze the volume and pressure rises. · Pada suhu tetap p ∝ 1/V — tekan volume dan tekanan naik.
| English | Bahasa Indonesia |
|---|---|
| equation of state/ɪˈkweɪʒn ɒv steɪt/ | persamaan keadaan |
| ideal gas/aɪˈdɪəl ɡæs/ | gas ideal |
| pressure/ˈpreʃə/ | tekanan |
| thermodynamic temperature/ˌθɜːməʊdaɪˈnæmɪk ˈtemprɪtʃə/ | suhu termodinamika |
| volume/ˈvɒljuːm/ | volume |
| molar gas constant/ˈməʊlə ɡæs ˈkɒnstənt/ | konstanta gas molar |
| Boltzmann constant/ˈbɒltsmən ˈkɒnstənt/ | konstanta Boltzmann |
| Boyle's law/bɔɪlz lɔː/ | Hukum Boyle |
| Charles's law/ˈtʃɑːlzɪz lɔː/ | Hukum Charles |
| pressure law/ˈpreʃə lɔː/ | hukum tekanan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The kinetic theory 分子动理论 explains a gas's large-scale behaviour from the random motion 无规则运动 of its molecules.
For an ideal gas:
These assumptions become poor at very high pressure (molecular volume matters) or very low temperature (intermolecular forces matter).
As the exam asks it. "State two (or three) basic assumptions of the kinetic theory": choose from the molecules are in continuous random motion; the volume of the molecules is negligible compared with the volume of the gas; there are no forces between the molecules except during collisions; the collisions are perfectly elastic; the time of a collision is negligible compared with the time between collisions. "State what is meant by an elastic collision": one in which the total kinetic energy is conserved (as well as momentum). "Use the assumptions to suggest why the gas at the surface of a star, at very high pressure, is not ideal": the molecules are so close together that their own volume is not negligible compared with the gas volume, and intermolecular forces act between them — the two assumptions that fail when a gas is compressed.
Worked example. A balloon contains $0.40\ \text{mol}$ of hydrogen ($m = 3.34 \times 10^{-27}\ \text{kg}$ per molecule) in $9.8 \times 10^{-3}\ \text{m}^{3}$. Estimate the average separation of the molecules, the gravitational force between neighbouring molecules, and comment on the kinetic-theory assumption this tests.
$N = 0.40 \times 6.02 \times 10^{23} = 2.4 \times 10^{23}$ molecules, so each occupies $V/N = 4.1 \times 10^{-26}\ \text{m}^{3}$ and the average separation is $d = (V/N)^{1/3} = 3.4 \times 10^{-9}\ \text{m}$. The gravitational force between two molecules is $F = \dfrac{Gm^{2}}{d^{2}} = \dfrac{6.67 \times 10^{-11} \times (3.34 \times 10^{-27})^{2}}{(3.4 \times 10^{-9})^{2}} = 6 \times 10^{-47}\ \text{N}$ — some $10^{20}$ times smaller than a molecule's own weight ($mg = 3.3 \times 10^{-26}\ \text{N}$). Forces between molecules really are negligible except during collisions, as the theory assumes.
Take a cubic box of side $L$ with $N$ molecules, each of mass $m$. Look at one molecule moving along the $x$-axis with velocity $u_{1}$.
In 3-D, by symmetry $\langle u_{x}^{2} \rangle = \tfrac{1}{3} \langle c^{2} \rangle$, where $\langle c^{2} \rangle$ is the mean-square speed 均方速率. So
"Explain how molecular movement causes the pressure exerted by a gas" (3 marks). The molecules move randomly and collide with the walls of the container; at each collision a molecule's momentum changes (it rebounds), so by Newton's second law the wall exerts a force on it, and by the third law it exerts an equal force on the wall; the very many collisions each second produce a steady total force on the wall, and the pressure is that force per unit area. The syllabus phrase "a simple model considering one-dimensional collisions" means the derivation above: it is sufficient to follow one molecule bouncing between two opposite faces and then average.
The density form. Since $Nm$ is the total mass of the gas and $Nm/V$ is its density $\rho$, the same result reads $p = \tfrac{1}{3}\rho\langle c^{2}\rangle$ — the version to use when a question gives a density instead of $N$ and $m$.
Worked example. An ideal gas at a pressure of $1.6 \times 10^{5}\ \text{Pa}$ has a density of $1.9\ \text{kg m}^{-3}$. Show that the r.m.s. speed of its molecules is about $500\ \text{m s}^{-1}$.
$\langle c^{2}\rangle = \dfrac{3p}{\rho} = \dfrac{3 \times 1.6 \times 10^{5}}{1.9} = 2.53 \times 10^{5}\ \text{m}^{2}\ \text{s}^{-2}$, so $c_{\text{r.m.s.}} = \sqrt{2.53 \times 10^{5}} = 503\ \text{m s}^{-1} \approx 500\ \text{m s}^{-1}$. In a "show that", keep an extra figure ($503$) before comparing with the value given.
The square root of $\langle c^{2} \rangle$ is the root-mean-square 均方根 (r.m.s.) speed:
It is a useful single measure of how fast the molecules move, slightly larger than the mean speed (squaring weights fast molecules more).
p ∝ 1/V
At constant temperature, pressure is inversely proportional to volume — squash the gas and the pressure rises. · Pada suhu tetap, tekanan berbanding terbalik dengan volume — tekan gas dan tekanannya akan naik.
| English | Bahasa Indonesia |
|---|---|
| mean-square speed/miːn skweə spiːd/ | kecepatan rata-rata kuadrat |
| kinetic theory/kɪˈnetɪk ˈθɪəri/ | teori kinetik |
| random motion/ˈrændəm ˈməʊʃn/ | gerak acak |
| intermolecular/ˌɪntəməˈlekjʊlə/ | antarmolekul |
| elastic collision/ɪˈlæstɪk kəˈlɪʒn/ | tabrakan elastis |
| momentum/məʊˈmentəm/ | momentum |
| impulse/ˈɪmpʌls/ | impuls |
| mean square/miːn skweə/ | rata-rata kuadrat |
| root-mean-square/ruːt miːn skweə/ | akar rata-rata kuadrat |
Compare the two expressions for $pV$:
Set them equal, cancel $N$, and multiply by $\tfrac{3}{2}$:
The right side is the average translational kinetic energy 平动动能 $\langle E_{\text{k}} \rangle$ of one molecule. So
This is a key result: the average translational kinetic energy of an ideal-gas molecule depends only on the thermodynamic temperature, not on the type of gas or its pressure.
Worked example. Find the root-mean-square speed of oxygen molecules at $300\ \text{K}$. (Mass of one $\text{O}_{2}$ molecule $= 5.3 \times 10^{-26}\ \text{kg}$, $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$.)
From $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$, the mean-square speed is $\langle c^{2}\rangle = 3kT/m$:
Worked example. The surface of the Moon reaches about $400\ \text{K}$ in sunlight. Calculate the r.m.s. speed of hydrogen molecules ($m = 3.34 \times 10^{-27}\ \text{kg}$) at this temperature, and use the Moon's escape speed of $2.4\ \text{km s}^{-1}$ to suggest why the Moon has no hydrogen atmosphere.
$c_{\text{r.m.s.}} = \sqrt{\dfrac{3kT}{m}} = \sqrt{\dfrac{3 \times 1.38 \times 10^{-23} \times 400}{3.34 \times 10^{-27}}} = 2.2 \times 10^{3}\ \text{m s}^{-1}$. This r.m.s. speed is close to the escape speed, and the speeds are spread widely about it, so a large fraction of the molecules move faster than $2.4\ \text{km s}^{-1}$ at any moment and escape; over time the hydrogen is lost.
Worked example. The gas at the surface of a star is mainly hydrogen atoms ($m = 1.67 \times 10^{-27}\ \text{kg}$) with an r.m.s. speed of $9300\ \text{m s}^{-1}$. Find the temperature of the surface.
$T = \dfrac{m\langle c^{2}\rangle}{3k} = \dfrac{1.67 \times 10^{-27} \times 9300^{2}}{3 \times 1.38 \times 10^{-23}} = 3.5 \times 10^{3}\ \text{K}$.
Worked example. For a fixed sample of gas, a graph of $pV$ against $kT$ is a straight line through the origin of gradient $7.2 \times 10^{22}$. When $pV = 270\ \text{J}$ the r.m.s. speed of the molecules is $1900\ \text{m s}^{-1}$. Find the number of molecules and the mass of one molecule in u.
From $pV = NkT$ the gradient is $N = 7.2 \times 10^{22}$. From $pV = \tfrac{1}{3}Nm\langle c^{2}\rangle$: $m = \dfrac{3pV}{N\langle c^{2}\rangle} = \dfrac{3 \times 270}{7.2 \times 10^{22} \times 1900^{2}} = 3.1 \times 10^{-27}\ \text{kg} = \dfrac{3.1 \times 10^{-27}}{1.66 \times 10^{-27}} = 1.9\ \text{u}$ — hydrogen molecules, within the rounding.
Comparing gases and samples. At the same temperature every gas has the same average kinetic energy per molecule, so $\langle c^{2}\rangle \propto 1/m$: hydrogen ($M_{\text{m}} = 2$) and oxygen ($M_{\text{m}} = 32$) have $\dfrac{c_{\text{H}}}{c_{\text{O}}} = \sqrt{\dfrac{32}{2}} = 4$. Two samples at the same $T$ — X with $N$ molecules of mass $m$ in volume $V$, Y with $2N$ molecules of mass $2m$ in volume $2V$ — have the same pressure ($p = NkT/V$ and both $N$ and $V$ double), the same average kinetic energy per molecule, but Y's molecules have half the mean-square speed ($3kT/2m$) and Y has twice the internal energy (twice as many molecules).
For an ideal gas the molecules are point particles with no intermolecular potential energy and (in this simple model) no rotation or vibration. So the internal energy 内能 is just the total kinetic energy 动能:
So the internal energy of an ideal gas is proportional to the thermodynamic temperature — doubling $T$ doubles $U$.
As the exam asks it. "Use one of the basic assumptions to explain what can be deduced about the potential energy of the molecules": there are no forces between the molecules (except in collisions), so there is no potential energy associated with their separation — the random-motion energy is entirely kinetic. "Explain why the internal energy of an ideal gas is directly proportional to thermodynamic temperature" (2 marks): the internal energy is the sum of the kinetic and potential energies of the molecules; the potential energy is zero (no intermolecular forces), and the average kinetic energy of a molecule is $\tfrac{3}{2}kT$, proportional to $T$; so the total, $U = \tfrac{3}{2}NkT$, is proportional to $T$. "Derive $\tfrac{3}{2}kT$" (2 marks): equate $pV = NkT$ with $pV = \tfrac{1}{3}Nm\langle c^{2}\rangle$, cancel $N$, and rearrange to $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$ — show the cancelling and the factor of $\tfrac{3}{2}$.
Worked example. A sample of $0.26\ \text{m}^{3}$ of an ideal gas is at $2.0 \times 10^{5}\ \text{Pa}$ and $290\ \text{K}$. Find the number of molecules, the average translational kinetic energy of a molecule, and the internal energy of the gas.
$N = \dfrac{pV}{kT} = \dfrac{2.0 \times 10^{5} \times 0.26}{1.38 \times 10^{-23} \times 290} = 1.3 \times 10^{25}$. $\langle E_{\text{k}}\rangle = \tfrac{3}{2}kT = 1.5 \times 1.38 \times 10^{-23} \times 290 = 6.0 \times 10^{-21}\ \text{J}$. $U = N\langle E_{\text{k}}\rangle = 1.3 \times 10^{25} \times 6.0 \times 10^{-21} = 7.8 \times 10^{4}\ \text{J}$ — or directly $U = \tfrac{3}{2}pV = 7.8 \times 10^{4}\ \text{J}$, a neat check.
Doubling $p$ at fixed $T$ (by squeezing the gas to half its volume) does not change $\langle E_{\text{k}} \rangle$ — that depends only on $T$. There are more wall collisions per second, but each molecule has the same average kinetic energy.
| English | Bahasa Indonesia |
|---|---|
| internal energy/ɪnˈtɜːnl ˈenədʒi/ | energi dalam |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| average translational kinetic energy/ˈævrɪdʒ trænˈsleɪʃənl kɪˈnetɪk ˈenədʒi/ | energi kinetik translasi rata-rata |
| translational kinetic energy/trænˈsleɪʃənl kɪˈnetɪk ˈenədʒi/ | energi kinetik translasi |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| mole | the amount of substance containing a number of particles equal to the Avogadro constant |
| Avogadro constant | the number of atoms or molecules in one mole of a substance, $6.02 \times 10^{23}\ \text{mol}^{-1}$ |
| ideal gas | a gas that obeys $pV = nRT$ (equivalently $pV \propto T$, $T$ thermodynamic) at all values of $p$, $V$ and $T$ |
| Boltzmann constant | the gas constant per molecule, $k = R/N_{\text{A}}$ |
| elastic collision | a collision in which the total kinetic energy is conserved |
| mean-square speed | the mean of the squares of the speeds of the molecules, $\langle c^{2}\rangle$ |
| root-mean-square speed | the square root of the mean-square speed, $c_{\text{r.m.s.}} = \sqrt{\langle c^{2}\rangle}$ |
| average translational kinetic energy of a molecule | $\tfrac{1}{2}m\langle c^{2}\rangle = \tfrac{3}{2}kT$, depending only on thermodynamic temperature |
| internal energy of an ideal gas | the total kinetic energy of its molecules, $\tfrac{3}{2}NkT$, since there is no potential energy |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The internal energy 内能 $U$ of a system is the sum of:
For a real solid, liquid or gas, both parts matter. In the ideal-gas 理想气体 model the intermolecular 分子间 forces are ignored, so the molecular potential energy is zero and the internal energy is purely kinetic.
The two-mark definition. The internal energy of a system is the sum of the random distribution of the kinetic and potential energies of its molecules (or atoms). Both marks need "sum of kinetic and potential energies" and "random" (or "of the molecules") — leaving out "random" describes the energy of a moving object, not its internal energy. For an ideal gas, "with reference to kinetic and potential energy": the internal energy is the total kinetic energy of the molecules only, because there are no intermolecular forces and therefore no potential energy.
Two key points:
Raising an object's temperature raises the random kinetic energy of its molecules, and so raises its internal energy.
For an ideal gas every molecule has average translational kinetic energy $\tfrac{3}{2} k T$ (Topic 15). With zero intermolecular potential energy, the total internal energy is
So the internal energy of an ideal gas is directly proportional to the thermodynamic temperature 热力学温度. Doubling $T$ doubles $U$. This is only exact for an ideal gas.
During a phase change 相变 (melting or boiling) of a real substance, $U$ rises because the molecular potential energy rises (bonds breaking), even though the temperature stays constant.
Describing a change in internal energy — always name both energies. A three-mark "describe and explain, with reference to molecular kinetic and potential energies" answer says what happens to each:
Worked example. A fixed mass of ideal gas is heated at constant pressure. Sketch the variation of its internal energy $U$ with its volume $V$.
At constant pressure $V \propto T$ (Charles's law) and for an ideal gas $U \propto T$, so $U \propto V$: a straight line through the origin. The origin is on the line because at absolute zero both $V$ (extrapolated) and $U$ are zero.
Energi dalam $U$ dari suatu sistem adalah jumlah dari:
Untuk zat padat, cair atau gas nyata, kedua bagian tersebut penting. Dalam model gas ideal, gaya antarmolekul diabaikan, sehingga energi potensial molekuler nol dan energi dalam murni kinetik.
Definisi dua poin. Energi dalam suatu sistem adalah jumlah distribusi acak dari energi kinetik dan potensial dari molekul-molekulnya (atau atom-atomnya). Kedua poin memerlukan "jumlah energi kinetik dan potensial" dan "acak" (atau "dari molekul-molekulnya") — mengabaikan "acak" menggambarkan energi benda yang bergerak, bukan energi dalamnya. Untuk gas ideal, "dengan referensi ke energi kinetik dan potensial": energi dalam adalah total kinetik molekul saja, karena tidak ada gaya antarmolekul dan karenanya tidak ada energi potensial.
Dua poin kunci:
Menaikkan suhu suatu benda meningkatkan energi kinetik acak molekulnya, dan sehingga menaikkan energi dalamnya.
Untuk gas ideal setiap molekul memiliki energi kinetik translasi rata-rata $\tfrac{3}{2} k T$ (Topik 15). Dengan nol energi potensial antarmolekul, total energi dalam adalah
Jadi energi dalam gas ideal berbanding lurus dengan suhu termodinamika. Menguatkan $T$ menggandakan $U$. Ini hanya tepat untuk gas ideal.
Selama perubahan fase (melebur atau mendidih) dari zat nyata, $U$ meningkat karena energi potensial molekuler meningkat (pemutusan ikatan), meskipun suhu tetap konstan.
Mendeskripsikan perubahan energi dalam — selalu sebutkan kedua energi. Jawaban tiga poin "deskripsikan dan jelaskan, dengan referensi ke energi kinetik dan potensial molekuler" menyatakan apa yang terjadi pada setiap:
Contoh dikerjakan. Massa tetap gas ideal dipanaskan pada tekanan konstan. Sketsakan variasi energi dalamnya $U$ terhadap volumenya $V$.
Pada tekanan tetap $V \propto T$ (hukum Charles) dan untuk gas ideal $U \propto T$, maka $U \propto V$: garis lurus melalui titik asal. Titik asal berada pada garis kerana pada sifar mutlak kedua-dua $V$ (dilepaskan ke luar) dan $U$ adalah sifar.
Internal energy is the total random kinetic + potential energy of the molecules. Heat the gas and the whole speed distribution shifts to higher energy. · Energi dalam adalah total energi kinetik + potensial acak dari molekul-molekulnya. Panaskan gas dan seluruh distribusi kecepatan bergeser ke energi yang lebih tinggi.
| English | Bahasa Indonesia |
|---|---|
| internal energy/ɪnˈtɜːnl ˈenədʒi/ | energi dalam |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| molecule/ˈmɒlɪkjuːl/ | molekul |
| translational/trænˈsleɪʃənl/ | translasi |
| rotational/rəʊˈteɪʃənl/ | rotasi |
| vibrational/vaɪˈbreɪʃənl/ | vibrasi |
| ideal-gas/aɪˈdɪəl ɡæs/ | gas ideal |
| intermolecular/ˌɪntəməˈlekjʊlə/ | antarmolekul |
| ideal gas/aɪˈdɪəl ɡæs/ | gas ideal |
| temperature/ˈtemprɪtʃə/ | suhu |
| pressure/ˈpreʃə/ | tekanan |
| volume/ˈvɒljuːm/ | volume |
| amount of substance/əˈmaʊnt ɒv ˈsʌbstəns/ | kuantiti bahan |
| thermodynamic temperature/ˌθɜːməʊdaɪˈnæmɪk ˈtemprɪtʃə/ | suhu termodinamika |
| phase change/feɪz tʃeɪndʒ/ | perubahan fase |
| latent heat/ˈleɪtənt hiːt/ | kalor laten |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
When a gas changes volume against an outside pressure, mechanical work is done. At constant pressure $p$ with a small volume change $\Delta V$, the size of the work is
Worked example. A gas at a constant pressure of $1.0 \times 10^{5}\ \text{Pa}$ expands from $2.0 \times 10^{-3}\ \text{m}^{3}$ to $5.0 \times 10^{-3}\ \text{m}^{3}$. Find the work done by the gas.
Worked example. An ideal gas of mass $0.35\ \text{kg}$ is heated at a constant pressure of $2.0 \times 10^{5}\ \text{Pa}$; its internal energy rises by $7600\ \text{J}$ and its volume increases by $1.2 \times 10^{-2}\ \text{m}^{3}$. Find the work done by the gas and the thermal energy supplied. The gas is then heated at constant volume until its internal energy rises by the same $7600\ \text{J}$; explain why less thermal energy is needed.
Work done by the gas $= p\Delta V = 2.0 \times 10^{5} \times 1.2 \times 10^{-2} = 2400\ \text{J}$, so the work done on it is $W = -2400\ \text{J}$. First law: $q = \Delta U - W = 7600 - (-2400) = 1.0 \times 10^{4}\ \text{J}$. At constant volume no work is done ($\Delta V = 0$, $W = 0$), so the whole of the thermal energy goes into internal energy: $q = \Delta U = 7600\ \text{J}$. The extra $2400\ \text{J}$ at constant pressure was the work the gas did pushing back the surroundings as it expanded.
Worked example. An aluminium block of volume $3.612 \times 10^{-3}\ \text{m}^{3}$ is heated from $0\ ^{\circ}\text{C}$ to $40\ ^{\circ}\text{C}$; its volume increases by $1.0 \times 10^{-5}\ \text{m}^{3}$ against atmospheric pressure ($1.0 \times 10^{5}\ \text{Pa}$). Compare the work it does on the atmosphere with the thermal energy it receives ($m = 9.75\ \text{kg}$, $c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$).
Work done by the block $= p\Delta V = 1.0 \times 10^{5} \times 1.0 \times 10^{-5} = 1.0\ \text{J}$; thermal energy $q = mc\Delta T = 9.75 \times 900 \times 40 = 3.5 \times 10^{5}\ \text{J}$. The work is about $3 \times 10^{-6}$ of the heating, so for a solid $\Delta U \approx q$: solids and liquids barely expand, and $p\Delta V$ only matters for a gas.
This syllabus writes the first law as $\Delta U = q + W$, where $W$ is the work done on the gas and $q$ is the energy put in by heating.
Watch which form a question wants:
At constant volume ($\Delta V = 0$), no work is done.
Ketika gas mengubah volume terhadap tekanan luar, kerja mekanik dilakukan. Pada tekanan konstan $p$ dengan perubahan volume kecil $\Delta V$, besar kerjanya adalah
Contoh terpecahkan. Gas pada tekanan konstan $1.0 \times 10^{5}\ \text{Pa}$ berekspansi dari $2.0 \times 10^{-3}\ \text{m}^{3}$ ke $5.0 \times 10^{-3}\ \text{m}^{3}$. Hitunglah kerja yang dilakukan oleh gas tersebut.
Contoh terpecahkan. Gas ideal bermassa $0.35\ \text{kg}$ dipanaskan pada tekanan konstan $2.0 \times 10^{5}\ \text{Pa}$; energi dalamnya naik sebesar $7600\ \text{J}$ dan volumenya bertambah $1.2 \times 10^{-2}\ \text{m}^{3}$. Tentukan kerja yang dilakukan oleh gas dan energi termal yang disuplai. Kemudian gas dipanaskan pada volume konstan hingga energi dalamnya naik sebesar $7600\ \text{J}$ yang sama; jelaskan mengapa energi termal yang dibutuhkan lebih sedikit.
Kerja yang dilakukan oleh gas $= p\Delta V = 2.0 \times 10^{5} \times 1.2 \times 10^{-2} = 2400\ \text{J}$, sehingga kerja yang dilakukan padanya adalah $W = -2400\ \text{J}$. Hukum pertama: $q = \Delta U - W = 7600 - (-2400) = 1.0 \times 10^{4}\ \text{J}$. Pada volume konstan tidak ada kerja yang dilakukan ($\Delta V = 0$, $W = 0$), sehingga seluruh energi termal masuk ke energi dalam: $q = \Delta U = 7600\ \text{J}$. Tambahan $2400\ \text{J}$ pada tekanan konstan adalah kerja yang dilakukan gas untuk mendorong lingkungan saat berekspansi.
Contoh terpecahkan. Balok aluminium bervolume $3.612 \times 10^{-3}\ \text{m}^{3}$ dipanaskan dari $0\ ^{\circ}\text{C}$ ke $40\ ^{\circ}\text{C}$; volumenya bertambah $1.0 \times 10^{-5}\ \text{m}^{3}$ melawan tekanan atmosfer ($1.0 \times 10^{5}\ \text{Pa}$). Bandingkan kerja yang dilakukannya pada atmosfer dengan energi termal yang diterimanya ($m = 9.75\ \text{kg}$, $c = 900\ \text{J kg}^{-1}\ \text{K}^{-1}$).
Kerja yang dilakukan balok $= p\Delta V = 1.0 \times 10^{5} \times 1.0 \times 10^{-5} = 1.0\ \text{J}$; energi termal $q = mc\Delta T = 9.75 \times 900 \times 40 = 3.5 \times 10^{5}\ \text{J}$. Kerja ini sekitar $3 \times 10^{-6}$ dari pemanasan, sehingga untuk padatan $\Delta U \approx q$: padat dan cair hampir tidak berekspansi, dan $p\Delta V$ hanya relevan untuk gas.
Silabus ini menuliskan hukum pertama sebagai $\Delta U = q + W$, di mana $W$ adalah kerja yang dilakukan pada gas dan $q$ adalah energi yang dimasukkan melalui pemanasan.
Perhatikan bentuk pertanyaan yang diminta:
Pada volume konstan ($\Delta V = 0$), tidak ada kerja yang dilakukan.
| English | Bahasa Indonesia |
|---|---|
| piston/ˈpɪstn/ | piston |
| first law of thermodynamics/fɜːst lɔː ɒv ˌθɜːməʊdaɪˈnæmɪks/ | hukum pertama termodinamika |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | hukum kekekalan energi |
The first law of thermodynamics 热力学第一定律 says that energy is conserved when heat and work pass between a system and its surroundings:
where $\Delta U$ is the rise in internal energy, $q$ is the energy added by heating (positive in, negative out), and $W$ is the work done on the gas (positive when compressed). This is conservation of energy 能量守恒 for a gas.
As the exam asks it. "State the first law of thermodynamics, identifying any symbols" (2 marks): the increase in internal energy of a system, $\Delta U$, is equal to the sum of the thermal energy transferred to the system by heating, $q$, and the work done on the system, $W$: $\Delta U = q + W$. Every symbol must be defined with its direction — "to the system" and "on the system" are what make the signs mean something. "State two ways in which the first law says the internal energy of a system may be changed": by heating (thermal energy transferred into or out of the system) and by doing work on or by the system.
Worked example. A gas absorbs $500\ \text{J}$ of heat while it expands and does $200\ \text{J}$ of work on its surroundings. Find the change in its internal energy.
The gas does work, so the work done on it is $W = -200\ \text{J}$:
Explaining with the first law. A three-mark "use the first law to explain" answer has three steps: say what $q$ is (and its sign), say what $W$ is (and its sign), then combine them for $\Delta U$ — and, if asked, say what the change in internal energy means for the molecules.
$\Delta U$ is fixed by the change of state (for an ideal gas, by the change in temperature). The same $\Delta U$ can come from different mixes of $q$ and $W$:
For an ideal gas, $\Delta U = \tfrac{3}{2} n R \Delta T$ — it depends only on $\Delta T$.
| Process | What stays constant | $\Delta U$ | $W$ (on gas) | $q$ |
|---|---|---|---|---|
| Isothermal | $T$ | $0$ | $W$ | $-W$ |
| Constant volume | $V$ | $\tfrac{3}{2}n R \Delta T$ | $0$ | $\Delta U$ |
| Constant pressure | $p$ | $\tfrac{3}{2}n R \Delta T$ | $-p \Delta V$ | $\Delta U - W$ |
| Adiabatic | (no heat) | varies | $W$ | $0$ |
Read each row with the first law $\Delta U = q + W$:
So for the same rise in internal energy, heating at constant pressure needs more thermal energy than heating at constant volume, by exactly the work $p\Delta V$ the gas does while expanding — the constant-volume gas keeps every joule; the constant-pressure gas hands some back to the surroundings.
A sample of ideal gas at temperature $T$ with internal energy $U$ goes through:
Step 1 ($T \to 3T$): $U = \tfrac{3}{2}nRT$, so $U \to 3U$, giving $\Delta U_{1} = 2U$. $W_{1} = +W$. So $q_{1} = \Delta U_{1} - W_{1} = 2U - W$.
Step 2 ($3T \to 2T$, constant volume): $\Delta U_{2} = -U$. $W_{2} = 0$. So $q_{2} = -U$ (heat flows out).
Check: total $\Delta U = 2U - U = U$, taking the gas from $T$ to $2T$ ($U \to 2U$) — consistent.
A fixed mass of ideal gas is taken round the cycle ABCDA: A to B at constant volume $V_{1}$ (pressure $p_{1} \to p_{2}$), B to C at constant pressure $p_{2}$ (volume $V_{1} \to V_{2}$), C to D at constant volume, D to A at constant pressure $p_{1}$. Complete a table of the signs of $q$, $W$ and $\Delta U$ for each leg, and explain the internal-energy change from B to C.
| Leg | Process | $W$ (on gas) | $\Delta U$ | $q$ |
|---|---|---|---|---|
| A → B | constant volume, pressure rises | $0$ (no volume change) | $+$ (temperature rises, $pV$ larger) | $+$ ($q = \Delta U$) |
| B → C | constant pressure, expands | $-$ (gas does work $p_{2}(V_{2} - V_{1})$) | $+$ ($T \propto V$ at constant $p$) | $+$ (and larger than $\Delta U$) |
| C → D | constant volume, pressure falls | $0$ | $-$ | $-$ (thermal energy leaves) |
| D → A | constant pressure, compressed | $+$ (work done on the gas $p_{1}(V_{2} - V_{1})$) | $-$ | $-$ (larger in size than $\Delta U$) |
From B to C the gas expands at constant pressure, so its temperature rises ($V/T$ constant) and its internal energy increases; it does work on the surroundings ($W$ negative), so the thermal energy supplied must cover both: $q = \Delta U - W$, larger than the rise in internal energy. Round the whole cycle $\Delta U = 0$ (the gas returns to its starting state), so the net thermal energy in equals the net work done by the gas — the area enclosed by the rectangle, $(p_{2} - p_{1})(V_{2} - V_{1})$.
For constant-volume heating of an ideal gas, $q = \Delta U = \tfrac{3}{2} n R \Delta T$. So the molar heat capacity 热容 at constant volume is $\tfrac{3}{2} R$ for a monatomic 单原子 ideal gas. (You are not required to use the symbol $C_V$, but the result $q = \tfrac{3}{2} n R \Delta T$ for constant-volume heating is.)
If heat is supplied during a phase change at constant pressure (e.g. boiling water), the temperature stays constant but the internal energy still rises (the latent heat 潜热 separates the molecules), and the gas does expansion work. The first law still holds: $\Delta U = q + W$.
Hukum pertama termodinamika menyatakan bahwa energi kekal ketika panas dan kerja berpindah antara sistem dan lingkungannya:
di mana $\Delta U$ adalah kenaikan energi dalam, $q$ adalah energi yang ditambahkan melalui pemanasan (positif masuk, negatif keluar), dan $W$ adalah kerja yang dilakukan pada gas (positif saat dikompresi). Ini adalah kekekalan energi untuk gas.
Seperti yang diminta dalam ujian. "Nyatakan hukum pertama termodinamika, identifikasi setiap simbol" (2 poin): peningkatan energi dalam suatu sistem, $\Delta U$, sama dengan jumlah energi termal yang ditransfer ke sistem melalui pemanasan, $q$, dan kerja yang dilakukan pada sistem, $W$: $\Delta U = q + W$. Setiap simbol harus didefinisikan beserta arahnya — "ke sistem" dan "pada sistem" adalah hal yang membuat tanda memiliki makna. "Nyatakan dua cara bagaimana hukum pertama menyatakan energi dalam sistem dapat berubah": melalui pemanasan (energi termal ditransfer masuk atau keluar dari sistem) dan melalui melakukan kerja pada atau oleh sistem.
Contoh terpecahkan. Gas menyerap $500\ \text{J}$ kalor saat berekspansi dan melakukan $200\ \text{J}$ kerja pada lingkungannya. Tentukan perubahan energi dalamnya.
Gas melakukan kerja, sehingga kerja yang dilakukan padanya adalah $W = -200\ \text{J}$:
Penjelasan menggunakan hukum pertama. Jawaban "gunakan hukum pertama untuk menjelaskan" bernilai tiga poin memiliki tiga langkah: sebutkan apa itu $q$ (dan tandanya), sebutkan apa itu $W$ (dan tandanya), lalu gabungkan keduanya untuk $\Delta U$ — dan, jika diminta, jelaskan apa arti perubahan energi dalam bagi molekul-molekulnya.
$\Delta U$ ditentukan oleh perubahan wujud (untuk gas ideal, oleh perubahan suhu). Nilai $\Delta U$ yang sama dapat berasal dari campuran berbeda $q$ dan $W$:
Untuk gas ideal, $\Delta U = \tfrac{3}{2} n R \Delta T$ — ia hanya bergantung pada $\Delta T$.
| Proses | Apa yang tetap konstan | $\Delta U$ | $W$ (pada gas) | $q$ |
|---|---|---|---|---|
| Isotermal | $T$ | $0$ | $W$ | $-W$ |
| Volume konstan | $V$ | $\tfrac{3}{2}n R \Delta T$ | $0$ | $\Delta U$ |
| Tekanan konstan | $p$ | $\tfrac{3}{2}n R \Delta T$ | $-p \Delta V$ | $\Delta U - W$ |
| Adiabatik | (tanpa panas) | bervariasi | $W$ | $0$ |
Baca setiap baris menggunakan hukum pertama $\Delta U = q + W$:
Jadi untuk kenaikan energi dalam yang sama, pemanasan pada tekanan konstan membutuhkan lebih banyak energi termal daripada pemanasan pada volume konstan, tepat sebesar kerja $p\Delta V$ yang dilakukan gas saat berekspansi — gas volume konstan menyimpan setiap joule; gas tekanan konstan mengembalikan sebagian ke lingkungan.
Sebuah sampel gas ideal pada suhu $T$ dengan energi dalam $U$ mengalami:
Langkah 1 ($T \to 3T$): $U = \tfrac{3}{2}nRT$, sehingga $U \to 3U$, menghasilkan $\Delta U_{1} = 2U$. $W_{1} = +W$. Jadi $q_{1} = \Delta U_{1} - W_{1} = 2U - W$.
Langkah 2 ($3T \to 2T$, volume konstan): $\Delta U_{2} = -U$. $W_{2} = 0$. Jadi $q_{2} = -U$ (panas mengalir keluar).
Periksa: total $\Delta U = 2U - U = U$, membawa gas dari $T$ ke $2T$ ($U \to 2U$) — konsisten.
Sejumlah tetap gas ideal dialirkan melalui siklus ABCDA: A ke B pada volume konstan $V_{1}$ (tekanan $p_{1} \to p_{2}$), B ke C pada tekanan konstan $p_{2}$ (volume $V_{1} \to V_{2}$), C ke D pada volume konstan, D ke A pada tekanan konstan $p_{1}$. Lengkapi tabel tanda $q$, $W$, dan $\Delta U$ untuk setiap kaki, dan jelaskan perubahan energi dalam dari B ke C.
| Kaki | Proses | $W$ (pada gas) | $\Delta U$ | $q$ |
|---|---|---|---|---|
| A → B | volume konstan, tekanan naik | $0$ (tidak ada perubahan volume) | $+$ (suhu naik, $pV$ lebih besar) | $+$ ($q = \Delta U$) |
| B → C | tekanan konstan, berekspansi | $-$ (gas melakukan kerja $p_{2}(V_{2} - V_{1})$) | $+$ ($T \propto V$ pada $p$ konstan) | $+$ (dan lebih besar dari $\Delta U$) |
| C → D | volume konstan, tekanan turun | $0$ | $-$ | $-$ (energi termal keluar) |
| D → A | tekanan konstan, dikompresi | $+$ (kerja dilakukan pada gas $p_{1}(V_{2} - V_{1})$) | $-$ | $-$ (lebih besar ukurannya dari $\Delta U$) |
Dari B ke C gas memuai pada tekanan konstan, sehingga suhunya naik ($V/T$ tetap) dan energi dalamnya meningkat; gas melakukan kerja pada lingkungan ($W$ negatif), sehingga energi termal yang disuplai harus menutupi keduanya: $q = \Delta U - W$, lebih besar daripada kenaikan energi dalam. Untuk siklus penuh $\Delta U = 0$ (gas kembali ke keadaan awal), maka energi termal masuk bersih sama dengan kerja yang dilakukan oleh gas — yaitu luas yang terkurung oleh persegi panjang, $(p_{2} - p_{1})(V_{2} - V_{1})$.
Untuk pemanasan volume konstan pada gas ideal, $q = \Delta U = \tfrac{3}{2} n R \Delta T$. Maka kapasitas kalor molar pada volume konstan adalah $\tfrac{3}{2} R$ untuk gas ideal monatomik. (Anda tidak diwajibkan menggunakan simbol $C_V$, tetapi hasil $q = \tfrac{3}{2} n R \Delta T$ untuk pemanasan volume konstan berlaku).
Jika panas disuplai selama perubahan fase pada tekanan konstan (misalnya merebus air), suhu tetap konstan namun energi dalam tetap meningkat (panas latent memisahkan molekul-molekulnya), dan gas melakukan kerja ekspansi. Hukum pertama tetap berlaku: $\Delta U = q + W$.
Push the piston in and you do work on the gas (W = pΔV); the first law says that work plus the heat added equals the rise in internal energy. · Dorong piston ke dalam dan Anda melakukan usaha pada gas (W = pΔV); hukum pertama menyatakan bahwa usaha ditambah kalor yang ditambahkan sama dengan kenaikan energi dalam.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| internal energy | the sum of the random distribution of the kinetic and potential energies of the molecules of a system |
| internal energy of an ideal gas | the total random kinetic energy of its molecules, since with no intermolecular forces there is no potential energy |
| first law of thermodynamics | the increase in internal energy of a system equals the thermal energy transferred to the system by heating plus the work done on the system: $\Delta U = q + W$ |
| work done on a gas at constant pressure | $W = p\Delta V$ in size; positive when the gas is compressed, negative when it expands (the gas then does work on its surroundings) |
| thermal energy transfer (heating) | energy transferred to or from a system because of a temperature difference; $q$ is positive when it enters |
| isothermal change | a change at constant temperature, so for an ideal gas $\Delta U = 0$ and $q = -W$ |
| adiabatic change | a change with no thermal energy transfer, $q = 0$, so $\Delta U = W$ |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| energi dalam | jumlah distribusi acak energi kinetik dan potensial dari molekul-molekul suatu sistem |
| energi dalam gas ideal | total energi kinetik acak dari molekul-molekulnya, karena tanpa gaya antarmolekul tidak ada energi potensial |
| hukum pertama termodinamika | kenaikan energi dalam suatu sistem sama dengan energi termal yang ditransfer ke sistem melalui pemanasan ditambah kerja yang dilakukan pada sistem: $\Delta U = q + W$ |
| kerja yang dilakukan pada gas pada tekanan konstan | $W = p\Delta V$ dalam besaran; positif ketika gas dikompresi, negatif ketika memuai (gas kemudian melakukan kerja pada lingkungannya) |
| perpindahan energi termal (pemanasan) | energi yang ditransfer masuk atau keluar dari suatu sistem karena perbedaan suhu; $q$ bernilai positif ketika masuk |
| perubahan isothermal | perubahan pada suhu konstan, sehingga untuk gas ideal $\Delta U = 0$ dan $q = -W$ |
| perubahan adiabatik | perubahan tanpa perpindahan energi termal, $q = 0$, sehingga $\Delta U = W$ |
| English | Bahasa Indonesia |
|---|---|
| isothermal/ˌaɪsəˈθɜːml/ | isotermal |
| adiabatic/ˌædiəˈbætɪk/ | adiabatik |
| heat capacity/hiːt kəˈpæsɪti/ | kapasiti haba |
| monatomic/ˌmɒnəˈtɒmɪk/ | monatomik |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A particle moves with simple harmonic motion 简谐运动 (SHM) when its acceleration 加速度 is:
The defining equation is
where $x$ is the displacement 位移 from equilibrium and $\omega$ is a positive constant, the angular frequency 角频率. The minus sign means "directed back towards equilibrium".
The two-mark definition. Simple harmonic motion is motion in which the acceleration is proportional to the displacement from a fixed point and is always directed towards that point (in the opposite direction to the displacement). Both halves are needed. "The significance of the minus sign": the acceleration is in the opposite direction to the displacement — always towards the equilibrium position. "State what is meant by the frequency of the oscillations": the number of complete oscillations per unit time.
Many systems do this near a stable equilibrium: a mass on a spring 弹簧, a pendulum 单摆 (small swing), a floating block pushed down, the charge on a capacitor 电容器 in an LC circuit, atoms in a solid.
So $T = 2\pi/\omega$ and $f = \omega/(2\pi)$ — given any one of $\omega$, $f$, $T$ you can find the others.
Worked example. A mass on a spring oscillates with SHM of amplitude $0.050\ \text{m}$ and frequency $2.5\ \text{Hz}$. Find its maximum acceleration.
The angular frequency is $\omega = 2\pi f = 2\pi \times 2.5 = 15.7\ \text{rad s}^{-1}$. The acceleration is largest at the extremes, where $|a| = \omega^{2}x_{0}$:
The exam's favourite four-mark derivation, and the recipe is always the same: (1) displace the object by $x$ from equilibrium; (2) find the resultant restoring force, which comes out as $F = -(\text{constant}) \times x$; (3) apply Newton's second law, $a = F/m = -(\text{constant}/m)\,x$; (4) this has the form $a = -\omega^{2}x$, so the motion is simple harmonic with $\omega^{2} = \text{constant}/m$, and $T = 2\pi/\omega$. Say in words why the force points back towards equilibrium and why it is proportional to $x$ — those are the marks, not the algebra.
Worked example. A mass $m$ hangs from a spring of spring constant $k$. Show that, when displaced and released, it moves with simple harmonic motion, and find the period.
At equilibrium the spring tension balances the weight. Displace the mass a distance $x$ downwards: the tension increases by $kx$ (Hooke's law), so there is a resultant force $kx$ upwards, towards equilibrium. Newton's second law: $a = -\dfrac{k}{m}x$. This is $a = -\omega^{2}x$ with $\omega^{2} = k/m$, so the motion is simple harmonic and $T = 2\pi/\omega = 2\pi\sqrt{m/k}$. (A trolley held between two identical springs has both pulling it back, $F = -2kx$, so $\omega^{2} = 2k/m$.)
Worked example. A cuboidal block of cross-sectional area $A$ and mass $m$ floats in a liquid of density $\rho$ with its base a depth $h$ below the surface. It is pushed down a small distance $x$ and released. Show that it oscillates with simple harmonic motion and find the period in terms of $h$ and $g$.
When the block is pushed down by $x$ an extra volume $Ax$ is submerged, so the upthrust 浮力 increases by the weight of that extra liquid, $\rho g A x$; this extra force acts upwards, towards the equilibrium position. Hence $a = -\dfrac{\rho g A}{m}x$: proportional to $x$ and opposite in direction, so the motion is simple harmonic with $\omega^{2} = \rho g A/m$. At equilibrium the upthrust on the submerged volume $Ah$ equals the weight, $\rho g A h = mg$, so $m = \rho A h$ and $\omega^{2} = g/h$: $T = 2\pi\sqrt{h/g}$ — the same form as a pendulum of length $h$.
Worked example. A U-tube of cross-sectional area $A$ contains liquid of density $\rho$; the column in each arm has length $L$. The liquid is displaced so that one surface is $x$ above its equilibrium level and the other $x$ below. Show that the liquid oscillates with simple harmonic motion.
The two surfaces now differ in height by $2x$, so a column of liquid of height $2x$ is unbalanced: the restoring force on the liquid is the weight of that column, $2x\rho g A$, directed so as to level the surfaces. The mass being moved is the whole liquid, $2L\rho A$. So $a = -\dfrac{2x\rho g A}{2L\rho A} = -\dfrac{g}{L}x$: simple harmonic, with $\omega^{2} = g/L$ and $T = 2\pi\sqrt{L/g}$. Note that $\rho$ and $A$ cancel — the period depends on neither the liquid nor the tube.
Set it swinging, then change the start angle — the time for one swing stays the same (that is what makes it a good clock). Now make the string longer, or move to the Moon, and watch the period change. · Biarkan berayun, lalu ubah sudut awal — waktu untuk satu ayunan tetap sama (itulah yang menjadikannya jam yang baik). Sekarang buat tali lebih panjang, atau pindah ke Bulan, dan perhatikan perubahan periodenya.
| English | Bahasa Indonesia |
|---|---|
| simple harmonic motion/ˈsɪmpl hɑːˈmɒnɪk ˈməʊʃn/ | gerak harmonik sederhana |
| spring/sprɪŋ/ | pegas |
| pendulum/ˈpendjʊləm/ | bandul |
| acceleration/əkˌseləˈreɪʃn/ | percepatan |
| displacement/dɪˈspleɪsmənt/ | pindahan |
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | kesetimbangan |
| angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ | frekuensi sudut |
| frequency/ˈfriːkwənsi/ | frekuensi |
| capacitor/kəˈpæsɪtə/ | kapasitor |
| amplitude/ˈæmplɪtjuːd/ | amplitudo |
| period/ˈpɪərɪəd/ | tempoh |
| phase difference/feɪz ˈdɪfrəns/ | selisih fase |
| upthrust/ˈʌpθrʌst/ | gaya apung |
If the particle starts at $x = 0$ moving in the positive direction at $t = 0$, then
Differentiating once gives the velocity 速度:
where $v_{0} = x_{0} \omega$ is the maximum speed (as the particle passes through equilibrium).
Differentiating again gives the acceleration:
which is the SHM defining equation again. (If the particle instead starts at the extreme position $x = x_{0}$ at $t = 0$, use $x = x_{0} \cos(\omega t)$. Choose the one that fits the start conditions.)
A useful relation that does not use time:
Worked example. The same oscillation has amplitude $x_{0} = 0.050\ \text{m}$ and angular frequency $\omega = 15.7\ \text{rad s}^{-1}$. Find the speed when the displacement is $x = 0.030\ \text{m}$.
Reading the $v$–$x$ graph. Its width gives the amplitude ($x = \pm x_{0}$ where $v = 0$) and its height the maximum speed ($v = \pm\omega x_{0}$ at $x = 0$), so $\omega = v_{0}/x_{0}$ and the period follows. Asked to sketch $a$ against $x$ on the same axes, draw the straight line through the origin with negative gradient $-\omega^{2}$, reaching $\mp\omega^{2}x_{0}$ at $x = \pm x_{0}$.
Worked example. The velocity of an oscillating block varies with time as $v = 0.56\sin(5.1t)$, with $v$ in $\text{m s}^{-1}$ and $t$ in seconds. Find the amplitude, the period and the maximum acceleration.
Compare with $v = v_{0}\sin\omega t$: $v_{0} = 0.56\ \text{m s}^{-1}$ and $\omega = 5.1\ \text{rad s}^{-1}$. Since $v_{0} = \omega x_{0}$, the amplitude is $x_{0} = 0.56/5.1 = 0.11\ \text{m}$; $T = 2\pi/\omega = 1.2\ \text{s}$; $a_{\text{max}} = \omega^{2}x_{0} = \omega v_{0} = 5.1 \times 0.56 = 2.9\ \text{m s}^{-2}$. (Because this $v$ is a sine, the block was at an extreme at $t = 0$: its displacement is a cosine.)
SHM as the shadow of circular motion. A ball moving round a circle of radius $R$ at angular speed $\omega$, lit from the side, casts a shadow that moves back and forth with simple harmonic motion of amplitude $R$ and the same $\omega$: the shadow's displacement is $x = R\cos\omega t$ (if it starts at an extreme), its maximum speed is $R\omega$ and its maximum acceleration $R\omega^{2}$. For a circle of diameter $0.46\ \text{m}$ and $\omega = 1.9\ \text{rad s}^{-1}$ the shadow has amplitude $0.23\ \text{m}$, $v_{0} = 0.44\ \text{m s}^{-1}$ and $a_{\text{max}} = 0.83\ \text{m s}^{-2}$ — greatest in size at the two ends of its path and always directed back towards the centre.
For $x = x_{0}\sin\omega t$:
A straight line through the origin with negative gradient $-\omega^{2}$. So you can read $\omega$ from the graph: gradient $= -\omega^{2}$, so $\omega = \sqrt{|\text{gradient}|}$, then $T = 2\pi / \omega$. This is a common exam pattern.
Worked example. A graph of the height $h$ of an object of mass $36\ \text{kg}$ undergoing vertical simple harmonic motion shows $h$ varying between $2\ \text{m}$ and $10\ \text{m}$, with one complete oscillation every $4.0\ \text{s}$. Find the amplitude, the angular frequency, the maximum speed, the maximum acceleration and the maximum kinetic energy.
Amplitude $x_{0} = (10 - 2)/2 = 4.0\ \text{m}$ (half the peak-to-peak range; the equilibrium height is $6\ \text{m}$). $\omega = 2\pi/T = 2\pi/4.0 = 1.57\ \text{rad s}^{-1}$. $v_{\text{max}} = \omega x_{0} = 6.3\ \text{m s}^{-1}$; $a_{\text{max}} = \omega^{2}x_{0} = 9.9\ \text{m s}^{-2}$; $E_{\text{K,max}} = \tfrac{1}{2}mv_{\text{max}}^{2} = \tfrac{1}{2} \times 36 \times 6.28^{2} = 7.1 \times 10^{2}\ \text{J}$. Reading the amplitude as the full height range instead of half of it is the commonest slip here.
Worked example. A small object rests on a horizontal platform that oscillates vertically with simple harmonic motion of amplitude $6.0\ \text{mm}$. Find the highest frequency at which the object stays in contact with the platform throughout, and state where in the motion contact is first lost.
The object leaves the platform when the platform's downward acceleration exceeds $g$ — at the top of the oscillation, where the acceleration is greatest and directed downwards: the platform then falls away faster than the object can. The limit is $\omega^{2}x_{0} = g$: $\omega = \sqrt{9.81/0.0060} = 40\ \text{rad s}^{-1}$, so $f = \omega/2\pi = 6.4\ \text{Hz}$. Below this the normal contact force at the top is small but positive; above it, zero.
| English | Bahasa Indonesia |
|---|---|
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A simple harmonic oscillator keeps swapping energy between two forms:
With no damping 阻尼, the total energy is constant (this is conservation of energy 能量守恒).
Using $v_{\text{max}} = \omega x_{0}$:
Two key facts: the total energy is proportional to the square of the amplitude (doubling $x_{0}$ gives four times the energy), and to $\omega^{2}$.
Using $v^{2} = \omega^{2}(x_{0}^{2} - x^{2})$:
So $E_{\text{K}}$ is a downward parabola (peak at $x = 0$, zero at $x = \pm x_{0}$) and $E_{\text{P}}$ is an upward parabola (zero at $x = 0$, largest at $x = \pm x_{0}$). Their sum is constant.
Worked example. A $0.20\ \text{kg}$ mass oscillates with amplitude $x_{0} = 0.050\ \text{m}$ and angular frequency $\omega = 15.7\ \text{rad s}^{-1}$. Find the total energy of the oscillation.
The total energy equals the maximum kinetic energy, as the mass passes through $x = 0$ at $v_{\text{max}} = \omega x_{0}$:
This stays constant, swapping between kinetic and potential form twice each cycle.
Describing the interchange without calculation. At an extreme of the motion the oscillator is momentarily at rest: its kinetic energy is zero and its potential energy is a maximum. As it moves towards equilibrium the restoring force does work on it, so potential energy is converted to kinetic energy; at the equilibrium position the kinetic energy is a maximum and the potential energy a minimum. Beyond it the kinetic energy is converted back to potential energy until the other extreme. The total stays constant, and the exchange happens twice per oscillation, so the energy graphs against time have twice the frequency of the displacement graph.
Worked example. A pendulum bob of mass $0.81\ \text{kg}$ swings with small oscillations of amplitude $32\ \text{mm}$ and period $1.9\ \text{s}$. Find the total energy of the oscillation, and state what happens to it if the amplitude falls to $16\ \text{mm}$.
$\omega = 2\pi/1.9 = 3.31\ \text{rad s}^{-1}$; $E = \tfrac{1}{2}m\omega^{2}x_{0}^{2} = \tfrac{1}{2} \times 0.81 \times 3.31^{2} \times 0.032^{2} = 4.5 \times 10^{-3}\ \text{J}$. Halving the amplitude quarters the energy ($E \propto x_{0}^{2}$), to $1.1 \times 10^{-3}\ \text{J}$; the difference has been dissipated as thermal energy by air resistance.
Watch energy swap between kinetic and potential as the oscillator moves — fastest (max KE) at the centre, still (max PE) at the ends. · Perhatikan pertukaran energi antara kinetik dan potensial saat osilator bergerak — tercepat (KE maksimum) di tengah, diam (PE maksimum) di ujung-ujung.
| English | Bahasa Indonesia |
|---|---|
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| damping/ˈdæmpɪŋ/ | redaman |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | hukum kekekalan energi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A resistive force (friction 摩擦力, drag 阻力, air resistance 空气阻力) causes damping — the amplitude shrinks over time as energy is lost as heat. Three named cases:
The amplitude shrinks slowly over many cycles (a light damping 轻阻尼 case). The system still oscillates near its natural frequency, but each cycle is smaller than the last. A car's suspension is light-to-medium damped, so bumps die away but the ride stays smooth.
The least damping that brings the system back to equilibrium without overshooting and without oscillating — a critical damping 临界阻尼 case. It returns in the shortest time. A galvanometer 检流计 or analogue voltmeter 电压表 is critically damped so the needle settles quickly.
So much resistance that the system returns slowly, with no oscillation, but more slowly than the critical case — a heavy damping 过阻尼 case. A door with a strong closer is heavily damped.
On a displacement–time graph: light damping is a wave whose size dies away smoothly; critical damping returns quickly with no overshoot; heavy damping returns slowly.
As the exam asks it. Damping is the decrease in the amplitude of an oscillation caused by a resistive force that dissipates the oscillator's energy. In light damping the system oscillates with an amplitude that decreases gradually (exponentially) with time; the period is almost unchanged. In critical damping the system returns to equilibrium in the shortest possible time without oscillating. In heavy damping it returns to equilibrium slowly, without oscillating. Sketching them: the light-damping curve is a sine wave inside a shrinking envelope; the critical and heavy curves both fall to zero without crossing the axis, the heavy one more slowly.
Worked example. A bar magnet hangs from a spring with one pole inside a coil that is connected through a switch to a resistor. The magnet is set oscillating. Explain why the oscillations die away more quickly when the switch is closed.
The moving magnet changes the magnetic flux linking the coil, so an e.m.f. is induced (Faraday's law); with the switch closed a current flows through the resistor and dissipates energy as heat. By Lenz's law the induced current opposes the motion of the magnet — an extra resistive force. That energy comes from the oscillation, so the amplitude decreases faster: the circuit adds damping. With the switch open there is an e.m.f. but no current, so no extra energy loss.
| English | Bahasa Indonesia |
|---|---|
| air resistance/eə rɪˈzɪstəns/ | hambatan udara |
| friction/ˈfrɪkʃn/ | gesekan |
| drag/dræɡ/ | menggeser |
| light damping/laɪt ˈdæmpɪŋ/ | redaman ringan |
| critical damping/ˈkrɪtɪkl ˈdæmpɪŋ/ | redaman kritis |
| galvanometer/ˌɡælvəˈnɒmɪtə/ | galvanometer |
| voltmeter/ˈvəʊltmiːtə/ | voltmeter |
| heavy damping/ˈhevi ˈdæmpɪŋ/ | redaman berat |
A forced oscillation 受迫振动 is driven by an outside periodic force at a frequency $f_{\text{d}}$ chosen by the experimenter. The system then oscillates at this driving frequency 驱动频率 $f_{\text{d}}$, not at its own natural frequency. A plot of amplitude against $f_{\text{d}}$ is a resonance curve 共振曲线 with a peak.
Resonance 共振 happens when the driving frequency equals the system's natural frequency 固有频率 $f_{0}$. At resonance the amplitude is largest and the energy transfer from the driver is most efficient.
Examples:
The peak's shape depends on damping: lighter damping → a sharper, higher peak; heavier damping → a broader, lower peak, shifted slightly to lower frequency.
The two-mark definition. Resonance is the condition in which a system is forced to oscillate at its natural frequency, and the amplitude of the oscillation is a maximum (the transfer of energy from the driver is greatest). Both parts — driving frequency equal to natural frequency, and maximum amplitude — are needed.
Worked example. A ball on a stretched string has a natural frequency of $4.0\ \text{Hz}$. The string is driven by a vibration generator whose frequency is increased slowly from $0$ to $10\ \text{Hz}$. Sketch the variation of the amplitude of the ball with the driving frequency, and state the effect of adding damping.
The curve starts at a small, non-zero amplitude at low frequency, rises to a sharp peak at $4.0\ \text{Hz}$, and falls away towards zero at high frequency; the peak is at the natural frequency, and only there is the amplitude a maximum. More damping makes the peak lower and broader and shifts it slightly to a lower frequency; the amplitude far from resonance hardly changes.
Drive the swing at different frequencies. Far from its natural frequency it barely moves; tune them to match and the amplitude explodes — resonance, the same effect that can shake a bridge apart. · Gerakkan ayunan pada berbagai frekuensi. Jauh dari frekuensi alaminya ia hampir tidak bergerak; sesuaikan agar sesuai dan amplitudonya meledak — resonansi, efek yang sama yang dapat menggoyangkan jembatan hingga hancur.
| English | Bahasa Indonesia |
|---|---|
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | frekuensi alami |
| forced oscillation/fɔːst ˌɒsɪˈleɪʃn/ | osilasi terdesak |
| resonance/ˈrezənəns/ | resonansi |
| driving frequency/ˈdraɪvɪŋ ˈfriːkwənsi/ | frekuensi penggerak |
| resonance curve/ˈrezənəns kɜːv/ | kurva resonansi |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| simple harmonic motion | motion in which the acceleration is proportional to the displacement from a fixed point and is always directed towards that point |
| amplitude | the maximum displacement from the equilibrium position |
| period | the time for one complete oscillation |
| frequency | the number of complete oscillations per unit time |
| angular frequency | $\omega = 2\pi f = 2\pi/T$; the rate at which the phase of the oscillation changes |
| phase difference | the fraction of a cycle, expressed as an angle, by which one oscillation leads or lags another |
| damping | the reduction in amplitude of an oscillation caused by a resistive force that dissipates its energy |
| critical damping | the damping that returns a displaced system to equilibrium in the shortest time without oscillation |
| natural frequency | the frequency at which a system oscillates when displaced and left to oscillate freely |
| resonance | the forced oscillation of a system at its natural frequency, at which the amplitude is a maximum |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
An electric field 电场 is a region where a charge feels a force 力 from other charges. The electric field strength 电场强度 $E$ at a point is the force per unit positive charge on a small positive test charge 检验电荷 placed there:
Unit: $\text{N C}^{-1}$ (the same as $\text{V m}^{-1}$, as we will see). $E$ is a vector 矢量, pointing the way the force acts on a positive charge. The force on a charge $q$ is
opposite to the field if $q$ is negative.
The two-mark definition. Electric field strength at a point is the force per unit positive charge acting on a small (test) charge placed at that point. Three words carry the marks: force per unit charge (not "force on a charge"), positive (which fixes the direction), and small or test (so the charge does not disturb the field it is measuring). The multiple-choice version offers "force per unit charge acting on a small mass" and "force per unit mass" as distractors: a field is defined by a charge, never by a mass.
Worked example. A proton accelerates at $2.00\ \text{m s}^{-2}$ in an electric field, with no other force acting. Find the field strength. ($m_{\text{p}} = 1.67 \times 10^{-27}\ \text{kg}$, $e = 1.60 \times 10^{-19}\ \text{C}$.)
$qE = ma$, so
The same rearrangement the other way round gives the acceleration of an electron in a field of $1500\ \text{V m}^{-1}$: $a = eE/m_{\text{e}} = (1.60 \times 10^{-19})(1500)/(9.11 \times 10^{-31}) = 2.6 \times 10^{14}\ \text{m s}^{-2}$, enormous because the electron's mass is so small. Gravity ($9.81\ \text{m s}^{-2}$) is negligible beside it, which is why exam questions about electrons between plates ignore weight.
field lines 场线 point the way the force acts on a positive test charge.
lines start on positive charges and end on negative charges (or go to infinity).
lines never cross; closer lines mean a stronger field.
every line carries an arrow; lines meet the surface of a conductor at right angles.
What a field line represents. Asked "state what is represented by an electric field line", give both halves: its direction shows the direction of the force on a positive charge placed there, and the spacing of the lines shows the strength of the field (closer lines, stronger field). A sketch is marked on exactly these points: arrows drawn, lines not crossing, even spacing where the field is uniform, and radial lines for a point charge or a sphere.
Examples: a positive point charge 点电荷 has radial lines pointing out; a negative one has lines pointing in; two opposite charges (a dipole 偶极子) have lines curving from + to −; two parallel charged plates give a uniform field 匀强场 of equally spaced parallel lines.
The charge on an isolated 孤立的 conductor 导体 sits on its outer surface, spread evenly when the sphere is on its own. Outside the sphere the field lines are radial, evenly spaced and pointing outwards (for positive charge), exactly the pattern of a point charge at the centre. Inside a hollow (or solid) conductor the field is zero: the fields of all the surface charges cancel everywhere inside, so no field line enters. Two graphs follow, and both are exam favourites. The field strength $E$ is zero out to the radius $R$, jumps to its surface value $E_0$ and then falls as $1/r^{2}$; the potential $V$ is constant inside (a zero field means a zero potential gradient) and then falls as $1/r$.
Reading the graphs. Given an $E$–$x$ graph for a charged sphere, the radius is the distance at which $E$ jumps from zero, and the charge comes from any point on the curve: $Q = 4\pi\varepsilon_{0} r^{2} E$. Given "$E_0$ at the surface", the value at $2R$ is $E_0/4$ and at $3R$ is $E_0/9$: sketch the curve through those points, starting at $E_0$ on the surface line, never from the origin.
Sebuah medan listrik adalah wilayah di mana muatan merasakan gaya dari muatan lain. Kekuatan medan listrik $E$ pada suatu titik adalah gaya per satuan muatan positif pada sebuah muatan uji positif kecil yang ditempatkan di sana:
Satuan: $\text{N C}^{-1}$ (sama dengan $\text{V m}^{-1}$, seperti yang akan kita lihat). $E$ adalah vektor, menunjukkan arah gaya bekerja pada muatan positif. Gaya pada muatan $q$ adalah
berlawanan dengan medan jika $q$ negatif.
Definisi bernilai dua poin. Kekuatan medan listrik pada suatu titik adalah gaya per satuan muatan positif yang bekerja pada muatan kecil (uji) yang ditempatkan di titik tersebut. Tiga kata membawa nilai: gaya per satuan muatan (bukan "gaya pada muatan"), positif (yang menentukan arah), dan kecil atau uji (agar muatan tidak mengganggu medan yang diukur). Versi pilihan ganda menawarkan "gaya per satuan muatan yang bekerja pada massa kecil" dan "gaya per satuan massa" sebagai pengecoh: medan didefinisikan oleh muatan, bukan oleh massa.
Contoh terpecahkan. Sebuah proton dipercepat pada $2.00\ \text{m s}^{-2}$ dalam medan listrik, tanpa gaya lain yang bekerja. Temukan kuat medan. ($m_{\text{p}} = 1.67 \times 10^{-27}\ \text{kg}$, $e = 1.60 \times 10^{-19}\ \text{C}$.)
$qE = ma$, jadi
Pengaturan ulang yang sama secara terbalik memberikan percepatan elektron dalam medan $1500\ \text{V m}^{-1}$: $a = eE/m_{\text{e}} = (1.60 \times 10^{-19})(1500)/(9.11 \times 10^{-31}) = 2.6 \times 10^{14}\ \text{m s}^{-2}$, sangat besar karena massa elektron sangat kecil. Gravitasi ($9.81\ \text{m s}^{-2}$) dapat diabaikan dibandingkan dengan itu, itulah sebabnya soal ujian tentang elektron antara pelat mengabaikan beratnya.
garis medan menunjukkan arah gaya yang bekerja pada muatan uji positif.
garis dimulai dari muatan positif dan berakhir pada muatan negatif (atau menuju tak hingga).
garis tidak pernah saling berpotongan; garis yang lebih rapat berarti medan yang lebih kuat.
setiap garis membawa panah; garis bertemu permukaan konduktor secara tegak lurus.
Apa yang diwakili oleh garis medan. Saat ditanya "nyatakan apa yang diwakili oleh garis medan listrik", berikan kedua bagian: arah-nya menunjukkan arah gaya pada muatan positif yang diletakkan di sana, dan jarak antar garis menunjukkan kekuatan medan (garis lebih rapat, medan lebih kuat). Sketsa diberi nilai tepat pada poin-poin ini: panah digambar, garis tidak bersilangan, jarak merata di mana medan seragam, dan garis radial untuk muatan titik atau bola.
Contoh: muatan titik positif memiliki garis radial mengarah keluar; yang negatif memiliki garis mengarah masuk; dua muatan berlawanan (dipol) memiliki garis melengkung dari + ke −; dua pelat bermuatan sejajar menghasilkan medan seragam berupa garis sejajar dengan jarak sama.
Muatan pada konduktor terisolasi berada di permukaan luarnya, tersebar merata ketika bola tersebut berdiri sendiri. Di luar bola, garis medannya radial, berjarak merata, dan mengarah keluar (untuk muatan positif), persis seperti pola muatan titik di pusat. Di dalam konduktor berongga (atau solid), medannya nol: medan dari semua muatan permukaan saling meniadakan di mana saja di dalam, sehingga tidak ada garis medan yang masuk. Dua grafik mengikuti, dan keduanya merupakan favorit soal ujian. Kuat medan $E$ adalah nol hingga jari-jari $R$, melompat ke nilai permukaannya $E_0$, dan kemudian turun sebagai $1/r^{2}$; potensial $V$ adalah konstan di dalam (medan nol berarti gradien potensial nol) dan kemudian turun sebagai $1/r$.
Membaca grafik. Diberikan grafik $E$–$x$ untuk bola bermuatan, jari-jari adalah jarak di mana $E$ melompat dari nol, dan muatan didapat dari titik manapun pada kurva: $Q = 4\pi\varepsilon_{0} r^{2} E$. Diberikan "$E_0$ di permukaan", nilai pada $2R$ adalah $E_0/4$ dan pada $3R$ adalah $E_0/9$: gambar kurva melalui titik-titik tersebut, dimulai dari $E_0$ pada garis permukaan, bukan dari titik asal.
E ∝ Q / r²
A charge sets up a radial field — out for +, in for −, obeying the inverse-square law. · Muatan membentuk medan radial — keluar untuk +, masuk untuk −, mematuhi hukum kuadrat terbalik.
| English | Bahasa Indonesia |
|---|---|
| electric field/ɪˈlektrɪk fiːld/ | medan listrik |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Between two parallel plates a distance $d$ apart with potential difference 电势差 $V$ between them, the field is uniform (apart from edge effects) with size
It points from the higher-potential plate to the lower one. The unit $\text{V m}^{-1}$ comes straight from this and equals $\text{N C}^{-1}$.
Worked example. Two parallel plates $5.0\ \text{mm}$ apart have a p.d. of $200\ \text{V}$ between them. Find the field strength, and the force on an electron in the gap. ($e = 1.6 \times 10^{-19}\ \text{C}$.)
Which changes the field? Two things only: $E = V/d$ rises if the p.d. is increased or the plates are moved closer. A resistor in series with the supply changes nothing: no current flows once the plates are charged, so there is no p.d. across the resistor and the plates sit at the full supply voltage. Between plates at $+800\ \text{V}$ and $+1300\ \text{V}$ the field points from the higher potential to the lower; two positive potentials change nothing about the direction rule.
A charge $q$ in a uniform field feels a constant force $F = qE$, so a constant acceleration $a = qE/m$ — just like a mass in a uniform gravitational field.
Two forces, not one. A charged oil drop "held stationary" between horizontal plates is the exam's favourite equilibrium 平衡: both an electric force and its weight act on it (the multiple-choice distractor is "electric force only"), and they are equal and opposite. For a negative drop the electric force is opposite to the field, so the top plate must be positive to hold it up.
Worked example. An oil drop of mass $2.6 \times 10^{-15}\ \text{kg}$ carries a charge of $-4.8 \times 10^{-19}\ \text{C}$ and is held stationary in a vacuum between horizontal plates $2.0\ \text{cm}$ apart. Find the p.d. between the plates, and say which plate is positive.
The drop is negative, so the force on it is against the field; for the force to be upwards the field must point down, so the top plate is positive. Notice that the charge is $3e$, three excess electrons, which is how Millikan showed that charge comes in multiples of $e$.
Worked example. Two parallel plates in a vacuum are $0.041\ \text{m}$ apart with a p.d. of $250\ \text{V}$ between them. An electron is released from rest at the negative plate. Find (a) the field strength, (b) the force on the electron and its acceleration, (c) the time it takes to reach the positive plate, (d) its kinetic energy on arrival. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$.)
(a) $E = V/d = 250/0.041 = 6.1 \times 10^{3}\ \text{V m}^{-1}$. (b) $F = eE = (1.60 \times 10^{-19})(6.1 \times 10^{3}) = 9.8 \times 10^{-16}\ \text{N}$; $a = F/m_{\text{e}} = 1.07 \times 10^{15}\ \text{m s}^{-2}$, towards the positive plate. (c) From rest with constant acceleration, $d = \tfrac{1}{2} a t^{2}$, so $t = \sqrt{2d/a} = \sqrt{2 \times 0.041 / (1.07 \times 10^{15})} = 8.8 \times 10^{-9}\ \text{s}$. (d) Energy method, no kinematics needed: $E_{\text{k}} = qV = (1.60 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$, so $v = \sqrt{2E_{\text{k}}/m_{\text{e}}} = 9.4 \times 10^{6}\ \text{m s}^{-1}$. (Check: $v = at = (1.07 \times 10^{15})(8.8 \times 10^{-9}) = 9.4 \times 10^{6}\ \text{m s}^{-1}$.)
Worked example (deflection). An electron travelling horizontally at $2.0 \times 10^{7}\ \text{m s}^{-1}$ enters the field between two horizontal plates $18\ \text{mm}$ apart with a p.d. of $400\ \text{V}$. The plates are $30\ \text{mm}$ long. Find the deflection 偏转 of the electron by the time it leaves the field.
The horizontal motion is unaffected: time in the field $t = 0.030/(2.0 \times 10^{7}) = 1.5 \times 10^{-9}\ \text{s}$. Vertically, $a = eE/m_{\text{e}} = eV/(m_{\text{e}} d) = (1.60 \times 10^{-19})(400)/[(9.11 \times 10^{-31})(0.018)] = 3.9 \times 10^{15}\ \text{m s}^{-2}$, so the deflection is $y = \tfrac{1}{2} a t^{2} = \tfrac{1}{2}(3.9 \times 10^{15})(1.5 \times 10^{-9})^{2} = 4.4 \times 10^{-3}\ \text{m}$, less than the half-gap of $9\ \text{mm}$, so the electron does leave the field. Its vertical velocity on exit is $at = 5.9 \times 10^{6}\ \text{m s}^{-1}$, so it leaves at $\tan\theta = 5.9/20$, about $16°$; the path is a parabola, and after the plates (no field) it travels in a straight line in the direction it had on exit.
Antara dua pelat sejajar berjarak $d$ dengan beda potensial $V$ di antaranya, medannya seragam (kecuali efek tepi) dengan ukuran
Garisnya mengarah dari pelat berpotensial tinggi ke pelat berpotensial rendah. Satuan $\text{V m}^{-1}$ berasal langsung dari hal ini dan setara dengan $\text{N C}^{-1}$.
Contoh terpecahkan. Dua pelat sejajar $5.0\ \text{mm}$ berjarak memiliki beda potensial $200\ \text{V}$ di antaranya. Temukan kuat medan, dan gaya pada elektron di celah. ($e = 1.6 \times 10^{-19}\ \text{C}$.)
Apa yang mengubah medan? Hanya dua hal: $E = V/d$ meningkat jika beda potensial ditingkatkan atau pelat digerakkan lebih dekat. Resistor seri dengan sumber tidak mengubah apa pun: tidak ada arus mengalir setelah pelat terisi, sehingga tidak ada beda potensial melintasi resistor dan pelat berada pada tegangan sumber penuh. Antara pelat pada $+800\ \text{V}$ dan $+1300\ \text{V}$, medan mengarah dari potensi lebih tinggi ke yang lebih rendah; dua potensi positif tidak mengubah aturan arah.
Muatan $q$ dalam medan seragam mengalami gaya konstan $F = qE$, sehingga percepatan konstan $a = qE/m$ — sama seperti massa dalam medan gravitasi seragam.
Dua gaya, bukan satu. Tetesan minyak "diam" di antara pelat horizontal adalah kesetimbangan favorit dalam ujian: kedua gaya listrik dan beratnya bekerja padanya (pilihan pengecoh pilihan ganda adalah "hanya gaya listrik"), dan keduanya sama besar serta berlawanan arah. Untuk tetesan negatif, gaya listrik berlawanan dengan medan, sehingga pelat atas harus positif untuk menahannya agar tetap di atas.
Contoh terarah. Sebuah tetesan minyak bermassa $2.6 \times 10^{-15}\ \text{kg}$ membawa muatan $-4.8 \times 10^{-19}\ \text{C}$ dan diam di ruang hampa di antara dua pelat horizontal $2.0\ \text{cm}$ jaraknya. Tentukan beda potensial antara pelat-pelat tersebut, dan sebutkan pelat mana yang positif.
Tetesan bermuatan negatif, sehingga gayanya berlawanan dengan medan; agar gayanya ke atas, medan harus mengarah ke bawah, sehingga pelat atas positif. Perhatikan bahwa muatannya $3e$, yaitu tiga elektron berlebih, inilah yang ditunjukkan Millikan bahwa muatan datang dalam kelipatan $e$.
Contoh terarah. Dua pelat sejajar dalam ruang hampa berjarak $0.041\ \text{m}$ dengan beda potensial $250\ \text{V}$ di antara keduanya. Sebuah elektron dilepaskan dari keadaan diam di pelat negatif. Tentukan (a) kuat medan, (b) gaya pada elektron dan percepatannya, (c) waktu yang dibutuhkan untuk mencapai pelat positif, (d) energi kinetiknya saat tiba. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$.)
(a) $E = V/d = 250/0.041 = 6.1 \times 10^{3}\ \text{V m}^{-1}$. (b) $F = eE = (1.60 \times 10^{-19})(6.1 \times 10^{3}) = 9.8 \times 10^{-16}\ \text{N}$; $a = F/m_{\text{e}} = 1.07 \times 10^{15}\ \text{m s}^{-2}$, menuju pelat positif. (c) Dari keadaan diam dengan percepatan konstan, $d = \tfrac{1}{2} a t^{2}$, sehingga $t = \sqrt{2d/a} = \sqrt{2 \times 0.041 / (1.07 \times 10^{15})} = 8.8 \times 10^{-9}\ \text{s}$. (d) Metode energi, tidak perlu kinematika: $E_{\text{k}} = qV = (1.60 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$, sehingga $v = \sqrt{2E_{\text{k}}/m_{\text{e}}} = 9.4 \times 10^{6}\ \text{m s}^{-1}$. (Periksa: $v = at = (1.07 \times 10^{15})(8.8 \times 10^{-9}) = 9.4 \times 10^{6}\ \text{m s}^{-1}$.)
Contoh terarah (pembiasan). Sebuah elektron bergerak secara horizontal dengan kecepatan $2.0 \times 10^{7}\ \text{m s}^{-1}$ memasuki medan di antara dua pelat horizontal $18\ \text{mm}$ jaraknya dengan beda potensial $400\ \text{V}$. Pelat-pelat tersebut memiliki panjang $30\ \text{mm}$. Tentukan pembiasan elektron hingga ia meninggalkan medan tersebut.
Gerakan horizontal tidak terpengaruh: waktu di dalam medan $t = 0.030/(2.0 \times 10^{7}) = 1.5 \times 10^{-9}\ \text{s}$. Secara vertikal, $a = eE/m_{\text{e}} = eV/(m_{\text{e}} d) = (1.60 \times 10^{-19})(400)/[(9.11 \times 10^{-31})(0.018)] = 3.9 \times 10^{15}\ \text{m s}^{-2}$, sehingga pembiasannya adalah $y = \tfrac{1}{2} a t^{2} = \tfrac{1}{2}(3.9 \times 10^{15})(1.5 \times 10^{-9})^{2} = 4.4 \times 10^{-3}\ \text{m}$, kurang dari separuh celah $9\ \text{mm}$, sehingga elektron dapat meninggalkan medan. Kecepatan vertikalnya saat keluar adalah $at = 5.9 \times 10^{6}\ \text{m s}^{-1}$, sehingga ia keluar pada sudut $\tan\theta = 5.9/20$, sekitar $16°$; lintasannya berbentuk parabola, dan setelah melewati pelat (tanpa medan) ia bergerak lurus searah dengan arah saat keluar.
Follow how a charge behaves between parallel plates. · Ikuti perilaku muatan antara pelat sejajar.
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
For two point charges $Q_{1}$ and $Q_{2}$ a distance $r$ apart in free space, each feels a force of size
This is Coulomb's law 库仑定律. Here $\varepsilon_{0} = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ is the permittivity of free space 真空电容率, and $1/(4\pi\varepsilon_{0}) \approx 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$. The force is along the line joining the charges: repulsive for like charges, attractive for opposite charges.
Stating the law in words. "State Coulomb's law" wants a sentence, not a formula: the (electric) force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation. Both proportionalities are needed, and the charges must be point charges (or spheres treated as point charges at their centres) in a vacuum.
Worked example. Find the electrostatic force between point charges of $+2.0\ \text{nC}$ and $+3.0\ \text{nC}$ placed $4.0\ \text{cm}$ apart. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
Worked example. In a hydrogen atom the proton and the electron may be treated as point charges $120\ \text{pm}$ apart. Find the electric force between them, and compare it with their gravitational attraction. ($m_{\text{p}} = 1.67 \times 10^{-27}\ \text{kg}$, $m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)
Gravity: $F_{\text{G}} = G m_{\text{p}} m_{\text{e}}/r^{2} = (6.67 \times 10^{-11})(1.67 \times 10^{-27})(9.11 \times 10^{-31})/(1.20 \times 10^{-10})^{2} = 7.0 \times 10^{-48}\ \text{N}$, about $10^{39}$ times smaller. Inside atoms, gravity is irrelevant.
Worked example. Two identical oil droplets in a vacuum have their centres $3.8 \times 10^{-6}\ \text{m}$ apart. Each carries the same charge, and they repel with a force of $1.0 \times 10^{-15}\ \text{N}$. Find the charge on each droplet, and the number of excess electrons it carries.
that is $1.3 \times 10^{-18}/1.60 \times 10^{-19} \approx 8$ electrons. Identical objects with the same sign of charge always repel; do not write "attract" for the direction.
Worked example. A charged sphere X is fixed on an insulating stand. A second sphere Y, of mass $2.0\ \text{g}$, hangs beside it on an insulating thread and settles in equilibrium with the thread at $12°$ to the vertical, its centre $5.0\ \text{cm}$ from the centre of X on the same horizontal level. Find the electric force on Y and, if the spheres carry equal charges, the charge on each.
Three forces act on Y: its weight $mg$ (down), the tension 张力 $T$ (along the thread) and the electric force $F$ (horizontal, away from X because like charges repel). Resolving, $T\cos\theta = mg$ and $T\sin\theta = F$, so
Then $F = Q^{2}/(4\pi\varepsilon_{0} r^{2})$ gives $Q = \sqrt{4\pi\varepsilon_{0} F r^{2}} = \sqrt{(4.2 \times 10^{-3})(0.050)^{2}/(8.99 \times 10^{9})} = 3.4 \times 10^{-8}\ \text{C}$. A closed vector triangle of the three forces is an equally good method; the exam accepts either, but it must show all three forces.
Worked example (helium). A helium atom may be modelled as a nucleus of charge $+2e$ with two electrons in diametrically opposite circular orbits of radius $170\ \text{pm}$. Find the resultant force 合力 on one electron, and hence its orbital speed.
Two forces act on the electron: attraction to the nucleus, at distance $r$, and repulsion from the other electron, at distance $2r$ on the far side:
towards the nucleus. This resultant is the centripetal force, $F = m_{\text{e}} v^{2}/r$, so $v = \sqrt{F r/m_{\text{e}}} = \sqrt{(1.4 \times 10^{-8})(1.70 \times 10^{-10})/(9.11 \times 10^{-31})} = 1.6 \times 10^{6}\ \text{m s}^{-1}$. The trap is to forget the second electron, or to put it at distance $r$ rather than $2r$.
A spherical conductor with total charge $Q$ gives, at any point outside, the same field as a point charge $Q$ at its centre (measure $r$ from the centre). Inside a hollow charged conductor the field is zero, so the conductor is an equipotential 等势面.
Untuk dua muatan titik $Q_{1}$ dan $Q_{2}$ yang berjarak $r$ di ruang bebas, masing-masing merasakan gaya sebesar
Ini adalah Hukum Coulomb. Di sini $\varepsilon_{0} = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ adalah permitivitas ruang bebas, dan $1/(4\pi\varepsilon_{0}) \approx 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$. Gaya ini sepanjang garis penghubung kedua muatan: tolak-menolak untuk muatan sejenis, tarik-menarik untuk muatan berbeda jenis.
Menyatakan hukum dengan kata-kata. "Nyatakan Hukum Coulomb" meminta sebuah kalimat, bukan rumus: gaya (listrik) antara dua muatan titik berbanding lurus dengan hasil kali muatan-muatan tersebut dan berbanding terbalik dengan kuadrat jarak pisahnya. Kedua perbandingan proporsionalitas diperlukan, dan muatan-muatannya harus berupa muatan titik (atau bola yang diperlakukan sebagai muatan titik di pusatnya) dalam ruang hampa.
Contoh terarah. Tentukan gaya elektrostatik antara dua muatan titik $+2.0\ \text{nC}$ dan $+3.0\ \text{nC}$ yang diletakkan berjarak $4.0\ \text{cm}$. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
Contoh terarah. Dalam atom hidrogen, proton dan elektron dapat diperlakukan sebagai muatan titik $120\ \text{pm}$ jaraknya. Tentukan gaya listrik di antara keduanya, dan bandingkan dengan tarikan gravitasinya. ($m_{\text{p}} = 1.67 \times 10^{-27}\ \text{kg}$, $m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)
Gravitasi: $F_{\text{G}} = G m_{\text{p}} m_{\text{e}}/r^{2} = (6.67 \times 10^{-11})(1.67 \times 10^{-27})(9.11 \times 10^{-31})/(1.20 \times 10^{-10})^{2} = 7.0 \times 10^{-48}\ \text{N}$, sekitar $10^{39}$ kali lebih kecil. Di dalam atom, gravitasi dapat diabaikan.
Contoh terarah. Dua tetesan minyak identik dalam ruang hampa berjarak $3.8 \times 10^{-6}\ \text{m}$ antar pusatnya. Masing-masing membawa muatan yang sama, dan mereka saling tolak dengan gaya $1.0 \times 10^{-15}\ \text{N}$. Tentukan muatan pada setiap tetesan, dan jumlah elektron berlebih yang dimilikinya.
yaitu $1.3 \times 10^{-18}/1.60 \times 10^{-19} \approx 8$ elektron. Benda identik dengan tanda muatan yang sama selalu tolak-menolak; jangan tulis "tarik-menarik" untuk arahnya.
Contoh terarah. Sebuah bola X bermuatan tetap di atas tiang isolator. Bola Y kedua, bermassa $2.0\ \text{g}$, tergantung di sampingnya pada benang isolator dan setimbang dengan benang membentuk sudut $12°$ terhadap vertikal, pusatnya $5.0\ \text{cm}$ dari pusat X pada ketinggian horizontal yang sama. Tentukan gaya listrik pada Y dan, jika bola-bola tersebut bermuatan sama, muatan pada masing-masing.
Tiga gaya bekerja pada Y: beratnya $mg$ (ke bawah), tegangan $T$ (sepanjang benang) dan gaya listrik $F$ (horizontal, menjauhi X karena muatan sejenis tolak-menolak). Resolusi komponen, $T\cos\theta = mg$ dan $T\sin\theta = F$, sehingga
Kemudian $F = Q^{2}/(4\pi\varepsilon_{0} r^{2})$ menghasilkan $Q = \sqrt{4\pi\varepsilon_{0} F r^{2}} = \sqrt{(4.2 \times 10^{-3})(0.050)^{2}/(8.99 \times 10^{9})} = 3.4 \times 10^{-8}\ \text{C}$. Segitiga vektor tertutup dari ketiga gaya merupakan metode yang sama baiknya; ujian menerima keduanya, tetapi harus menampilkan semua tiga gaya.
Contoh terarah (helium). Atom helium dapat dimodelkan sebagai inti bermuatan $+2e$ dengan dua elektron dalam orbit melingkar berlawanan diameter berjari-jari $170\ \text{pm}$. Tentukan resultan gaya pada salah satu elektron, dan karenanya kecepatannya dalam orbit.
Dua gaya bekerja pada elektron: tarik-menarik ke inti, pada jarak $r$, dan tolak-menolak dari elektron lain, pada jarak $2r$ di sisi seberang:
menuju inti. Resultan ini adalah gaya sentripetal, $F = m_{\text{e}} v^{2}/r$, sehingga $v = \sqrt{F r/m_{\text{e}}} = \sqrt{(1.4 \times 10^{-8})(1.70 \times 10^{-10})/(9.11 \times 10^{-31})} = 1.6 \times 10^{6}\ \text{m s}^{-1}$. Jebakanannya adalah melupakan elektron kedua, atau meletakkannya pada jarak $r$ bukan $2r$.
Konduktor bola dengan total muatan $Q$ memberikan, pada setiap titik di luar, medan yang sama dengan muatan titik $Q$ di pusatnya (ukur $r$ dari pusat). Di dalam konduktor berongga bermuatan, medannya nol, sehingga konduktor tersebut adalah equipotensial.
F ∝ Qq / r²
Like charges repel, unlike attract — and the force follows 1/r². · Muatan sejenis tolak-menolak, muatan berlawanan tarik-menarik — dan gayanya mengikuti 1/r².
| English | Bahasa Indonesia |
|---|---|
| force/fɔːs/ | gaya |
| electric field strength/ɪˈlektrɪk fiːld streŋθ/ | kekuatan medan listrik |
| test charge/test tʃɑːdʒ/ | muatan uji |
| vector/ˈvektə/ | vektor |
| electron/ɪˈlektrɒn/ | elektron |
| field line/fiːld laɪn/ | garis medan |
| infinity/ɪnˈfɪnɪti/ | tak hingga |
| conductor/kənˈdʌktə/ | konduktor |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Coulomb's law gives the force between two charges. Divide it by the test charge ($E = F/q$) and you are left with the field of the source charge $Q$ alone. The field at distance $r$ from a point charge $Q$ is
It points out from a positive $Q$, in towards a negative $Q$, and falls as $1/r^{2}$ — just like gravitational field 重力场 strength, except gravity is always attractive. For several charges, add the fields as a vector sum 矢量和.
Worked example. The Earth may be treated as a uniform conducting sphere of radius $6.37 \times 10^{6}\ \text{m}$ carrying a total charge of $-4.80 \times 10^{5}\ \text{C}$ spread over its surface. Find the electric field strength at the surface, and compare it with the gravitational field strength there. ($M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}$.)
directed towards the centre (the charge is negative). Gravity: $g = GM/R^{2} = 9.83\ \text{N kg}^{-1}$, so $E/g \approx 11$; in symbols $E/g = Q/(4\pi\varepsilon_{0} G M)$, which shows that the ratio depends only on the charge-to-mass ratio of the sphere. A charged sphere of mass $m$ and charge $q$ would need $qE = mg$ to float in this field: $q/m = g/E = 0.093\ \text{C kg}^{-1}$.
Worked example. Point charges of $+4.0\ \text{nC}$ and $+1.0\ \text{nC}$ are $30\ \text{cm}$ apart. Where on the line joining them is the resultant field zero?
Between the charges the two fields point in opposite directions; call the distance from the $4.0\ \text{nC}$ charge $x$. Zero resultant needs equal magnitudes:
twice as far from the larger charge. For opposite charges the fields between them point the same way and never cancel: the zero lies outside, beyond the smaller charge.
Hukum Coulomb memberikan gaya antara dua muatan. Bagi dengan muatan uji ($E = F/q$) dan Anda akan mendapatkan medan dari muatan sumber $Q$ sendirian. Medan pada jarak $r$ dari muatan titik $Q$ adalah
Medan ini mengarah keluar dari muatan positif $Q$, masuk menuju muatan negatif $Q$, dan berkurang sebanding dengan $1/r^{2}$ — persis seperti kekuatan medan gravitasi, kecuali gravitasi selalu menarik. Untuk beberapa muatan, jumlahkan medan sebagai jumlah vektor.
Contoh terpecahkan. Bumi dapat diperlakukan sebagai bola konduktor seragam berjari-jari $6.37 \times 10^{6}\ \text{m}$ yang membawa total muatan $-4.80 \times 10^{5}\ \text{C}$ yang tersebar di permukaannya. Temukan kekuatan medan listrik di permukaan, dan bandingkan dengan kekuatan medan gravitasi di sana. ($M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}$.)
terarah menuju pusat (muatannya negatif). Gravitasi: $g = GM/R^{2} = 9.83\ \text{N kg}^{-1}$, sehingga $E/g \approx 11$; dalam simbol $E/g = Q/(4\pi\varepsilon_{0} G M)$, yang menunjukkan bahwa rasio hanya bergantung pada rasio muatan-terhadap-massa bola. Bola bermuatan dengan massa $m$ dan muatan $q$ akan membutuhkan $qE = mg$ untuk melayang di medan ini: $q/m = g/E = 0.093\ \text{C kg}^{-1}$.
Contoh terpecahkan. Muatan titik $+4.0\ \text{nC}$ dan $+1.0\ \text{nC}$ terpisah sejauh $30\ \text{cm}$. Di mana pada garis penghubung keduanya resultan medannya nol?
Antara kedua muatan, kedua medan mengarah berlawanan arah; sebut jarak dari muatan $4.0\ \text{nC}$ adalah $x$. Resultan nol memerlukan besar yang sama:
dua kali lebih jauh dari muatan yang lebih besar. Untuk muatan berlawanan tanda, medan di antara mereka mengarah sama dan tidak pernah saling meniadakan: titik nol berada di luar, melewati muatan yang lebih kecil.
Change the charge. Field lines point away from positive and toward negative, and crowd together where the field is strongest. · Ubah muatannya. Garis-garis medan mengarah jauh dari positif dan menuju negatif, serta merapat di tempat medan paling kuat.
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Electric potential 电势 $V$ at a point is the work done per unit positive charge in bringing a small positive test charge from infinity 无穷远 to that point:
Unit: $\text{V}$. The potential is zero at infinity. For a positive source charge $V > 0$ everywhere outside; for a negative source charge $V < 0$.
The two-mark definition. Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to the point. The marks are for work done per unit (positive) charge and from infinity to the point. The potential of a positive charge is positive because work must be done on a positive test charge to push it in against the repulsion; near a negative charge the field does the work, so the potential is negative. "Explain why the potential near an isolated proton is positive" is answered in exactly those words.
Note the $1/r$ here (compared with $1/r^{2}$ for the field). $V$ is a scalar 标量; for several charges, add the potentials (with sign).
Worked example. Find the electric potential $4.0\ \text{cm}$ from a point charge of $+3.0\ \text{nC}$. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
Because potential is a scalar, the potential from several charges is simply their sum (each with its own sign) — no directions to resolve.
The exam's favourite structured question puts two charged spheres X and Y some distance apart and draws the potential $V$ along the line joining their centres (or asks you to reason from it). Everything follows from two facts: potential is a scalar that adds with sign, and the field is the negative gradient of the graph.
Worked example. Two isolated charged metal spheres X and Y have their centres $1.2\ \text{m}$ apart in a vacuum. The potential along the line between them is positive everywhere and has a minimum at $x = 0.50\ \text{m}$ from the centre of X. State three conclusions, and then find the force on a proton held at $x = 0.60\ \text{m}$, where the gradient of the graph is $+180\ \text{V m}^{-1}$.
Conclusions: (1) both spheres are positively charged, because the potential is positive everywhere; (2) the electric field is zero at $x = 0.50\ \text{m}$, where the gradient is zero; (3) the charge on Y is larger, because the zero-field point lies closer to X; in fact $Q_{\text{Y}}/Q_{\text{X}} = (0.70/0.50)^{2} = 2.0$. At $x = 0.60\ \text{m}$: $E = -180\ \text{V m}^{-1}$ (pointing towards X, since $V$ rises towards Y), so $F = eE = (1.60 \times 10^{-19})(180) = 2.9 \times 10^{-17}\ \text{N}$ towards X. Released, the proton accelerates towards X, its acceleration increasing as the gradient steepens; it needs an external force to hold it still.
Describing the motion. "A positively charged particle is placed at P and released. Describe and explain its motion." The examiner wants: the direction (towards lower potential, down the slope of the $V$ graph); that the force, and so the acceleration, is not constant (it follows the gradient, which changes along the line); and that the particle speeds up throughout, since the force stays in the direction of motion. If P is the zero-field point the particle stays at rest, in unstable equilibrium: a nudge either way sends it off.
The field equals the negative potential gradient 电势梯度:
Between parallel plates $V$ changes evenly with position, giving $E = V/d$ as before. The minus sign means the field points towards lower potential. For a point charge, $-\dfrac{dV}{dr} = \dfrac{Q}{4\pi\varepsilon_{0} r^{2}} = E$.
The one-line relationship. "State the relationship between electric field and electric potential": the field strength is equal to the negative of the potential gradient, $E = -\Delta V/\Delta x$. On a $V$–$x$ graph $E$ is minus the gradient of a tangent; on a $V$–$r$ graph for a point charge the gradient is steepest close to the charge, where the field is strongest, and the sign of $E$ is fixed by "the field points towards lower potential". The uniform-field version $E = V/d$ is the same statement with a constant gradient.
Equipotentials. An equipotential surface joins points of equal potential; no work is done moving a charge along it, so the field is always at right angles to it. Around a point charge the equipotentials are concentric spheres, growing further apart as $V$ falls; between parallel plates they are planes parallel to the plates. The surface of a conductor is an equipotential, which is why field lines meet it at right angles.
A charge $q$ at a point of potential $V$ has electric potential energy 电势能 $E_{\text{P}} = qV$. For two point charges $Q$ and $q$ a distance $r$ apart:
In both cases $E_{\text{P}} \to 0$ as $r \to \infty$.
Worked example. The proton and the electron in a hydrogen atom are $5.3 \times 10^{-11}\ \text{m}$ apart. Find the electric potential energy of the pair.
The negative sign is part of the answer: the charges are opposite, so energy ($4.3 \times 10^{-18}\ \text{J}$) must be supplied to pull them apart to infinity. (The electron also has kinetic energy of half that size, so the energy needed to ionise 电离 the atom is $13.6\ \text{eV}$; the "worked-example pattern" below shows why.)
Charge through a p.d. A charge $q$ moved through a potential difference $V$ changes its potential energy by $qV$, which is why an electron accelerated from rest through $250\ \text{V}$ arrives with $250\ \text{eV}$ of kinetic energy whatever the shape of the field. For an MCQ about "moving P a distance $x$ along the field lines", the work done is $qEx$, and its potential energy falls if it moves the way the force pushes it.
An electron 电子 orbits a nucleus 原子核 of charge $+Ze$ at distance $r$. The Coulomb attraction provides the centripetal force 向心力:
The total energy is kinetic plus potential:
which is negative (a bound state).
The two field theories look alike:
| Quantity | Gravitational | Electric |
|---|---|---|
| Source | mass $M$ (always positive) | charge $Q$ (can be ±) |
| Field strength | $g = GM/r^{2}$ | $E = Q/(4\pi\varepsilon_{0} r^{2})$ |
| Force on test object | $F = mg$ | $F = qE$ |
| Potential | $\phi = -GM/r$ | $V = Q/(4\pi\varepsilon_{0} r)$ |
| PE of two | $-GMm/r$ | $Qq/(4\pi\varepsilon_{0} r)$ |
| Nature | always attractive | attractive or repulsive |
The minus sign in the gravitational potential 引力势 reflects that gravity is always attractive; the electric potential takes the sign of the source charge.
Similarity and difference (a standard two-marker). Both potentials are proportional to $1/r$, both are zero at infinity, and both are scalars; but gravitational potential is always negative (the force is always attractive) whereas electric potential takes the sign of the charge and can be positive. For the fields: both obey an inverse-square law and both are drawn as radial lines around a point source, but gravitational field lines only ever point inwards (attraction), while electric field lines point outwards from a positive charge and inwards to a negative one.
Potensial listrik $V$ di suatu titik adalah usaha per satuan muatan positif untuk memindahkan muatan uji positif kecil dari tak hingga ke titik tersebut:
Unit: $\text{V}$. Potensial nol di tak hingga. Untuk muatan sumber positif $V > 0$ di mana pun di luar; untuk muatan sumber negatif $V < 0$.
Definisi dua poin. Potensial listrik di suatu titik adalah usaha per satuan muatan positif untuk memindahkan muatan uji kecil dari tak hingga ke titik tersebut. Poin diberikan untuk usaha per satuan (positif) muatan dan dari tak hingga ke titik. Potensial muatan positif bernilai positif karena usaha harus dilakukan pada muatan uji positif untuk mendorongnya masuk melawan tolakan; dekat muatan negatif, medan melakukan usaha, sehingga potensialnya negatif. "Jelaskan mengapa potensial di dekat proton terisolasi bernilai positif" dijawab tepat dengan kata-kata tersebut.
Perhatikan $1/r$ di sini (dibandingkan dengan $1/r^{2}$ untuk medan). $V$ adalah skalar; untuk beberapa muatan, jumlahkan potensialnya (dengan tandanya).
Contoh terpecahkan. Temukan potensial listrik $4.0\ \text{cm}$ dari muatan titik $+3.0\ \text{nC}$. ($1/(4\pi\varepsilon_{0}) = 9.0 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2}$.)
Karena potensial adalah skalar, potensial dari beberapa muatan hanyalah jumlah mereka (masing-masing dengan tandanya sendiri) — tidak ada komponen arah yang perlu diuraikan.
Soal terstruktur favorit ujian menempatkan dua bola bermuatan X dan Y pada jarak tertentu dan menggambar potensial $V$ sepanjang garis penghubung pusatnya (atau meminta Anda menyimpulkannya). Semua hal mengikuti dari dua fakta: potensial adalah skalar yang dijumlahkan dengan tanda, dan medan adalah gradien negatif dari grafik.
Contoh terpecahkan. Dua bola logam bermuatan terisolasi X dan Y berjarak $1.2\ \text{m}$ satu sama lain di ruang hampa. Potensial sepanjang garis di antara keduanya positif di mana saja dan memiliki minimum pada $x = 0.50\ \text{m}$ dari pusat X. Nyatakan tiga kesimpulan, lalu temukan gaya pada proton yang dijaga pada $x = 0.60\ \text{m}$, di mana gradien grafiknya adalah $+180\ \text{V m}^{-1}$.
Kesimpulan: (1) kedua bola bermuatan positif, karena potensial positif di mana-mana; (2) medan listrik nol pada $x = 0.50\ \text{m}$, di mana gradiennya nol; (3) muatan pada Y lebih besar, karena titik nol-medan berada lebih dekat ke X; sebenarnya $Q_{\text{Y}}/Q_{\text{X}} = (0.70/0.50)^{2} = 2.0$. Pada $x = 0.60\ \text{m}$: $E = -180\ \text{V m}^{-1}$ (mengarah ke X, karena $V$ naik menuju Y), sehingga $F = eE = (1.60 \times 10^{-19})(180) = 2.9 \times 10^{-17}\ \text{N}$ ke arah X. Jika dilepaskan, proton dipercepat ke arah X, percepatannya meningkat saat gradien semakin curam; diperlukan gaya eksternal untuk menahannya tetap diam.
Mendeskripsikan gerak. "Sebuah partikel bermuatan positif diletakkan di P dan dilepaskan. Jelaskan dan uraikan geraknya." Penilai menginginkan: arah (menuju potensial lebih rendah, menurun sepanjang grafik $V$); bahwa gaya, dan karenanya percepatan, tidak konstan (mengikuti gradien yang berubah sepanjang garis); dan bahwa partikel mempercepat sepanjang waktu, karena gaya tetap searah dengan gerak. Jika P adalah titik nol-medan, partikel tetap diam dalam kesetimbangan tidak stabil: dorongan kecil ke arah manapun akan menjauhkannya.
Medan sama dengan negatif gradien potensial:
Di antara pelat sejajar $V$ berubah secara merata terhadap posisi, menghasilkan $E = V/d$ seperti sebelumnya. Tanda minus berarti medan mengarah ke potensial lebih rendah. Untuk muatan titik, $-\dfrac{dV}{dr} = \dfrac{Q}{4\pi\varepsilon_{0} r^{2}} = E$.
Hubungan satu baris. "Nyatakan hubungan antara medan listrik dan potensial listrik": kekuatan medan sama dengan negatif dari gradien potensial, $E = -\Delta V/\Delta x$. Pada grafik $V$–$x$ $E$ adalah negatif dari gradien sebuah tangen; pada grafik $V$–$r$ untuk muatan titik, gradien paling curam dekat muatan, di mana medan terkuat, dan tanda $E$ ditentukan oleh "medan mengarah ke potensial lebih rendah". Versi medan seragam $E = V/d$ adalah pernyataan yang sama dengan gradien konstan.
Isipotensial. Permukaan isipotensial menghubungkan titik-titik dengan potensial yang sama; tidak ada usaha yang dilakukan memindahkan muatan di atasnya, sehingga medan selalu tegak lurus terhadapnya. Di sekitar muatan titik, isipotensial berupa bola-bola konsentris yang makin renggang saat $V$ turun; di antara pelat sejajar mereka berupa bidang paralel terhadap pelat. Permukaan konduktor adalah isipotensial, itulah sebabnya garis-garis medan meetinya secara tegak lurus.
Muatan $q$ pada titik dengan potensial $V$ memiliki energi potensial listrik $E_{\text{P}} = qV$. Untuk dua muatan titik $Q$ dan $q$ yang berjarak $r$:
Dalam kedua kasus $E_{\text{P}} \to 0$ saat $r \to \infty$.
Contoh terpecahkan. Proton dan elektron dalam atom hidrogen berjarak $5.3 \times 10^{-11}\ \text{m}$. Hitung energi potensial listrik pasangan tersebut.
Tanda negatif merupakan bagian jawaban: muatan berlawanan, sehingga energi ($4.3 \times 10^{-18}\ \text{J}$) harus disuplai untuk menarik mereka hingga tak terhingga. (Elektron juga memiliki energi kinetik setengah besarnya, sehingga energi yang dibutuhkan untuk mengionisasi atom adalah $13.6\ \text{eV}$; pola "contoh terpecahkan" di bawah ini menunjukkan alasannya.)
Muatan melalui beda potensial. Muatan $q$ yang dipindahkan melalui beda potensial $V$ mengubah energi potensialnya sebesar $qV$, inilah mengapa elektron yang dipercepat dari keadaan diam melewati $250\ \text{V}$ tiba dengan $250\ \text{eV}$ energi kinetik apa pun bentuk medannya. Untuk soal pilihan ganda tentang "menggerakkan P sejauh $x$ sepanjang garis-garis medan", usaha yang dilakukan adalah $qEx$, dan energi potensialnya turun jika bergerak searah dorongan gaya.
Sebuah elektron mengorbit inti bermuatan $+Ze$ pada jarak $r$. Tarikan Coulomb menyediakan gaya sentripetal:
Energi total adalah kinetik ditambah potensial:
yang bernilai negatif (keadaan terikat).
Kedua teori medan tampak serupa:
| Besaran | Gravitasi | Listrik |
|---|---|---|
| Sumber | massa $M$ (selalu positif) | muatan $Q$ (bisa ±) |
| Kekuatan medan | $g = GM/r^{2}$ | $E = Q/(4\pi\varepsilon_{0} r^{2})$ |
| Gaya pada benda uji | $F = mg$ | $F = qE$ |
| Potensial | $\phi = -GM/r$ | $V = Q/(4\pi\varepsilon_{0} r)$ |
| PE dua | $-GMm/r$ | $Qq/(4\pi\varepsilon_{0} r)$ |
| Sifat | selalu tarik-menarik | tarik-menarik atau tolak-menolak |
Tanda minus dalam potensial gravitasi mencerminkan bahwa gravitasi selalu menarik; potensial listrik mengambil tanda dari muatan sumber.
Kesamaan dan perbedaan (soal standar dua poin). Kedua potensial berbanding lurus dengan $1/r$, keduanya nol di tak terhingga, dan keduanya adalah besaran skalar; namun potensial gravitasi selalu negatif (gaya selalu menarik) sedangkan potensial listrik mengambil tanda muatan dan bisa bernilai positif. Untuk medan: keduanya mematuhi hukum kuadrat terbalik dan digambarkan sebagai garis radial di sekitar sumber titik, tetapi garis medan gravitasi hanya pernah mengarah ke dalam (tarikan), sementara garis medan listrik mengarah keluar dari muatan positif dan masuk ke muatan negatif.
V = kQ / r
Potential ∝ 1/r around a charge — steep near it, flattening out. · Potensi ∝ 1/r di sekeliling muatan — curam berhampiran itu, rata keluar.
| English | Bahasa Indonesia |
|---|---|
| point charge/pɔɪnt tʃɑːdʒ/ | muatan titik |
| dipole/ˈdaɪpəʊl/ | dipol |
| uniform field/ˈjuːnɪfɔːm fiːld/ | medan seragam |
| isolated/ˈaɪsəleɪtɪd/ | terisolasi |
| potential gradient/pəˈtenʃl ˈɡreɪdɪənt/ | gradien potensial |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | beda potensial |
| gravitational field/ˌɡrævɪˈteɪʃənl fiːld/ | medan gravitasi |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| parabolic/ˌpærəˈbɒlɪk/ | parabolik |
| projectile/prəˈdʒektaɪl/ | proyektil |
| cathode-ray tube/ˈkæθəʊd reɪ tjuːb/ | tabung sinar katode |
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | kesetimbangan |
| deflection/dɪˈflekʃn/ | deviasi |
| Coulomb's law/ˈkuːlɒmz lɔː/ | Hukum Coulomb |
| permittivity of free space/ˌpɜːmɪˈtɪvəti ɒv friː speɪs/ | permitivitas ruang hampa |
| tension/ˈtenʃn/ | tegangan |
| nucleus/ˈnjuːklɪəs/ | inti sel |
| resultant force/rɪˈzʌltənt fɔːs/ | resultan gaya |
| centripetal force/senˈtrɪpɪtl fɔːs/ | kekuatan sentripetal |
| equipotential/ˌiːkwɪpəˈtenʃl/ | equipotensial |
| vector sum/ˈvektə sʌm/ | jumlah vektor |
| electric potential/ɪˈlektrɪk pəˈtenʃl/ | potensi listrik |
| scalar/ˈskeɪlə/ | skalar |
| tangent/ˈtændʒənt/ | tangens |
| negative potential gradient/ˈneɡətɪv pəˈtenʃl ˈɡreɪdɪənt/ | gradien potensial negatif |
| electric potential energy/ɪˈlektrɪk pəˈtenʃl ˈenədʒi/ | energi potensial listrik |
| bound/baʊnd/ | terikat |
| ionise/ˈaɪənaɪz/ | ionisasi |
| gravitational potential/ˌɡrævɪˈteɪʃənl pəˈtenʃl/ | potensial gravitasi |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| electric field | a region in which a charge experiences an (electric) force |
| electric field strength | the force per unit positive charge acting on a small (test) charge placed at the point |
| electric field line | a line whose direction shows the direction of the force on a positive charge; the spacing of the lines shows the field strength |
| Coulomb's law | the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation |
| electric potential | the work done per unit positive charge in bringing a small test charge from infinity to the point |
| field and potential | the electric field strength is equal to the negative of the potential gradient: $E = -\Delta V/\Delta x$ |
| electric potential energy (of two charges) | the work done in bringing the charges from infinity to their separation; $Qq/(4\pi\varepsilon_{0} r)$ |
| equipotential | a surface (or line) on which every point has the same potential; the field is at right angles to it |
| uniform field | a field with the same strength and direction at every point, shown by parallel, evenly spaced field lines |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| medan listrik | daerah di mana sebuah muatan mengalami gaya (listrik) |
| kuat medan listrik | gaya per satuan muatan positif yang bekerja pada suatu muatan kecil (uji) yang diletakkan pada titik tersebut |
| garis medan listrik | garis yang arahnya menunjukkan arah gaya pada muatan positif; kerapatan garis menunjukkan kekuatan medan |
| Hukum Coulomb | gaya antara dua muatan titik berbanding lurus dengan hasil kali muatan-muatan tersebut dan berbanding terbalik dengan kuadrat jarak pisahnya |
| potensial listrik | usaha yang dilakukan per satuan muatan positif untuk membawa muatan uji kecil dari tak terhingga ke titik tersebut |
| medan dan potensial | kuat medan listrik sama dengan negatifnya gradien potensial: $E = -\Delta V/\Delta x$ |
| energi potensial listrik (dua muatan) | usaha yang dilakukan untuk membawa muatan-muatan tersebut dari tak terhingga hingga terpisah sejauh tertentu; $Qq/(4\pi\varepsilon_{0} r)$ |
| ekuipotensial | permukaan (atau garis) di mana setiap titiknya memiliki potensial yang sama; medan tegak lurus terhadapnya |
| medan seragam | medan dengan kekuatan dan arah yang sama di setiap titik, ditunjukkan oleh garis-garis medan yang sejajar dan berjarak sama |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A capacitor 电容器 stores charge. The simplest one is two parallel conductor 导体 plates with an insulator 绝缘体 (a dielectric 电介质, or just vacuum 真空 / air) between them. Connected to a battery, charge $+Q$ builds up on one plate and $-Q$ on the other, with a potential difference 电势差 $V$ across the gap.
How the plates become charged. Connect the plates to a battery and electrons flow, through the battery, from the plate joined to the positive terminal 接线端 to the plate joined to the negative terminal. One plate is left with charge $+Q$ and the other gains $-Q$: the charges are always equal and opposite, so the capacitor as a whole stays neutral. The flow stops when the p.d. between the plates equals the e.m.f. of the battery. Nothing crosses the gap; the insulator is what keeps the charge separated.
The capacitance 电容 $C$ of any capacitor (or any isolated conductor) is
This applies to:
Unit: farad 法拉 (F) $= \text{C V}^{-1}$. A farad is huge, so real capacitors run from $\text{pF}$ to $\text{mF}$.
Capacitance is constant for a given capacitor (set by its size and dielectric). Doubling the charge doubles the voltage, so $C = Q/V$ stays the same.
The two-mark definitions. For a parallel-plate capacitor: the charge on one plate per unit potential difference between the plates. For an isolated sphere (or any isolated conductor): the charge on the conductor per unit potential. The examiner looks for "charge per unit p.d." (or "per unit potential") and, for the plates, that the charge is the charge on one plate and the p.d. is that between the plates. "The charge stored" is accepted; "the total charge on both plates" is not, because that total is zero.
What sets the capacitance. For parallel plates the capacitance is proportional to the plate area and inversely proportional to the separation $x$ ($C \propto 1/x$: halve the gap and the capacitance doubles), and it rises when a dielectric replaces the air. A question that states "$C$ is inversely proportional to $x$" and then moves the plates apart at constant charge is asking for $V = Q/C \propto x$: the p.d. rises in proportion. At constant p.d. (still connected to the supply) it is the charge that falls instead.
Worked example. A $100\ \mu\text{F}$ capacitor is charged to $12\ \text{V}$. Find the charge stored.
Worked example. An isolated metal sphere of radius $15\ \text{cm}$ is charged to a potential of $9.0 \times 10^{3}\ \text{V}$. Find its capacitance and its charge. It is then discharged to earth through a $120\ \text{M}\Omega$ resistor: find the time constant of the discharge.
The time constant is $RC = (1.2 \times 10^{8})(1.67 \times 10^{-11}) = 2.0 \times 10^{-3}\ \text{s}$: even through an enormous resistance the sphere loses its charge in milliseconds, because its capacitance is so small. Notice that $4\pi\varepsilon_{0} = 1/(8.99 \times 10^{9})$, so $C = r/(8.99 \times 10^{9})$ with $r$ in metres.
Capacitors in parallel share the same p.d. $V$. The total charge is the sum:
A parallel combination has larger capacitance than any one capacitor.
Capacitors in series carry the same charge $Q$. The total p.d. is the sum:
A series combination has smaller capacitance than any one capacitor.
Note: these rules are the opposite of those for resistors (resistors sum in series; capacitors sum in parallel), because $C = Q/V$ has $V$ on the bottom while $R = V/I$ has $I$ on the bottom.
"Derive an expression for the combined capacitance" is a show that question, and the mark scheme wants the physics stated, not just the algebra.
Two capacitors in series always give less than the smaller one; two equal capacitors $C$ give $C/2$ in series and $2C$ in parallel.
Worked example. A $6.0\ \mu\text{F}$ capacitor is in series with a parallel pair of $2.0\ \mu\text{F}$ and $4.0\ \mu\text{F}$. Find the total capacitance, and the charge on each capacitor when $12\ \text{V}$ is applied across the network.
Parallel pair: $2.0 + 4.0 = 6.0\ \mu\text{F}$. In series with $6.0\ \mu\text{F}$: $1/C = 1/6.0 + 1/6.0$, so $C = 3.0\ \mu\text{F}$. Total charge $Q = CV = (3.0 \times 10^{-6})(12) = 3.6 \times 10^{-5}\ \text{C}$: this is the charge on the $6.0\ \mu\text{F}$ capacitor (series, same charge), and it is shared by the parallel pair in the ratio of their capacitances (same p.d., $6.0\ \text{V}$ across each): the $2.0\ \mu\text{F}$ holds $1.2 \times 10^{-5}\ \text{C}$ and the $4.0\ \mu\text{F}$ holds $2.4 \times 10^{-5}\ \text{C}$. Check: they add to $3.6 \times 10^{-5}\ \text{C}$.
Worked example. Capacitors X and Y, each of capacitance $C$, are connected in series across a supply of voltage $V$; capacitors P and Q, each $C$, are connected in parallel across an identical supply. Compare the charge drawn from each supply and the energy stored in each arrangement.
Series: $C_{\text{s}} = C/2$, charge $Q_{\text{s}} = CV/2$, energy $\tfrac{1}{2} C_{\text{s}} V^{2} = CV^{2}/4$. Parallel: $C_{\text{p}} = 2C$, charge $2CV$, energy $CV^{2}$. The parallel arrangement draws four times the charge and stores four times the energy, because at the same supply voltage the energy is $\tfrac{1}{2} C V^{2}$ and the capacitance is four times larger. Asked which arrangement stores more energy from a given supply, the answer is always parallel.
Sebuah kapasitor menyimpan muatan. Yang paling sederhana adalah dua pelat konduktor sejajar dengan isolator (sebuah dielektrik, atau sekadar ruang hampa / udara) di antaranya. Terhubung ke baterai, muatan $+Q$ terkumpul pada satu pelat dan $-Q$ pada pelat lainnya, dengan beda potensial $V$ di celah.
Bagaimana pelat menjadi bermuatan. Hubungkan pelat ke baterai dan elektron mengalir, melalui baterai, dari pelat yang terhubung ke terminal positif ke pelat yang terhubung ke terminal negatif. Satu pelat tertinggal dengan muatan $+Q$ dan pelat lainnya memperoleh $-Q$: muatan-muatannya selalu sama besar dan berlawanan, sehingga kapasitor secara keseluruhan tetap netral. Aliran berhenti ketika beda potensial antara pelat sama dengan GGL baterai. Tidak ada yang melintasi celah; isolator lah yang menjaga muatan tetap terpisah.
Kapasitansi $C$ dari setiap kapasitor (atau konduktor terisolasi apa pun) adalah
Ini berlaku untuk:
Unit: farad (F) $= \text{C V}^{-1}$. Satu farad itu sangat besar, sehingga kapasitor nyata berkisar dari $\text{pF}$ hingga $\text{mF}$.
Kapasitansi adalah konstan untuk sebuah kapasitor tertentu (ditentukan oleh ukuran dan dielektriknya). Menggandakan muatan akan menggandakan tegangan, sehingga $C = Q/V$ tetap sama.
Definisi dua poin. Untuk kapasitor pelat sejajar: muatan pada satu pelat per satuan beda potensial antara pelat-pelat tersebut. Untuk bola terisolasi (atau konduktor terisolasi lainnya): muatan pada konduktor per satuan potensial. Penguji mencari "muatan per satuan p.d." (atau "per satuan potensial") dan, untuk pelat-pelat tersebut, bahwa muatan adalah muatan pada satu pelat dan p.d. adalah yang antara pelat-pelat tersebut. "Muatan yang tersimpan" diterima; "total muatan pada kedua pelat" tidak, karena totalnya adalah nol.
Apa yang menentukan kapasitansi. Untuk pelat sejajar, kapasitansinya berbanding lurus dengan luas pelat dan berbanding terbalik dengan jarak pemisahan $x$ ($C \propto 1/x$: setengahkan celah dan kapasitansinya menjadi dua kali), dan meningkat ketika dielektrik menggantikan udara. Soal yang menyatakan "$C$ berbanding terbalik dengan $x$" lalu memisahkan pelat pada muatan konstan meminta Anda untuk $V = Q/C \propto x$: p.d. naik sebanding. Pada p.d. konstan (masih terhubung ke sumber), justru muatan yang turun.
Contoh terpecahkan. Sebuah kapasitor $100\ \mu\text{F}$ diisi hingga $12\ \text{V}$. Temukan muatan yang tersimpan.
Contoh dikerjakan. Sebuah bola logam terisolasi berjari-jari $15\ \text{cm}$ diberi potensial $9.0 \times 10^{3}\ \text{V}$. Temukan kapasitansinya dan muatannya. Kemudian dibuang ke bumi melalui resistor $120\ \text{M}\Omega$: temukan konstanta waktu pembuangan.
Konstanta waktunya $RC = (1.2 \times 10^{8})(1.67 \times 10^{-11}) = 2.0 \times 10^{-3}\ \text{s}$: bahkan melalui resistansi yang sangat besar, bola kehilangan muatannya dalam milidetik, karena kapasitansinya sangat kecil. Perhatikan bahwa $4\pi\varepsilon_{0} = 1/(8.99 \times 10^{9})$, jadi $C = r/(8.99 \times 10^{9})$ dengan $r$ dalam meter.
Kapasitor paralel memiliki p.d. yang sama $V$. Total muatan adalah jumlah:
Kombinasi paralel memiliki kapasitansi yang lebih besar daripada salah satu kapasitor pun.
Kapasitor seri membawa muatan yang sama $Q$. Total p.d. adalah jumlah:
Kombinasi seri memiliki kapasitansi yang lebih kecil daripada salah satu kapasitor pun.
Catatan: aturan ini adalah kebalikan dari aturan untuk resistor (resistor menjumlahkan dalam seri; kapasitor menjumlahkan dalam paralel), karena $C = Q/V$ memiliki $V$ di bagian bawah sementara $R = V/I$ memiliki $I$ di bagian bawah.
"Turunkan ekspresi untuk kapasitansi gabungan" adalah soal buktikan, dan kunci jawaban menginginkan fisika yang dijelaskan, bukan hanya aljabar.
Dua kapasitor seri selalu menghasilkan kurang daripada yang terkecil; dua kapasitor identik $C$ menghasilkan $C/2$ dalam seri dan $2C$ dalam paralel.
Contoh dikerjakan. Sebuah kapasitor $6.0\ \mu\text{F}$ seri dengan pasangan paralel dari $2.0\ \mu\text{F}$ dan $4.0\ \mu\text{F}$. Temukan kapasitansi total, dan muatan pada setiap kapasitor ketika $12\ \text{V}$ diterapkan melintasi jaringan.
Pasangan paralel: $2.0 + 4.0 = 6.0\ \mu\text{F}$. Seri dengan $6.0\ \mu\text{F}$: $1/C = 1/6.0 + 1/6.0$, sehingga $C = 3.0\ \mu\text{F}$. Total muatan $Q = CV = (3.0 \times 10^{-6})(12) = 3.6 \times 10^{-5}\ \text{C}$: ini adalah muatan pada kapasitor $6.0\ \mu\text{F}$ (seri, muatan sama), dan dibagi oleh pasangan paralel sesuai rasio kapasitansinya (p.d. sama, $6.0\ \text{V}$ di setiap satu): kapasitor $2.0\ \mu\text{F}$ memegang $1.2 \times 10^{-5}\ \text{C}$ dan kapasitor $4.0\ \mu\text{F}$ memegang $2.4 \times 10^{-5}\ \text{C}$. Cek: mereka menjumlahkan ke $3.6 \times 10^{-5}\ \text{C}$.
Contoh dikerjakan. Kapasitor X dan Y, masing-masing berkapasitansi $C$, dihubungkan seri melintasi sumber tegangan $V$; kapasitor P dan Q, masing-masing $C$, dihubungkan paralel melintasi sumber yang identik. Bandingkan muatan yang diambil dari setiap sumber dan energi yang tersimpan dalam setiap susunan.
Seri: $C_{\text{s}} = C/2$, muatan $Q_{\text{s}} = CV/2$, energi $\tfrac{1}{2} C_{\text{s}} V^{2} = CV^{2}/4$. Paralel: $C_{\text{p}} = 2C$, muatan $2CV$, energi $CV^{2}$. Susunan paralel menarik empat kali lebih banyak muatan dan menyimpan empat kali lebih banyak energi, karena pada tegangan suplai yang sama energi adalah $\tfrac{1}{2} C V^{2}$ dan kapasitansinya empat kali lebih besar. Ditanya susunan mana yang menyimpan lebih banyak energi dari suplai tertentu, jawabannya selalu paralel.
Q = C·V
Charge stored is proportional to voltage — the gradient is the capacitance C. · Muatan yang tersimpan berbanding lurus dengan tegangan — gradiennya adalah kapasitansi C.
| English | Bahasa Indonesia |
|---|---|
| capacitor/kəˈpæsɪtə/ | kapasitor |
| conductor/kənˈdʌktə/ | konduktor |
| insulator/ˈɪnsjuːleɪtə/ | isolator |
| dielectric/ˌdaɪɪˈlektrɪk/ | dielektrik |
| vacuum/ˈvækjuːm/ | vakum |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | beda potensial |
| terminal/ˈtɜːmɪnl/ | terminal |
| capacitance/kəˈpæsɪtəns/ | kapasitansi |
| farad/ˈfæræd/ | farad |
| resistor/rɪˈzɪstə/ | resistor |
| resistance/rɪˈzɪstəns/ | penolakan |
| Kirchhoff's second law/ˈkɜːkhɒfs ˈsekənd lɔː/ | Hukum Kirchhoff kedua |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ | peluruhan eksponensial |
| time constant/taɪm ˈkɒnstənt/ | konstanta waktu |
| intercept/ˌɪntəˈsept/ | intersep |
| rectifier/ˈrektɪfaɪə/ | penyearah |
| peak/piːk/ | puncak |
| ripple/ˈrɪpl/ | riak |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Charging a capacitor from $0$ to $Q$ needs work, because each extra bit of charge is pushed against the p.d. already there. When the charge is $q$, the p.d. is $V(q) = q/C$, so adding a small charge $dq$ needs work $V\,dq$. The total work is
Using $V = Q/C$, this is the energy 能量 stored:
Worked example. Find the energy stored in a $100\ \mu\text{F}$ capacitor charged to $12\ \text{V}$.
Worked example. The graph of charge $Q$ against p.d. $V$ for a capacitor is a straight line through the origin passing through $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$. Find the capacitance and the energy stored at $10\ \text{V}$, and then the extra energy stored when the p.d. is raised to $12\ \text{V}$.
The gradient of $Q$ against $V$ is $C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$. Energy is the area under the line: $\tfrac{1}{2}QV = \tfrac{1}{2}(1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$. At $12\ \text{V}$: $\tfrac{1}{2}CV^{2} = \tfrac{1}{2}(1.2 \times 10^{-4})(12)^{2} = 8.6 \times 10^{-3}\ \text{J}$, so the extra energy is $2.6 \times 10^{-3}\ \text{J}$. Do not find the extra energy from the extra charge as $\tfrac{1}{2}(\Delta Q)V$: energy is not proportional to charge, so subtract the two areas.
A plot of $V$ against $Q$ is a straight line through the origin with gradient $1/C$. The energy stored is the area under the line up to a given charge $Q$, which is the triangle $\tfrac{1}{2} Q V$. The factor $\tfrac{1}{2}$ is there because the average p.d. during charging is $V/2$ (it grows from zero to $V$), not $V$.
Connect a capacitor $C$ to an ideal battery of e.m.f. $V$ through a wire. The capacitor stores $\tfrac{1}{2} C V^{2}$, but the battery supplies charge $Q = CV$ at e.m.f. $V$, giving out $QV = CV^{2}$. The other half is lost as heat in the wire — whatever the wire's resistance.
A fully charged capacitor X (capacitance $C$, p.d. $V$, charge $Q = CV$) is disconnected from its supply and connected across an uncharged capacitor Y of capacitance $3C$. Charge flows until the two p.d.s are equal. Two principles settle everything:
Worked example. A $220\ \mu\text{F}$ capacitor charged to $9.0\ \text{V}$ is connected across an uncharged $440\ \mu\text{F}$ capacitor. Find the final p.d. and the energy lost.
$Q = CV = (220 \times 10^{-6})(9.0) = 1.98 \times 10^{-3}\ \text{C}$; total capacitance $660\ \mu\text{F}$, so $V' = 1.98 \times 10^{-3}/(660 \times 10^{-6}) = 3.0\ \text{V}$. Energy before: $\tfrac{1}{2}(220 \times 10^{-6})(9.0)^{2} = 8.9 \times 10^{-3}\ \text{J}$; after: $\tfrac{1}{2}(660 \times 10^{-6})(3.0)^{2} = 3.0 \times 10^{-3}\ \text{J}$; lost: $5.9 \times 10^{-3}\ \text{J}$, two-thirds of the original. A capacitor connected to another of capacitance $kC$ always loses the fraction $k/(k+1)$.
Mengisi kapasitor dari $0$ ke $Q$ memerlukan usaha, karena setiap tambahan muatan didorong melawan beda potensial (p.d.) yang sudah ada. Ketika muatan adalah $q$, p.d. adalah $V(q) = q/C$, sehingga menambahkan muatan kecil $dq$ memerlukan usaha $V\,dq$. Total usahanya adalah
Menggunakan $V = Q/C$, ini adalah energi yang tersimpan:
Contoh terpecahkan. Temukan energi yang tersimpan dalam kapasitor $100\ \mu\text{F}$ yang diisi hingga $12\ \text{V}$.
Contoh terpecahkan. Grafik muatan $Q$ terhadap p.d. $V$ untuk sebuah kapasitor adalah garis lurus melalui titik asal yang melewati $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$. Temukan kapasitas dan energi yang tersimpan pada $10\ \text{V}$, lalu cari energi tambahan yang tersimpan ketika p.d. dinaikkan menjadi $12\ \text{V}$.
Gradien $Q$ terhadap $V$ adalah $C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$. Energi adalah area di bawah garis: $\tfrac{1}{2}QV = \tfrac{1}{2}(1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$. Pada $12\ \text{V}$: $\tfrac{1}{2}CV^{2} = \tfrac{1}{2}(1.2 \times 10^{-4})(12)^{2} = 8.6 \times 10^{-3}\ \text{J}$, sehingga energi tambahan adalah $2.6 \times 10^{-3}\ \text{J}$. Jangan mencari energi tambahan dari muatan tambahan seperti $\tfrac{1}{2}(\Delta Q)V$: energi tidak sebanding dengan muatan, jadi kurangkan kedua area tersebut.
Plot $V$ terhadap $Q$ adalah garis lurus melalui titik asal dengan gradien $1/C$. Energi yang tersimpan adalah luas di bawah garis hingga muatan tertentu $Q$, yaitu segitiga $\tfrac{1}{2} Q V$. Faktor $\tfrac{1}{2}$ ada karena rata-rata p.d. selama pengisian adalah $V/2$ (meningkat dari nol hingga $V$), bukan $V$.
Hubungkan kapasitor $C$ ke baterai ideal dengan g.g.b. $V$ melalui kawat. Kapasitor menyimpan $\tfrac{1}{2} C V^{2}$, tetapi baterai menyediakan muatan $Q = CV$ pada g.g.b. $V$, menghasilkan $QV = CV^{2}$. Setengah lainnya hilang sebagai panas di kawat — berapapun resistansi kawatnya.
Kapasitor X yang terisi penuh (kapasitansi $C$, p.d. $V$, muatan $Q = CV$) dilepas dari sumbernya dan dihubungkan melintasi kapasitor Y tak bermuatan dengan kapasitansi $3C$. Muatan mengalir hingga kedua p.d.-nya setara. Dua prinsip menentukan semuanya:
Contoh terpecahkan. Kapasitor $220\ \mu\text{F}$ yang diisi hingga $9.0\ \text{V}$ dihubungkan melintasi kapasitor $440\ \mu\text{F}$ tak bermuatan. Temukan p.d. akhir dan energi yang hilang.
$Q = CV = (220 \times 10^{-6})(9.0) = 1.98 \times 10^{-3}\ \text{C}$; total kapasitansi $660\ \mu\text{F}$, sehingga $V' = 1.98 \times 10^{-3}/(660 \times 10^{-6}) = 3.0\ \text{V}$. Energi sebelum: $\tfrac{1}{2}(220 \times 10^{-6})(9.0)^{2} = 8.9 \times 10^{-3}\ \text{J}$; sesudah: $\tfrac{1}{2}(660 \times 10^{-6})(3.0)^{2} = 3.0 \times 10^{-3}\ \text{J}$; hilang: $5.9 \times 10^{-3}\ \text{J}$, dua pertiga dari awal. Kapasitor yang dihubungkan dengan kapasitor lain berkapasitansi $kC$ selalu kehilangan pecahan $k/(k+1)$.
E = ½C·V²
Stored energy grows with the square of the voltage. · Energi tersimpan tumbuh sesuai dengan kuadrat tegangan.
| English | Bahasa Indonesia |
|---|---|
| energy/ˈenədʒi/ | tenaga |
| dissipated/ˈdɪsɪpeɪtɪd/ | terdisipasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A capacitor $C$ charged to $V_{0}$ is connected through a switch to a resistor 电阻器 of resistance 电阻 $R$. When the switch closes at $t = 0$, the capacitor discharges.
By Kirchhoff's second law 基尔霍夫第二定律 around the loop, $V_{C} = V_{R}$. Using $V_{C} = Q/C$, $V_{R} = IR$ and $I = -dQ/dt$:
This is solved by an exponential decay 指数衰减 with time constant $RC$.
Charge $Q$, p.d. $V$ and current $I$ all decay exponentially with the same time constant:
with $I_{0} = V_{0}/R$.
Reading the three graphs. All three curves have the same shape and the same time constant, so one measurement of $\tau$ from any of them gives $RC$. The current is largest the instant the switch closes, $I_{0} = V_{0}/R$, because the full p.d. $V_{0}$ is then across the resistor; it falls as the p.d. falls. The area under the $I$–$t$ graph is the charge that has flowed, so the total area is $Q_{0} = CV_{0}$, and the charge that flows in the first time constant is $Q_{0}(1 - e^{-1}) = 0.63\,Q_{0}$. The gradient of the $Q$–$t$ graph is $-I$: steepest at the start, flattening as the current dies away. (During charging through a resistor the current has the same decaying shape, while $V$ and $Q$ rise as $V_{0}(1 - e^{-t/RC})$, the mirror image of the discharge curve.)
is the time constant 时间常数 (in seconds: $\Omega \cdot \text{F} = \text{s}$). It is the time for a decaying quantity to fall to $1/e \approx 0.37$ (about 37%) of its starting value. After $2\tau$ it is at about 13.5%; after $5\tau$, below 1%.
Worked example. A $100\ \mu\text{F}$ capacitor charged to $12\ \text{V}$ is discharged through a $47\ \text{k}\Omega$ resistor. Find the time constant and the voltage after one time constant.
After one time constant the voltage falls to $1/e$ of its start: $V = 12 \times 0.37 \approx 4.4\ \text{V}$.
Worked example. A $470\ \mu\text{F}$ capacitor is charged to $24\ \text{V}$ and then discharged through a $5.6\ \text{k}\Omega$ resistor. Find (a) the initial charge and energy, (b) the time constant and the initial current, (c) the p.d. after $4.0\ \text{s}$, (d) the time for the p.d. to fall to $6.0\ \text{V}$.
(a) $Q_{0} = CV_{0} = (470 \times 10^{-6})(24) = 1.1 \times 10^{-2}\ \text{C}$; $W = \tfrac{1}{2}CV_{0}^{2} = \tfrac{1}{2}(470 \times 10^{-6})(24)^{2} = 0.135\ \text{J}$. (b) $\tau = RC = (5.6 \times 10^{3})(470 \times 10^{-6}) = 2.6\ \text{s}$; $I_{0} = V_{0}/R = 24/5600 = 4.3\ \text{mA}$. (c) $V = V_{0} e^{-t/RC} = 24\, e^{-4.0/2.63} = 24 \times 0.219 = 5.3\ \text{V}$. (d) Rearrange with logarithms: $t = -RC \ln(V/V_{0}) = -2.63 \ln(6.0/24) = 2.63 \times 1.386 = 3.6\ \text{s}$.
Keep the time constant unrounded ($2.632\ \text{s}$) inside the calculation and round only the answers. The rearrangement in (d) is the one most often asked and most often botched: $\ln(V/V_{0})$ is negative, so $t$ comes out positive.
Worked example. From a discharge graph, the p.d. across a $2200\ \mu\text{F}$ capacitor falls from $6.0\ \text{V}$ to $2.2\ \text{V}$ in $4.5\ \text{s}$. Find the resistance of the resistor.
$2.2/6.0 = 0.367 \approx 1/e$, so $4.5\ \text{s}$ is one time constant: $R = \tau/C = 4.5/(2200 \times 10^{-6}) = 2.0 \times 10^{3}\ \Omega$. If the fall is not to $1/e$, use $RC = -t/\ln(V/V_{0})$; the answer is the same either way.
To find $\tau$ from a curve: read the time to fall to $1/e$ of the start. Or take logs: $\ln(V/V_{0}) = -t/(RC)$, so a plot of $\ln V$ against $t$ is a straight line with gradient $-1/(RC)$.
Using the logarithmic graph. In an experiment, plot $\ln V$ (or $\ln I$) on the $y$-axis against $t$. The points should lie on a straight line: its gradient is $-1/(RC)$, so $RC = -1/\text{gradient}$, and its intercept 截距 on the $\ln V$ axis is $\ln V_{0}$. A straight line is the test that the decay really is exponential; a curve means it is not. Take the gradient from two points far apart on the best-fit line, not from two data points, and quote $RC$ in seconds.
Given a discharge curve $V(t)$ or $Q(t)$:
A rectifier 整流器 (Topic 21) turns an alternating voltage into a series of positive humps. A capacitor connected across the load smooths them. Near each peak 峰值 the diodes conduct and the capacitor charges to the peak voltage; as the supply voltage falls away the diodes stop conducting and the capacitor discharges through the load resistor, holding the output up until the next hump rises above it and recharges the capacitor. The output therefore falls only a little between peaks; the rise and fall that remains is the ripple 纹波.
How much it falls depends on the time constant of the discharge, $RC$, compared with the time between peaks. A larger capacitance, or a larger load resistance (a smaller load current), gives a larger $RC$, a slower decay and a smaller ripple; a heavily loaded supply (small $R$) has a larger ripple. Asked to sketch the smoothed output, draw it touching each peak, decaying slightly between them, and never falling to zero.
Sebuah kapasitor $C$ yang diisi hingga $V_{0}$ dihubungkan melalui sakelar ke resistor dengan hambatan $R$. Ketika sakelar ditutup pada $t = 0$, kapasitor akan mengosongkan muatannya.
Dengan Hukum Kedua Kirchhoff di sekitar loop, $V_{C} = V_{R}$. Menggunakan $V_{C} = Q/C$, $V_{R} = IR$ dan $I = -dQ/dt$:
Ini diselesaikan oleh peluruhan eksponensial dengan konstanta waktu $RC$.
Muatan $Q$, p.d. $V$ dan arus $I$ semuanya meluruh secara eksponensial dengan konstanta waktu yang sama:
dengan $I_{0} = V_{0}/R$.
Membaca ketiga grafik. Ketiga kurva memiliki bentuk yang sama dan konstanta waktu yang sama, sehingga satu pengukuran $\tau$ dari salah satunya memberikan $RC$. Arus terbesar pada saat sakelar ditutup, $I_{0} = V_{0}/R$, karena seluruh beda potensial (b.p.) $V_{0}$ berada di atas resistor; arus menurun seiring b.p. menurun. Luas di bawah grafik $I$–$t$ adalah muatan yang telah mengalir, sehingga total luasnya adalah $Q_{0} = CV_{0}$, dan muatan yang mengalir dalam konstanta waktu pertama adalah $Q_{0}(1 - e^{-1}) = 0.63\,Q_{0}$. Gradien dari grafik $Q$–$t$ adalah $-I$: paling curam di awal, melandai saat arus mereda. (Selama pengisian melalui resistor, arus memiliki bentuk peluruhan yang sama, sementara $V$ dan $Q$ naik sebagai $V_{0}(1 - e^{-t/RC})$, cerminan dari kurva pembuangan.)
adalah konstanta waktu (dalam detik: $\Omega \cdot \text{F} = \text{s}$). Ini adalah waktu untuk besaran yang meluruh turun ke $1/e \approx 0.37$ (sekitar 37%) dari nilainya awal. Setelah $2\tau$ nilainya sekitar 13,5%; setelah $5\tau$, di bawah 1%.
Contoh soal terpecahkan. Kapasitor $100\ \mu\text{F}$ yang diisi hingga $12\ \text{V}$ dibuang melalui resistor $47\ \text{k}\Omega$. Tentukan konstanta waktu dan tegangan setelah satu konstanta waktu.
Setelah satu konstanta waktu, tegangan turun ke $1/e$ dari nilai awalnya: $V = 12 \times 0.37 \approx 4.4\ \text{V}$.
Contoh soal terpecahkan. Kapasitor $470\ \mu\text{F}$ diisi hingga $24\ \text{V}$ lalu dibuang melalui resistor $5.6\ \text{k}\Omega$. Tentukan (a) muatan dan energi awal, (b) konstanta waktu dan arus awal, (c) p.d. setelah $4.0\ \text{s}$, (d) waktu agar p.d. turun ke $6.0\ \text{V}$.
(a) $Q_{0} = CV_{0} = (470 \times 10^{-6})(24) = 1.1 \times 10^{-2}\ \text{C}$; $W = \tfrac{1}{2}CV_{0}^{2} = \tfrac{1}{2}(470 \times 10^{-6})(24)^{2} = 0.135\ \text{J}$. (b) $\tau = RC = (5.6 \times 10^{3})(470 \times 10^{-6}) = 2.6\ \text{s}$; $I_{0} = V_{0}/R = 24/5600 = 4.3\ \text{mA}$. (c) $V = V_{0} e^{-t/RC} = 24\, e^{-4.0/2.63} = 24 \times 0.219 = 5.3\ \text{V}$. (d) Susun ulang dengan logaritma: $t = -RC \ln(V/V_{0}) = -2.63 \ln(6.0/24) = 2.63 \times 1.386 = 3.6\ \text{s}$.
Pertahankan konstanta waktu tanpa pembulatan ($2.632\ \text{s}$) di dalam perhitungan dan bulatkan hanya jawabannya. Penyusunan ulang pada (d) adalah yang paling sering ditanyakan dan paling sering keliru: $\ln(V/V_{0})$ bernilai negatif, sehingga $t$ akan keluar positif.
Contoh soal terpecahkan. Dari grafik pembuangan, p.d. pada kapasitor $2200\ \mu\text{F}$ turun dari $6.0\ \text{V}$ ke $2.2\ \text{V}$ dalam waktu $4.5\ \text{s}$. Tentukan hambatan resistor.
$2.2/6.0 = 0.367 \approx 1/e$, sehingga $4.5\ \text{s}$ adalah satu konstanta waktu: $R = \tau/C = 4.5/(2200 \times 10^{-6}) = 2.0 \times 10^{3}\ \Omega$. Jika penurunan tidak menuju $1/e$, gunakan $RC = -t/\ln(V/V_{0})$; jawabannya tetap sama baik menggunakan cara mana pun.
Untuk menemukan $\tau$ dari kurva: baca waktu untuk turun ke $1/e$ dari nilai awal. Atau ambil logaritma: $\ln(V/V_{0}) = -t/(RC)$, sehingga plot $\ln V$ terhadap $t$ adalah garis lurus dengan gradien $-1/(RC)$.
Menggunakan grafik logaritma. Dalam suatu percobaan, plot $\ln V$ (atau $\ln I$) pada sumbu $y$ terhadap $t$. Titik-titik harus terletak pada garis lurus: gradiennya adalah $-1/(RC)$, sehingga $RC = -1/\text{gradient}$, dan intersepnya pada sumbu $\ln V$ adalah $\ln V_{0}$. Garis lurus adalah uji bahwa peluruhan benar-benar eksponensial; kurva berarti tidak. Ambil gradien dari dua titik yang jauh pada garis fit terbaik, bukan dari dua titik data, dan nyatakan $RC$ dalam satuan detik.
Diberikan kurva pembuangan $V(t)$ atau $Q(t)$:
Sebuah penyearah (Topik 21) mengubah tegangan bolak-balik menjadi serangkaian hump positif. Kapasitor yang dihubungkan melintang di atas beban meratakannya. Dekat setiap puncak, dioda menghantarkan arus dan kapasitor mengisi muatan hingga tegangan puncak; saat tegangan suplai mereda, dioda berhenti menghantarkan dan kapasitor membongkar muatan melalui resistor beban, menjaga output tetap tinggi hingga hump berikutnya naik melebihi level tersebut dan mengisi kembali kapasitor. Oleh karena itu, output hanya turun sedikit di antara puncak-puncak; kenaikan dan penurunan yang tersisa adalah riak.
Berapa banyak penurunan tergantung pada konstanta waktu pembuangan, $RC$, dibandingkan dengan waktu antar puncak. Kapasitansi yang lebih besar, atau hambatan beban yang lebih besar (arus beban yang lebih kecil), menghasilkan $RC$ yang lebih besar, peluruhan yang lebih lambat, dan riak yang lebih kecil; suplai yang dibebani berat ($R$ kecil) memiliki riak yang lebih besar. Saat diminta menggambar output yang diratakan, gambarlah menyentuh setiap puncak, meluruh sedikit di antaranya, dan tidak pernah turun ke nol.
The voltage rises (or decays) exponentially with time constant τ = RC. · Tegangan naik (atau meluruh) secara eksponensial dengan konstanta waktu τ = RC.
Q = Q₀·bᵗ
Charge decays exponentially through the resistor. · Muatan meluruh secara eksponensial melalui resistor.
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| capacitance (parallel-plate capacitor) | the charge on one plate per unit potential difference between the plates |
| capacitance (isolated conductor) | the charge on the conductor per unit potential |
| farad | one coulomb per volt |
| capacitors in series | the same charge on each; the p.d.s add; $1/C = 1/C_{1} + 1/C_{2}$ |
| capacitors in parallel | the same p.d. across each; the charges add; $C = C_{1} + C_{2}$ |
| energy stored | the work done in charging the capacitor, equal to the area under the potential–charge graph; $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2}$ |
| time constant | the product $RC$; the time for the charge (or p.d., or current) to fall to $1/e$ of its initial value |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| kapasitansi (kapasitor pelat sejajar) | muatan pada salah satu pelat per beda potensial satuan antar pelat |
| kapasitansi (konduktor terisolasi) | muatan pada konduktor per potensial satuan |
| farad | satu coulomb per volt |
| kapasitor seri | muatan yang sama pada masing-masing; p.d.s menjumlah; $1/C = 1/C_{1} + 1/C_{2}$ |
| kapasitor dalam paralel | tegangan yang sama di setiap kapasitor; muatan dijumlahkan; $C = C_{1} + C_{2}$ |
| energi tersimpan | usaha yang dilakukan saat mengisi kapasitor, sama dengan luas di bawah grafik potensial–muatan; $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2}$ |
| konstanta waktu | hasil kali $RC$; waktu yang dibutuhkan agar muatan (atau tegangan, atau arus) turun menjadi $1/e$ dari nilai awalnya |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A magnetic field 磁场 is a region where a moving charge (or a current 电流) feels a force 力. It is made by:
"State what is meant by a magnetic field." A region in which a force acts on a moving charge, a current-carrying conductor or a magnetic material. The mark scheme wants the word force and one of those three objects; "a region around a magnet" on its own scores nothing.
Patterns to know:
Drawing the wire's field. "Draw four field lines" around a straight wire is marked on three things: the lines are concentric circles centred on the wire; their spacing increases with distance (the field weakens with distance from the wire); and every line carries an arrow in the direction given by the right-hand grip rule (current into the page: clockwise; current out of the page: anticlockwise). Draw them as circles, not free-hand ovals.
For a flat circular coil the loops around the two sides of the coil reinforce through the centre, so the field there is at right angles to the plane of the coil and strongest at the centre; seen from one face the coil is a north pole (the field leaves it), from the other a south pole. A long solenoid is many such coils in a row: the field inside is uniform and parallel to the axis, and the lines emerge from one end (its N end) and return outside to the other.
An iron core 铁芯 inside a solenoid greatly increases the field, because the iron's atomic magnets line up and add to it. This is why electromagnets 电磁铁 and transformers 变压器 have iron cores.
The syllabus calls this a ferrous 含铁的 core: iron, steel or another magnetic material. Asked why a core increases the field, say that the core becomes magnetised 被磁化 by the solenoid's field and that its own field adds to the field of the current.
Move between magnetic arrangements and see how field patterns change. · Pindah antar susunan magnet dan lihat bagaimana pola medan berubah.
Connect current direction to the circular magnetic field around a wire. · Hubungkan arah arus dengan medan magnet melingkar di sekitar kawat.
| English | Bahasa Indonesia |
|---|---|
| magnetic field/mæɡˈnetɪk fiːld/ | medan magnet |
| bar magnet/bɑː ˈmæɡnɪt/ | magnet batang |
| field line/fiːld laɪn/ | garis medan |
| current/ˈkʌrənt/ | arus |
| force/fɔːs/ | gaya |
| permanent magnet/ˈpɜːmənənt ˈmæɡnɪt/ | magnet permanen |
| right-hand grip rule/raɪt hænd ɡrɪp ruːl/ | aturan genggaman tangan kanan |
| long solenoid/lɒŋ ˈsəʊlənɔɪd/ | solenaoid panjang |
| solenoid/ˈsəʊlənɔɪd/ | solenaoid |
| iron core/ˈaɪən kɔː/ | inti besi |
| electromagnet/ɪˌlektrəʊˈmæɡnɪt/ | elektromagnet |
| transformer/trænsˈfɔːmə/ | transformator |
| ferrous/ˈferəs/ | berbesi |
| magnetised/ˈmæɡnɪtaɪzd/ | ter magnetisasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A current $I$ in a wire of length $L$ in a magnetic field of flux density $B$ feels a force
where $\theta$ is the angle between the wire and the field. The force is largest when the wire is at right angles to the field ($F = BIL$) and zero when the wire is along the field.
Worked example. A wire of length $0.20\ \text{m}$ carries a current of $3.0\ \text{A}$ at right angles to a magnetic field of flux density $0.50\ \text{T}$. Find the force on it.
Two conditions for a force. A copper wire in a magnetic field feels a force only if (1) it carries a current, and (2) the current is not parallel to the field (some part of the wire is at right angles to $B$). A wire with no current, or one lying along the field lines, feels nothing: $\sin\theta = 0$.
Worked example (the current balance). A horizontal wire of length $5.0\ \text{cm}$ lies at right angles to the field between the poles of a magnet standing on a top-pan balance. With no current the balance reads $102.30\ \text{g}$; with a current of $2.5\ \text{A}$ it reads $102.86\ \text{g}$. Find the flux density.
By Newton's third law the force on the wire is matched by an equal and opposite force on the magnet, so the balance reading changes by the magnetic force: $F = \Delta m\, g = (0.56 \times 10^{-3})(9.81) = 5.5 \times 10^{-3}\ \text{N}$. Then $B = F/(IL) = 5.5 \times 10^{-3}/(2.5 \times 0.050) = 4.4 \times 10^{-2}\ \text{T}$ ($44\ \text{mT}$). The reading rose, so the force on the magnet is downwards and the force on the wire is upwards; reversing the current would make the reading fall by the same amount.
Worked example (forces on a coil). A rectangular coil PQRS with sides PS and QR of length $8.0\ \text{cm}$ carries a current of $1.2\ \text{A}$ in a horizontal field of $0.30\ \text{T}$, with the plane of the coil parallel to the field. Describe the forces on the coil.
Sides PS and QR are at right angles to the field, so each feels $F = BIL = 0.30 \times 1.2 \times 0.080 = 2.9 \times 10^{-2}\ \text{N}$; the currents in them run in opposite directions, so by Fleming's rule the forces are equal and opposite, one up and one down. Sides PQ and SR are parallel to the field and feel no force. The two forces form a couple 力偶 that turns the coil about its axis; as the coil turns the forces stay the same size but their perpendicular separation shrinks, so the turning effect falls to zero when the plane of the coil is at right angles to the field. This is the principle of the electric motor and of the moving-coil meter.
This equation also defines the magnetic flux density 磁通密度 $B$:
So $B$ is the force per unit current per unit length on a wire at right angles to the field. Unit: tesla 特斯拉, $\text{T} = \text{N A}^{-1}\ \text{m}^{-1}$.
The two-mark definitions. Magnetic flux density is the force per unit current per unit length acting on a (straight) conductor placed at right angles to the field. Both "per unit current" and "per unit length" are needed, and so is "at right angles" (or "perpendicular"): without it the definition is false, because the force also depends on the angle. The tesla is the flux density that produces a force of one newton on one metre of conductor carrying a current of one ampere at right angles to the field: $1\ \text{T} = 1\ \text{N A}^{-1}\ \text{m}^{-1}$.
Use the left hand (Fleming's left-hand rule 弗莱明左手定则): first finger = Field, second finger = Current, thumb = force (thrust). Hold the three at right angles.
A current in a magnetic field feels a force F = BIL at right angles to both — reverse the current or flip the magnet and the force jumps the other way. · Arus dalam medan magnet mengalami gaya F = BIL tegak lurus terhadap keduanya — balikkan arus atau balikkan magnet, dan gaya akan berbalik arah.
| English | Bahasa Indonesia |
|---|---|
| couple/ˈkʌpl/ | kopel |
| magnetic flux density/mæɡˈnetɪk flʌks ˈdensɪti/ | densitas fluks magnet |
| tesla/ˈteslə/ | tesla |
| Fleming's left-hand rule/ˈflemɪŋz left hænd ruːl/ | aturan tangan kiri Fleming |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A charge $Q$ moving at velocity 速度 $v$ through a field feels
with $\theta$ the angle between $v$ and $B$. Same left-hand rule (the second finger is the motion of a positive charge — reverse it for a negative charge). The force is largest when $v$ is at right angles to $B$, and zero when $v$ is along $B$.
Two situations with no force. A charged particle in a magnetic field feels no force when it is stationary (no $v$) or when it moves parallel or antiparallel to the field ($\sin\theta = 0$). Contrast the electric field, which pushes a stationary charge just as hard: that difference is the whole point of the velocity selector below.
A charge moving at right angles to a uniform field 匀强场 feels a force at right angles to both $v$ and $B$. This force does no work (always at right angles to the motion), so the kinetic energy 动能 and speed stay constant — the particle moves in a circle. Set the magnetic force equal to the centripetal force 向心力:
Worked example. A proton (mass $1.7 \times 10^{-27}\ \text{kg}$, charge $1.6 \times 10^{-19}\ \text{C}$) moves at $2.0 \times 10^{6}\ \text{m s}^{-1}$ at right angles to a $0.50\ \text{T}$ field. Find the radius of its circular path.
So the radius depends on the momentum 动量 $mv$. The period is
which does not depend on the speed — a faster particle goes in a bigger circle but takes the same time per turn. If $v$ also has a part along $B$, that part is unchanged, and the path is a helix 螺旋.
"Explain why the path is circular." The magnetic force is always at right angles to the velocity, so it changes the direction of motion but not the speed, and its magnitude $BQv$ is constant because $B$, $Q$ and $v$ are constant. A force of constant size that stays perpendicular to the velocity is exactly a centripetal force, so the path is a circle. Give all three ideas: perpendicular, constant magnitude, speed unchanged.
Worked example. Electrons moving at $1.7 \times 10^{7}\ \text{m s}^{-1}$ enter a uniform field of flux density $4.8\ \text{mT}$ at right angles to it. Find the radius of their path. The electrons are then replaced by positrons 正电子 (same mass, charge $+e$) moving at $3.4 \times 10^{7}\ \text{m s}^{-1}$ along the same initial line: describe how the path changes.
The positrons are deflected the opposite way (positive charge, so Fleming's rule with the second finger along the motion gives a force in the opposite direction), and since $r \propto v$ their radius is twice as large, $4.0\ \text{cm}$. Their period is the same ($T = 2\pi m/BQ$ has no $v$ in it): they go round a bigger circle at twice the speed.
A slab of conductor carrying current $I$, in a field $B$ at right angles to the current, develops a voltage across its faces — the Hall voltage 霍尔电压 $V_{\text{H}}$ (the Hall effect 霍尔效应).
The moving charges feel a magnetic force $BQv_{\text{d}}$ ($v_{\text{d}}$ is the drift velocity 漂移速度), so they build up on one face, making an electric field 电场 $E$ that opposes more build-up. At steady state $eE = Bev_{\text{d}}$, so $E = B v_{\text{d}}$. With $V_{\text{H}} = E w$ and $I = n e v_{\text{d}} w t$:
where $q$ is the carrier charge. A Hall probe 霍尔探头 uses this to measure $B$: pass a known current through a thin semiconductor 半导体 slab and read $V_{\text{H}}$ (largest when the slab is at right angles to $B$).
Deriving $V_{\text{H}} = BI/(ntq)$ (the exam asks for this in full). Take a slice of width $d$, across which the Hall voltage appears, and thickness $t$ along $B$, so the current passes through a cross-section $A = dt$. (1) At steady state the electric force on a carrier balances the magnetic force: $qE = Bqv$, so $E = Bv$. (2) The Hall field is uniform across the width, so $V_{\text{H}} = Ed = Bvd$. (3) The current is $I = nAvq = n\,dt\,vq$, so $v = I/(n\,dt\,q)$. Substituting: $V_{\text{H}} = Bd \cdot I/(n\,dt\,q) = BI/(ntq)$. Two things to notice: the voltage appears across the faces perpendicular to the magnetic force, not across the ends where the current enters; and it is larger for a semiconductor, whose number density $n$ is millions of times smaller than a metal's, which is why probes use a semiconductor slice.
Using a Hall probe. The reading is proportional to $B$ but also depends on the orientation: it is a maximum when the plane of the slice is at right angles to the field and zero when the field lies in the plane of the slice. So (1) rotate the probe until the reading is largest, and (2) calibrate 校准 it in a known field, since $V_{\text{H}} \propto B$ at fixed current.
Worked example. A Hall probe gives a maximum reading of $24\ \text{mV}$ in a uniform field of $32\ \text{mT}$. In a second field, with the probe again turned for a maximum, the reading is $15\ \text{mV}$. Find the second flux density.
$V_{\text{H}} \propto B$ (same probe, same current): $B = 32 \times 15/24 = 20\ \text{mT}$. If the probe had been left at some other angle the reading would be smaller and $B$ would be underestimated, which is why "rotate for a maximum" is part of the method.
A velocity selector 速度选择器 uses crossed electric and magnetic fields to let through only one speed. With the electric force $qE$ and magnetic force $qvB$ set to oppose each other, the net force is zero only when
Particles at speed $E/B$ go straight through; faster or slower ones are deflected.
Deriving and explaining the selector. The magnetic force on a positive ion moving through the crossed fields is $BQv$, at right angles to its motion; the electric field is applied so that the electric force $QE$ is in the opposite direction (the plates must be arranged so that the field points against the magnetic force). For the ion to pass straight through, $QE = BQv$, so $v = E/B$, independent of the charge and the mass. A faster ion has a larger magnetic force and bends towards the magnetic-force side; a slower one is pushed the other way by the unchanged electric force. Asked "explain how an electric field can make the particle reach Z", say which direction the electric force must have, that it must equal the magnetic force in size, and give $E = Bv$. This is velocity selection 速度选择.
F = BQv
The magnetic force on a charge is proportional to its speed (for a fixed field and charge). · Gaya magnetik pada sebuah muatan berbanding lurus dengan kecepatannya (untuk medan dan muatan tetap).
| English | Bahasa Indonesia |
|---|---|
| uniform field/ˈjuːnɪfɔːm fiːld/ | medan seragam |
| velocity/vəˈlɒsɪti/ | kecepatan (vektor) |
| electric field/ɪˈlektrɪk fiːld/ | medan listrik |
| velocity selector/vəˈlɒsɪti sɪˈlektə/ | selektor kecepatan |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| centripetal force/senˈtrɪpɪtl fɔːs/ | kekuatan sentripetal |
| momentum/məʊˈmentəm/ | momentum |
| helix/ˈhiːlɪks/ | heliks |
| positrons/ˈpɒzɪtrɒnz/ | positron |
| Hall effect/hɔːl ɪˈfekt/ | efek Hall |
| Hall voltage/hɔːl ˈvəʊltɪdʒ/ | tegangan Hall |
| drift velocity/drɪft vəˈlɒsɪti/ | kecepatan rambat |
| Hall probe/hɔːl prəʊb/ | probe Hall |
| semiconductor/ˌsemɪkənˈdʌktə/ | semikonduktor |
| calibrate/ˈkælɪbreɪt/ | kalibrasi |
| velocity selection/vəˈlɒsɪti sɪˈlekʃn/ | pemilihan kecepatan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Two long parallel wires each sit in the other's magnetic field. Using Fleming's left-hand rule: parallel currents (same direction) attract; antiparallel currents (opposite directions) repel. This is the basis of the SI definition of the ampere.
Explaining the force (a standard three-marker). (1) The current in P produces a magnetic field around it (circles centred on P). (2) At Q this field is at right angles to the current in Q, so Q, a current-carrying conductor in a magnetic field, feels a force $BIL$. (3) Fleming's left-hand rule gives its direction: towards P for currents in the same direction, away for opposite currents. By Newton's third law the force on P is equal and opposite, so both wires attract (or both repel) with the same force. Doubling the current in one wire doubles the force on both: the force on Q rises because $I_{\text{Q}}$ has doubled, the force on P because the field it sits in has doubled.
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The magnetic flux 磁通量 $\Phi$ through a flat area $A$ at right angles to $B$ is
If the area's normal is at angle $\theta$ to $B$, use $\Phi = B A \cos\theta$. Unit: weber 韦伯, $\text{Wb} = \text{T m}^{2}$. For a coil 线圈 of $N$ turns, the flux linkage 磁链 is $N\Phi = N B A$.
The two-mark definitions. Magnetic flux is the product of the magnetic flux density and the area (of the circuit) perpendicular to the field. Flux linkage is the product of the flux and the number of turns of the coil, $N\Phi$, in weber-turns. Say "perpendicular" (or "normal to the field"): $\Phi = BA$ is only true for the component of the area at right angles to $B$.
Worked example. A small coil of 64 turns and cross-sectional area $0.71\ \text{cm}^{2}$ is placed inside a long solenoid, on its axis, where the flux density is $2.5\ \text{mT}$. Find the flux linkage of the coil.
The flux through the coil uses the coil's own area, not the solenoid's: $\Phi = BA = (2.5 \times 10^{-3})(0.71 \times 10^{-4}) = 1.8 \times 10^{-7}\ \text{Wb}$, so the flux linkage is $N\Phi = 64 \times 1.8 \times 10^{-7} = 1.1 \times 10^{-5}\ \text{Wb}$. When the solenoid's current changes, the e.m.f. induced in the small coil is the rate of change of this linkage.
When the flux linkage through a circuit changes, an electromotive force 电动势 (e.m.f.) is induced — this is electromagnetic induction 电磁感应.
Faraday's law 法拉第定律: the induced e.m.f. equals the rate of change of flux linkage:
Worked example. A coil of $200$ turns and area $0.010\ \text{m}^{2}$ sits with its plane at right angles to a $0.50\ \text{T}$ field. The field falls steadily to zero in $0.20\ \text{s}$. Find the average induced e.m.f.
The flux linkage changes from $N\Phi = NBA = 200 \times 0.50 \times 0.010 = 1.0\ \text{Wb}$ to zero, so
Lenz's law 楞次定律: the induced e.m.f. acts to oppose the change that makes it. This is conservation of energy 能量守恒 — if it reinforced the change, energy would come from nothing. Combined:
Reading an e.m.f. off a flux graph. The induced e.m.f. is the gradient of the flux-linkage graph (with a minus sign for Lenz). A steadily rising flux gives a constant e.m.f.; a constant flux gives zero e.m.f. however large the flux is; a faster fall gives a larger e.m.f. of the opposite sign. "Sketch the variation with time of the e.m.f." is answered by differentiating the flux graph by eye, section by section.
Worked example. The 64-turn coil above sits in a solenoid whose flux density rises uniformly from $0$ to $2.5\ \text{mT}$ in $40\ \text{ms}$ and is then held steady. Find the e.m.f. induced in the coil during the rise, and afterwards.
During the rise the flux linkage grows steadily to $1.1 \times 10^{-5}\ \text{Wb}$: $\varepsilon = \Delta(N\Phi)/\Delta t = 1.1 \times 10^{-5}/0.040 = 2.8 \times 10^{-4}\ \text{V}$ ($0.28\ \text{mV}$), constant while the rise lasts. Afterwards the flux linkage is constant, so the e.m.f. is zero, even though the coil is still in a strong field: it is the change that induces.
Worked example (flux cutting). An aircraft with a wingspan 翼展 of $60\ \text{m}$ flies horizontally at $250\ \text{m s}^{-1}$ where the vertical component of the Earth's field is $45\ \mu\text{T}$. Find the e.m.f. between the wingtips.
In time $\Delta t$ the wing sweeps out an area $L v \Delta t$, cutting flux $B L v \Delta t$, so $\varepsilon = BLv = (45 \times 10^{-6})(60)(250) = 0.68\ \text{V}$. Only the component of $B$ perpendicular to the swept area counts (here the vertical one). There is a p.d. between the wingtips but no current, because there is no complete circuit; Fleming's rules give which tip is positive.
Electromagnetic braking, explained with Lenz's law. A vehicle's aluminium disc rotates between the poles of an electromagnet. The disc cuts the field lines, so an e.m.f. is induced in it; the disc is a conductor, so eddy currents flow; by Lenz's law these currents produce forces that oppose the motion of the disc, slowing the vehicle. The kinetic energy is converted to thermal energy in the disc. The braking is stronger at high speed (a larger rate of flux cutting, so larger currents) and fades as the vehicle slows; it cannot hold a stationary vehicle, because no motion means no induced current.
A magnet oscillating in a coil. A bar magnet on a spring oscillates in and out of a coil connected to a resistor. The changing flux linkage induces an e.m.f.; with the circuit complete a current flows and, by Lenz's law, the coil's field opposes the magnet's motion, so the oscillation is damped: energy leaves the oscillation as heat in the resistor and the amplitude decays. Open the switch and the e.m.f. is still induced but no current flows, so the coil no longer damps the motion.
From $\varepsilon = N\,d\Phi/dt$ with $\Phi = BA$: more turns $N$, a stronger $B$, a larger area $A$, or a faster change — each gives a larger induced e.m.f.
Finding the direction with Lenz's law. Push the N pole of a magnet towards a coil and the induced current makes the near face of the coil a north pole, to repel the approaching magnet; pull it away and the near face becomes a south pole, to attract it. In each case the person moving the magnet does work against that force, and that work is the source of the electrical energy: Lenz's law is conservation of energy in disguise. Two sentences to learn: the induced e.m.f. acts in such a direction as to produce effects that oppose the change producing it (Lenz), and the induced e.m.f. is proportional to the rate of change of flux linkage (Faraday).
Move the magnet through the coil — a current is induced only while the field is changing. Faster gives more current; flip the magnet to reverse it. · Gerakkan magnet melalui kumparan — arus terinduksi hanya saat medan berubah. Lebih cepat memberikan arus lebih besar; balik magnet untuk membaliknya.
| English | Bahasa Indonesia |
|---|---|
| coil/kɔɪl/ | kumparan |
| magnetic flux/mæɡˈnetɪk flʌks/ | fluks magnetik |
| electromagnetic induction/ɪˌlektrəʊməɡˈnetɪk ɪnˈdʌkʃn/ | induksi elektromagnetik |
| weber/ˈveɪbə/ | weber |
| flux linkage/flʌks ˈlɪŋkɪdʒ/ | linkage fluks |
| Lenz's law/ˈlentsɪz lɔː/ | Hukum Lenz |
| electromotive force/ɪˌlektrəʊˈməʊtɪv fɔːs/ | gaya gerak listrik |
| Faraday's law/ˈfærədeɪz lɔː/ | Hukum Faraday |
| conservation of energy/ˌkɒnsəˈveɪʃn ɒv ˈenədʒi/ | hukum kekekalan energi |
| wingspan/ˈwɪŋspæn/ | rentang sayap |
| galvanometer/ˌɡælvəˈnɒmɪtə/ | galvanometer |
| eddy currents/ˈedi ˈkʌrənts/ | arus eddy |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| magnetic field | a region in which a force acts on a moving charge, a current-carrying conductor or a magnetic material |
| magnetic flux density | the force per unit current per unit length on a straight conductor placed at right angles to the field |
| tesla | the flux density producing a force of 1 N on 1 m of conductor carrying 1 A at right angles to the field |
| magnetic flux | the product of the flux density and the area perpendicular to the field, $\Phi = BA$ |
| flux linkage | the product of the flux through a coil and its number of turns, $N\Phi$ |
| weber | one tesla metre squared |
| Faraday's law | the induced e.m.f. is proportional to the rate of change of flux linkage |
| Lenz's law | the induced e.m.f. acts in such a direction as to produce effects that oppose the change producing it |
| Hall voltage | the p.d. that develops across a current-carrying slice in a magnetic field, $V_{\text{H}} = BI/(ntq)$ |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
An alternating current 交流电 (a.c.) keeps reversing direction. Mains supply is sinusoidal a.c.: $I$ or $V$ follows a sine wave in time:
(For a purely resistive load the voltage 电压 and current 电流 are in phase, which is the case in this syllabus.)
The two-mark definitions. The frequency of an alternating current is the number of complete cycles per unit time (say "per second" or "per unit time", and "complete cycles" or "oscillations"). The period is the time for one complete cycle, and the peak value is the maximum value of the current or voltage during a cycle. Note that the mean value of a sinusoidal current over a cycle is zero, which is exactly why the r.m.s. value is needed to describe it.
Reading the equation. In $V = V_{0} \sin(\omega t)$ the number in front is the peak value and the number multiplying $t$ is $\omega = 2\pi f$; the angle $\omega t$ is in radians 弧度, so set the calculator to radians before evaluating. A supply written with $\cos$ instead of $\sin$ is the same wave starting at its peak rather than at zero.
Worked example. The output of a supply is $V = 320 \sin(100\pi t)$ (volts, seconds). Find the peak value, the frequency, the period and the r.m.s. value, and the first time after $t = 0$ at which $V = 160\ \text{V}$.
Peak $V_{0} = 320\ \text{V}$. $\omega = 100\pi\ \text{rad s}^{-1}$, so $f = \omega/2\pi = 50\ \text{Hz}$ and $T = 1/f = 0.020\ \text{s}$. $V_{\text{r.m.s.}} = 320/\sqrt{2} = 226\ \text{V}$. For $V = 160\ \text{V}$: $\sin(100\pi t) = 0.5$, so $100\pi t = \pi/6$ and $t = 1/600 = 1.7 \times 10^{-3}\ \text{s}$. (For $V = 18\cos(40\pi t)$ the same reading gives $V_{0} = 18\ \text{V}$, $f = 20\ \text{Hz}$, $T = 50\ \text{ms}$, and a sketch that starts at $+18\ \text{V}$.)
Same as for any wave (Topic 7), using a cathode-ray oscilloscope 示波器:
Arus bolak-balik (a.c.) terus-menerus membalik arah. Suplai jaringan menggunakan a.c. sinusoidal: $I$ atau $V$ mengikuti gelombang sinus terhadap waktu:
(Untuk beban murni resistif, tegangan dan arus sefasa, yang merupakan kasus dalam silabus ini.)
Definisi bernilai dua poin. Frekuensi arus bolak-balik adalah jumlah siklus lengkap per satuan waktu (katakan "per detik" atau "per satuan waktu", dan "siklus lengkap" atau "osilasi"). Periode adalah waktu untuk satu siklus lengkap, dan nilai puncak adalah nilai maksimum arus atau tegangan selama satu siklus. Perhatikan bahwa nilai rata-rata arus sinusoidal selama satu siklus adalah nol, yang tepat menjadi alasan mengapa nilai r.m.s. diperlukan untuk mendeskripsikannya.
Membaca persamaan. Dalam $V = V_{0} \sin(\omega t)$, angka di depan adalah nilai puncak dan angka yang mengalikan $t$ adalah $\omega = 2\pi f$; sudut $\omega t$ berada dalam radian, jadi atur kalkulator ke mode radian sebelum melakukan perhitungan. Suplai yang ditulis dengan $\cos$ alih-alih $\sin$ adalah gelombang yang sama yang dimulai dari puncaknya bukan dari nol.
Contoh terpecahkan. Output suplai adalah $V = 320 \sin(100\pi t)$ (volt, detik). Temukan nilai puncak, frekuensi, periode, dan nilai r.m.s., serta waktu pertama setelah $t = 0$ ketika $V = 160\ \text{V}$.
G.g.b. puncak $V_{0} = 320\ \text{V}$. $\omega = 100\pi\ \text{rad s}^{-1}$, sehingga $f = \omega/2\pi = 50\ \text{Hz}$ dan $T = 1/f = 0.020\ \text{s}$. $V_{\text{r.m.s.}} = 320/\sqrt{2} = 226\ \text{V}$. Untuk $V = 160\ \text{V}$: $\sin(100\pi t) = 0.5$, sehingga $100\pi t = \pi/6$ dan $t = 1/600 = 1.7 \times 10^{-3}\ \text{s}$. (Untuk $V = 18\cos(40\pi t)$ pembacaan yang sama menghasilkan $V_{0} = 18\ \text{V}$, $f = 20\ \text{Hz}$, $T = 50\ \text{ms}$, dan sketsa yang dimulai dari $+18\ \text{V}$.)
Sama seperti untuk gelombang apa pun (Topik 7), menggunakan osiloskop sinar katode:
I = a sin(bt)
AC is a sine wave — amplitude is the peak, b sets the frequency. · AC adalah gelombang sinus — amplitudo adalah puncak, b menetapkan frekuensi.
| English | Bahasa Indonesia |
|---|---|
| alternating current/ˈɔːltəneɪtɪŋ ˈkʌrənt/ | arus bolak-balik |
| current/ˈkʌrənt/ | arus |
| voltage/ˈvəʊltɪdʒ/ | tegangan |
| load/ləʊd/ | beban |
| period/ˈpɪərɪəd/ | tempoh |
| frequency/ˈfriːkwənsi/ | frekuensi |
| angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ | frekuensi sudut |
| peak value/piːk ˈvæljuː/ | nilai puncak |
| peak-to-peak value/piːk tə piːk ˈvæljuː/ | nilai puncak-ke-puncak |
| radians/ˈreɪdɪənz/ | radian |
| cathode-ray oscilloscope/ˈkæθəʊd reɪ ɒˈsɪləskəʊp/ | osiloskop sinar katoda |
| time-base/taɪm beɪs/ | basis waktu |
For a resistive load $R$, the instant power 功率 is $P(t) = I(t)^{2} R$. With $I = I_{0}\sin(\omega t)$:
This is always positive, with peak $I_{0}^{2} R$ and minimum zero, oscillating at twice the frequency of $I$. The mean of $\sin^{2}(\omega t)$ over a cycle is $\tfrac{1}{2}$, so the average power is
Average a.c. power in a resistor is half the peak power.
"Show by calculation that the mean power is half the peak power." A supply of peak value $12\ \text{V}$ drives a $680\ \Omega$ resistor. Peak power: $P_{0} = V_{0}^{2}/R = 12^{2}/680 = 0.212\ \text{W}$, the instantaneous power at the moment the voltage is at its peak. Mean power: use the r.m.s. value, $V_{\text{r.m.s.}} = 12/\sqrt{2} = 8.49\ \text{V}$, so $\langle P \rangle = V_{\text{r.m.s.}}^{2}/R = 8.49^{2}/680 = 0.106\ \text{W}$, exactly half. The two calculations must be shown separately; writing "$\tfrac{1}{2}$ of $0.212$" scores nothing, because that is the thing being shown. The $\tfrac{1}{2}$ is $(1/\sqrt{2})^{2}$: squaring the r.m.s. factor.
Untuk beban resistif $R$, daya sesaat adalah $P(t) = I(t)^{2} R$. Dengan $I = I_{0}\sin(\omega t)$:
Ini selalu positif, dengan puncak $I_{0}^{2} R$ dan minimum nol, bergetar pada dua kali frekuensi $I$. Rata-rata $\sin^{2}(\omega t)$ selama satu siklus adalah $\tfrac{1}{2}$, sehingga daya rata-rata adalah
Daya a.c. rata-rata pada resistor adalah setengah dari daya puncak.
"Tunjukkan dengan perhitungan bahwa daya rata-rata adalah setengah dari daya puncak." Suplai dengan nilai puncak $12\ \text{V}$ mendorong resistor $680\ \Omega$. Daya puncak: $P_{0} = V_{0}^{2}/R = 12^{2}/680 = 0.212\ \text{W}$, yaitu daya sesaat pada saat tegangan mencapai puncaknya. Daya rata-rata: gunakan nilai r.m.s., $V_{\text{r.m.s.}} = 12/\sqrt{2} = 8.49\ \text{V}$, sehingga $\langle P \rangle = V_{\text{r.m.s.}}^{2}/R = 8.49^{2}/680 = 0.106\ \text{W}$, tepat setengahnya. Kedua perhitungan harus ditunjukkan secara terpisah; menulis "$\tfrac{1}{2}$ dari $0.212$" tidak mendapat poin, karena itu adalah hal yang ingin dibuktikan. Faktor $\tfrac{1}{2}$ adalah $(1/\sqrt{2})^{2}$: kuadratkan faktor r.m.s.
| English | Bahasa Indonesia |
|---|---|
| power/ˈpaʊə/ | kuasa |
The r.m.s. current $I_{\text{r.m.s.}}$ is the steady direct current that would give the same average power in the same resistance 电阻 $R$. From $\langle P \rangle = I_{\text{r.m.s.}}^{2} R = \tfrac{1}{2} I_{0}^{2} R$:
Worked example. An a.c. supply has a peak voltage of $12\ \text{V}$. Find its r.m.s. voltage.
The $\sqrt{2}$ comes from the name root-mean-square 均方根: $I_{\text{r.m.s.}} = \sqrt{\langle I^{2} \rangle}$ and $\langle \sin^{2}\rangle = \tfrac{1}{2}$. (Only the sinusoidal case is needed.)
The two-mark definition. The r.m.s. value of an alternating current is the value of the direct (steady) current that would dissipate the same (mean) power in the same resistor. "By reference to the heating effect" means exactly this sentence: same resistor, same power (or same heating), direct current. A definition that only says "$I_{0}/\sqrt{2}$" scores nothing, because that formula is true only for a sine wave.
Worked example (non-sinusoidal). A current through a resistor is $+2.0\ \text{A}$ for the first half of each cycle and $-1.0\ \text{A}$ for the second half, a square wave 方波. Find its r.m.s. value and the mean power in a $10\ \Omega$ resistor.
Square the current: $4.0\ \text{A}^{2}$ for half the time and $1.0\ \text{A}^{2}$ for the other half, so the mean square is $(4.0 + 1.0)/2 = 2.5\ \text{A}^{2}$ and $I_{\text{r.m.s.}} = \sqrt{2.5} = 1.6\ \text{A}$. Mean power $= I_{\text{r.m.s.}}^{2} R = 2.5 \times 10 = 25\ \text{W}$. The sign of the current does not matter to the heating (it is squared away), and $I_{0}/\sqrt{2}$ would have given the wrong answer, $1.4\ \text{A}$: that shortcut is for sine waves only.
Quoted a.c. values are r.m.s. values. "$230\ \text{V}$ mains" means $V_{\text{r.m.s.}} = 230\ \text{V}$, with peak $V_{0} = 230\sqrt{2} \approx 325\ \text{V}$. Components must be rated for the peak, not the r.m.s. Average power then takes the d.c. form:
Worked example. A heater of resistance $50\ \Omega$ is connected to the $230\ \text{V}$ r.m.s. mains. Find the r.m.s. current and the average power dissipated.
Worked example. A sinusoidal supply of r.m.s. value $4.2\ \text{V}$ and frequency $50\ \text{kHz}$ is connected across a $150\ \Omega$ resistor. Find the peak current, the mean power, and write down an equation for the current.
$V_{0} = 4.2\sqrt{2} = 5.9\ \text{V}$, so $I_{0} = V_{0}/R = 5.9/150 = 0.040\ \text{A}$ (or $I_{\text{r.m.s.}} = 4.2/150 = 0.028\ \text{A}$ and then $\times\sqrt{2}$). Mean power $= V_{\text{r.m.s.}}^{2}/R = 4.2^{2}/150 = 0.12\ \text{W}$. With $\omega = 2\pi f = 2\pi \times 5.0 \times 10^{4} = 3.1 \times 10^{5}\ \text{rad s}^{-1}$: $I = 0.040 \sin(3.1 \times 10^{5} t)$ (amps, seconds). Keep the two families apart: peak values go into the equation and into component ratings; r.m.s. values go into power.
Arus r.m.s. $I_{\text{r.m.s.}}$ adalah arus searah stabil yang akan menghasilkan daya rata-rata yang sama pada resistansi yang sama $R$. Dari $\langle P \rangle = I_{\text{r.m.s.}}^{2} R = \tfrac{1}{2} I_{0}^{2} R$:
Contoh terpecahkan. Suplai a.c. memiliki tegangan puncak $12\ \text{V}$. Temukan tegangan r.m.s.-nya.
$\sqrt{2}$ berasal dari nama akar-kuadrat-rata: $I_{\text{r.m.s.}} = \sqrt{\langle I^{2} \rangle}$ dan $\langle \sin^{2}\rangle = \tfrac{1}{2}$. (Kasus sinusoidal saja yang dibutuhkan.)
Definisi bernilai dua. Nilai r.m.s. dari arus bolak-balik adalah nilai arus searah (stabil) yang akan menyebarkan daya yang sama (rata-rata) pada resistor yang sama. "Dengan merujuk pada efek pemanasan" berarti persis kalimat ini: resistor yang sama, daya yang sama (atau pemanasan yang sama), arus searah. Definisi yang hanya menyebutkan "$I_{0}/\sqrt{2}$" tidak mendapat nilai, karena rumus itu hanya benar untuk gelombang sinus.
Contoh terpecahkan (non-sinusoidal). Arus yang mengalir melalui sebuah resistor bernilai $+2.0\ \text{A}$ untuk setengah pertama setiap siklus dan $-1.0\ \text{A}$ untuk setengah keduanya, membentuk gelombang persegi. Temukan nilai r.m.s. dan daya rata-rata pada resistor $10\ \Omega$.
Kuadratkan arus: $4.0\ \text{A}^{2}$ selama separuh waktu dan $1.0\ \text{A}^{2}$ untuk separuh lainnya, sehingga rata-rata kuadrat adalah $(4.0 + 1.0)/2 = 2.5\ \text{A}^{2}$ dan $I_{\text{r.m.s.}} = \sqrt{2.5} = 1.6\ \text{A}$. Daya rata-rata $= I_{\text{r.m.s.}}^{2} R = 2.5 \times 10 = 25\ \text{W}$. Tanda arus tidak berpengaruh terhadap pemanasan (karena dikuadratkan), dan $I_{0}/\sqrt{2}$ akan menghasilkan jawaban yang salah, $1.4\ \text{A}$: jalan pintas itu hanya untuk gelombang sinus.
Nilai a.c. yang dikutip adalah nilai r.m.s. "Jaringan listrik $230\ \text{V}$" berarti $V_{\text{r.m.s.}} = 230\ \text{V}$, dengan puncak $V_{0} = 230\sqrt{2} \approx 325\ \text{V}$. Komponen harus dinilai untuk nilai puncak, bukan r.m.s. Daya rata-rata kemudian mengambil bentuk d.c.:
Contoh terpecahkan. Pemanas dengan hambatan $50\ \Omega$ terhubung ke mains r.m.s. $230\ \text{V}$. Temukan arus r.m.s. dan daya rata-rata yang dibuang.
Contoh terpecahkan. Suplai sinusoidal dengan nilai r.m.s. $4.2\ \text{V}$ dan frekuensi $50\ \text{kHz}$ dihubungkan melintasi resistor $150\ \Omega$. Temukan arus puncak, daya rata-rata, dan tuliskan persamaan untuk arus tersebut.
$V_{0} = 4.2\sqrt{2} = 5.9\ \text{V}$, sehingga $I_{0} = V_{0}/R = 5.9/150 = 0.040\ \text{A}$ (atau $I_{\text{r.m.s.}} = 4.2/150 = 0.028\ \text{A}$ dan kemudian $\times\sqrt{2}$). Daya rata-rata $= V_{\text{r.m.s.}}^{2}/R = 4.2^{2}/150 = 0.12\ \text{W}$. Dengan $\omega = 2\pi f = 2\pi \times 5.0 \times 10^{4} = 3.1 \times 10^{5}\ \text{rad s}^{-1}$: $I = 0.040 \sin(3.1 \times 10^{5} t)$ (ampere, detik). Pisahkan kedua keluarga ini: nilai puncak masuk ke dalam persamaan dan penilaian komponen; nilai r.m.s. masuk ke dalam perhitungan daya.
| English | Bahasa Indonesia |
|---|---|
| root-mean-square/ruːt miːn skweə/ | akar rata-rata kuadrat |
| resistance/rɪˈzɪstəns/ | penolakan |
| square wave/skweə weɪv/ | gelombang persegi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Rectification 整流 turns an alternating voltage into a one-direction (d.c.-like) voltage, using diodes 二极管 (which conduct in only one direction).
"State what is meant by rectification." The conversion of an alternating current (or voltage) into a direct current (or voltage): one that flows in one direction only, however much it varies. A diode conducts only when it is forward-biased 正向偏置, that is, when its anode (the flat end of the symbol's triangle) is more positive than its cathode (the bar); otherwise it is reverse-biased and behaves like an open switch. Treat it as ideal: zero resistance one way, infinite the other.
A single diode in series with the load passes only the positive half of each cycle; in the negative half the diode is reverse-biased 反向偏置 and no current flows. This is half-wave rectification 半波整流.
Output: positive half-waves with flat zero gaps. The mean output is $V_{0}/\pi \approx 0.32 V_{0}$. Drawback: half the input is wasted and the output is very uneven.
Completing the circuit. "Complete the diagram to produce half-wave rectification" needs one diode in series between the supply and the load, with the triangle pointing the way the output current must flow, and $V_{\text{OUT}}$ taken across the load. If a capacitor is to smooth the output it goes across the load (in parallel); a capacitor in series would block the d.c. altogether.
A bridge rectifier 桥式整流器 uses four diodes arranged so the current through the load always flows the same way, whichever a.c. terminal is positive — full-wave rectification 全波整流. On each half-cycle a different pair of diodes conducts, but the load always sees the same direction.
Explaining the bridge (the standard four-marker). When terminal P is positive, current leaves P, passes through the diode pointing away from P to the top of the load, flows down through the load and returns to Q through the diode pointing towards Q; the other two diodes are reverse-biased and carry nothing. When Q is positive the other pair conducts, but they are arranged so that the current still enters the load at the top: the load current is in the same direction in both half-cycles. Name the diodes in each half-cycle and say which end of the load is positive.
Completing a bridge. Given a bridge with diodes missing, remember the rule for every diode: conventional current flows through it in the direction of the triangle. Both diodes joined to the positive output terminal must point towards it; both joined to the negative output terminal must point away from it. A bridge with a diode the wrong way round either short-circuits the supply on one half-cycle or passes nothing.
Output: a continuous run of positive half-waves (no gaps), at twice the input frequency. The mean output is $2V_{0}/\pi \approx 0.64 V_{0}$ — double the half-wave value. It uses all the input and is smoother and easier to filter.
"State the difference between half-wave and full-wave rectification." In half-wave rectification only one half of each input cycle appears at the output and the other half is blocked (the output is zero for half the time); in full-wave rectification both halves appear, one of them inverted, so there are no gaps and the output has twice the input frequency. A sketch should show, for half-wave, positive humps separated by flat zero sections of equal length; for full-wave, positive humps joined at the zero line.
Penyearahan mengubah tegangan bolak-balik menjadi tegangan satu arah (mirip d.c.) menggunakan dioda (yang menghantarkan arus hanya dalam satu arah).
"Nyatakan apa yang dimaksud dengan penyearahan." Konversi arus bolak-balik (atau tegangan) menjadi arus searah (atau tegangan): yaitu arus yang mengalir hanya dalam satu arah, seberapa pun perubahannya. Dioda menghantarkan arus hanya ketika bias maju, yaitu ketika anodanya (ujung datar dari simbol segitiga) lebih positif daripada katodnya (garis); jika tidak, ia bias mundur dan berperilaku seperti saklar terbuka. Anggaplah sebagai ideal: hambatan nol satu arah, tak hingga di arah lain.
Satu dioda seri dengan beban hanya meneruskan setengah positif dari setiap siklus; pada setengah negatif, dioda tersebut bias mundur dan tidak ada arus yang mengalir. Ini disebut penyearahan gelombang setengah.
Output: gelombang setengah positif dengan celah nol yang datar. Output rata-rata adalah $V_{0}/\pi \approx 0.32 V_{0}$. Kekurangannya: separuh input terbuang dan output sangat tidak merata.
Melengkapi sirkuit. "Lengkapi diagram untuk menghasilkan penyearahan gelombang setengah" membutuhkan satu dioda seri antara suplai dan beban, dengan segitiga menunjuk ke arah aliran arus output, dan $V_{\text{OUT}}$ diambil di atas beban. Jika kapasitor digunakan untuk menapis output, ia dipasang di atas beban (paralel); kapasitor seri akan memblokir d.c. sepenuhnya.
Rectifier jembatan menggunakan empat dioda yang diatur sedemikian rupa sehingga arus melalui beban selalu mengalir ke satu arah yang sama, terlepas dari terminal a.c. mana yang positif — penyearahan gelombang penuh. Pada setiap setengah siklus, pasangan dioda berbeda yang menghantarkan arus, tetapi beban selalu melihat arah yang sama.
Menjelaskan jembatan (marker standar empat). Ketika terminal P positif, arus keluar dari P, melewati dioda yang menjauh dari P menuju atas beban, mengalir turun melalui beban dan kembali ke Q melalui dioda yang mengarah ke Q; dua dioda lainnya bias mundur dan tidak mengalirkan arus. Ketika Q positif, pasangan lainnya menghantarkan, tetapi mereka diatur sedemikian rupa agar arus tetap memasuki beban dari bagian atas: arus beban berada dalam arah yang sama pada kedua setengah siklus. Sebutkan nama dioda pada setiap setengah siklus dan sebutkan ujung mana dari beban yang positif.
Menyempurnakan jembatan. Diberikan sebuah jembatan dengan dioda yang hilang, ingat aturan untuk setiap dioda: arus konvensional mengalir melaluinya searah dengan segitiga. Kedua dioda yang terhubung ke terminal output positif harus mengarah menuju terminal tersebut; kedua dioda yang terhubung ke terminal output negatif harus mengarah menjauhi terminal tersebut. Jembatan dengan satu dioda terbalik akan menyebabkan korsleting pada sumber pada setengah siklus atau tidak mengalirkan apa pun.
Output: rentetan gelombang positif penuh yang berkelanjutan (tanpa jeda), pada frekuensi input dua kali lipat. Rata-rata output adalah $2V_{0}/\pi \approx 0.64 V_{0}$ — dua kali nilai gelombang separuh. Ini menggunakan seluruh input dan lebih halus serta lebih mudah difilter.
"Sebutkan perbedaan antara penyearahan gelombang separuh dan gelombang penuh." Dalam penyearahan gelombang separuh hanya satu setengah dari setiap siklus input yang muncul di output dan setengah lainnya diblokir (output nol selama setengah waktu); dalam penyearahan gelombang penuh kedua setengahnya muncul, salah satunya terbalik, sehingga tidak ada jeda dan output memiliki frekuensi dua kali lipatnya input. Sketsa harus menunjukkan, untuk gelombang separuh, hump positif yang dipisahkan oleh bagian datar nol yang panjangnya sama; untuk gelombang penuh, hump positif yang menyambung di garis nol.
Watch alternating input become a smoother direct output. · Perhatikan input bolak-balik berubah menjadi output searah yang lebih halus.
| English | Bahasa Indonesia |
|---|---|
| rectification/ˌrektɪfɪˈkeɪʃn/ | rektifikasi |
| diode/ˈdaɪəʊd/ | dioda |
| forward-biased/ˈfɔːwəd ˈbaɪəst/ | bias maju |
| reverse-biased/rɪˈvɜːs ˈbaɪəst/ | bias mundur |
| half-wave rectification/hɑːf weɪv ˌrektɪfɪˈkeɪʃn/ | penyearahan setengah-gelombang |
| full-wave rectification/fʊl weɪv ˌrektɪfɪˈkeɪʃn/ | penyearahan gelombang penuh |
| bridge rectifier/brɪdʒ ˈrektɪfaɪə/ | penyearah jembatan |
| time constant/taɪm ˈkɒnstənt/ | konstanta waktu |
| ripple/ˈrɪpl/ | riak |
A rectifier's output is still bumpy. To smooth it, put a capacitor 电容器 $C$ in parallel with the load $R$.
The output now sits near the peak with small dips. The size of the dips is the ripple 纹波 (this whole step is called smoothing 平滑).
Worked example. A half-wave rectifier fed from a $50\ \text{Hz}$ supply has a $470\ \mu\text{F}$ capacitor across its $1.2\ \text{k}\Omega$ load 负载. Estimate the fractional fall in the output between peaks, and say how it changes with a bridge rectifier.
The peaks are $T = 1/50 = 20\ \text{ms}$ apart (half-wave: one peak per cycle). The time constant is $RC = (1.2 \times 10^{3})(470 \times 10^{-6}) = 0.56\ \text{s}$. Between peaks the capacitor discharges to $V_{0} e^{-t/RC} = V_{0} e^{-0.020/0.56} = 0.965\,V_{0}$, a fall of about $3.5\%$. With full-wave rectification the peaks are $10\ \text{ms}$ apart, so the fall is only $1.8\%$; doubling the capacitance would halve it again. Because $t \ll RC$ the fall is approximately $V_{0}\, t/(RC)$: the ripple is proportional to the time between peaks and inversely proportional to both $R$ and $C$. (A network of capacitors across the output combines by the rules of Topic 19 before it is used here.)
In short, a large $RC$ compared with the time between peaks gives a smoother output.
Sketching the smoothed output. Draw the unsmoothed humps faintly first. The smoothed curve touches each peak, then falls along a gentle curve (steepest just after the peak) until the next hump rises to meet it, where it turns sharply upwards and follows the hump to the peak. It never falls to zero, and it never rises above the peak. For half-wave rectification the decay has to last through the missing half-cycle as well, so the ripple is about twice that of full-wave for the same $RC$. If the question changes $C$ or $R$, redraw on the same axes: a larger $RC$ hugs the peak line more closely; a smaller $RC$ sags further.
The smoothing capacitor reduces the ripple, giving a steadier d.c. voltage suitable for sensitive electronics.
Output penyearah masih bergelombang. Untuk menghaluskannya, pasanglah kapasitor $C$ secara paralel dengan beban $R$.
Output sekarang berada di dekat puncak dengan lekukan kecil. Ukuran lekukan tersebut adalah ripar (seluruh langkah ini disebut penghalusan).
Contoh soal terpecahkan. Penyearah gelombang separuh yang dialiri dari sumber $50\ \text{Hz}$ memiliki kapasitor $470\ \mu\text{F}$ yang dipasangkan melintasi beban $1.2\ \text{k}\Omega$. Perkirakan penurunan pecahan pada output antar-puncak, dan jelaskan bagaimana hal itu berubah dengan penyearah jembatan.
Puncak-puncaknya terpisah $T = 1/50 = 20\ \text{ms}$ (gelombang separuh: satu puncak per siklus). Konstanta waktunya adalah $RC = (1.2 \times 10^{3})(470 \times 10^{-6}) = 0.56\ \text{s}$. Antara puncak kapasitor kosong hingga $V_{0} e^{-t/RC} = V_{0} e^{-0.020/0.56} = 0.965\,V_{0}$, penurunan sekitar $3.5\%$. Dengan penyearahan gelombang penuh, puncak-puncaknya terpisah $10\ \text{ms}$, sehingga penurunannya hanya $1.8\%$; menggandakan kapasitansi akan memotongnya menjadi setengah lagi. Karena $t \ll RC$ penurunan tersebut kira-kira $V_{0}\, t/(RC)$: ripar sebanding dengan waktu antar-puncak dan berbanding terbalik dengan keduanya $R$ dan $C$. (Jaringan kapasitor melintasi output digabungkan sesuai aturan Topik 19 sebelum digunakan di sini.)
Singkatnya, konstanta waktu $RC$ yang besar dibandingkan dengan waktu antar-puncak menghasilkan output yang lebih halus.
Menggambar sketsa output yang dihaluskan. Gambarlah hump yang belum dihaluskan samar-samar terlebih dahulu. Kurva yang dihaluskan sentuh setiap puncak, lalu turun sepanjang kurva landai (paling curam tepat setelah puncak) hingga hump berikutnya naik untuk menemuinya, di mana ia berbelok tajam ke atas dan mengikuti hump menuju puncak. Ia tidak pernah turun ke nol, dan tidak pernah naik melebihi puncak. Untuk penyearahan gelombang separuh, peluruhan harus berlangsung melalui setengah siklus yang hilang juga, sehingga riparnya sekitar dua kali lipat dari gelombang penuh untuk konstanta waktu $RC$ yang sama. Jika soal mengubah $C$ atau $R$, gambar ulang pada sumbu yang sama: konstanta waktu $RC$ yang lebih besar akan lebih dekat menempel pada garis puncak; konstanta waktu $RC$ yang lebih kecil akan melengkung lebih jauh ke bawah.
Kapasitor penghalus mengurangi ripar, memberikan tegangan d.c. yang lebih stabil yang cocok untuk elektronik sensitif.
| English | Bahasa Indonesia |
|---|---|
| smoothing/ˈsmuːðɪŋ/ | penyaring |
| capacitor/kəˈpæsɪtə/ | kapasitor |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| alternating current | a current that reverses its direction periodically (varies sinusoidally with time about zero) |
| period | the time for one complete cycle |
| frequency | the number of complete cycles per unit time |
| peak value | the maximum value of the current or voltage in a cycle |
| r.m.s. value | the value of the direct (steady) current that would dissipate the same power in the same resistor |
| rectification | the conversion of an alternating current into a direct (one-direction) current |
| half-wave rectification | only one half of each cycle is passed to the output; the other half is blocked |
| full-wave rectification | both halves of each cycle appear at the output in the same direction |
| smoothing | using a capacitor across the load to reduce the variation (ripple) of a rectified output |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| arus bolak-balik | arus yang membalikkan arahnya secara berkala (bervariasi secara sinusoidal terhadap waktu di sekitar nol) |
| periode | waktu untuk satu siklus lengkap |
| frekuensi | jumlah siklus lengkap per satuan waktu |
| nilai puncak | nilai maksimum dari arus atau tegangan dalam satu siklus |
| nilai r.m.s. | nilai arus langsung (stabil) yang akan menghamburkan daya yang sama pada resistor yang sama |
| penyearahan | konversi arus bolak-balik menjadi arus langsung (satu arah) |
| penyearah gelombang setengah | hanya satu separuh dari setiap siklus yang dilewatkan ke output; separuh lainnya diblokir |
| penyearah gelombang penuh | kedua separuh dari setiap siklus muncul pada output dalam arah yang sama |
| penyaringan | menggunakan kapasitor di beban untuk mengurangi variasi (riak) dari output yang telah disearahkan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Electromagnetic radiation behaves like particles as well as like a wave. The particles of EM radiation are photons 光子 — small packets ("quanta" 量子) of EM energy that travel at the speed of light.
"State what is meant by a photon" (two marks). A photon is a quantum (a discrete packet) of energy of electromagnetic radiation. Both halves score: quantum or packet (or "discrete amount"), and of electromagnetic radiation (or "of light"). "A particle of light" alone is not enough. That radiation comes in such packets is what the syllabus calls its particulate nature 粒子性: energy is delivered in lumps of $hf$, never in smaller pieces.
A photon of frequency 频率 $f$ has energy
where $h = 6.63 \times 10^{-34}\ \text{J s}$ is the Planck constant 普朗克常量. Using $c = f\lambda$:
Worked example. Find the energy of a photon of green light of wavelength $500\ \text{nm}$. ($h = 6.63 \times 10^{-34}\ \text{J s}$, $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$.)
Higher-frequency (shorter-wavelength 波长) photons carry more energy: one $\gamma$-ray photon carries far more than one radio photon.
The electronvolt 电子伏特 (eV) is a handy energy unit on the atomic scale:
It is the kinetic energy 动能 an electron 电子 gains moving through a potential difference 电势差 of 1 V. For example, a visible photon ($\lambda \approx 500\ \text{nm}$) has energy $\approx 2.5\ \text{eV}$. To go eV → J multiply by $1.60 \times 10^{-19}$; J → eV divide.
Using the electronvolt. Photon energies, work functions and energy levels are all a few eV, so the exam quotes them that way and expects you to move between units without fuss: a $2.0\ \text{eV}$ work function is $2.0 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19}\ \text{J}$; a photon of $4.0 \times 10^{-19}\ \text{J}$ is $2.5\ \text{eV}$. A useful shortcut for wavelengths: $hc = 1.99 \times 10^{-25}\ \text{J m} = 1240\ \text{eV nm}$, so a $2.0\ \text{eV}$ photon has $\lambda = 1240/2.0 = 620\ \text{nm}$ (red) and a $400\ \text{nm}$ photon carries $3.1\ \text{eV}$.
A photon also carries momentum 动量:
It has zero rest mass but a non-zero momentum $E/c$. Radiation pressure (photons pushing on a surface) follows from this.
"Show that $p = h/\lambda$." Start from the two photon relations: $E = hf$ and $p = E/c$. Then $p = hf/c$, and since $c = f\lambda$, $f/c = 1/\lambda$, so $p = h/\lambda$. Give both starting equations and the wave equation; the mark is for the chain, not the result.
Worked example. A photon in free space has momentum $9.5 \times 10^{-28}\ \text{N s}$. Show that it is a photon of red light.
$\lambda = h/p = 6.63 \times 10^{-34}/9.5 \times 10^{-28} = 7.0 \times 10^{-7}\ \text{m} = 700\ \text{nm}$, which lies at the red end of the visible spectrum ($400$–$700\ \text{nm}$). The energy is $pc = 2.9 \times 10^{-19}\ \text{J} = 1.8\ \text{eV}$.
Radiation pressure 辐射压. A beam of intensity $I$ (power per unit area) falling on area $A$ delivers $IA/(hf)$ photons per second, each with momentum $h/\lambda$. Force is the rate of change of momentum. On a mirror each photon bounces back, so its momentum changes by $2p$ and the force is $F = 2IA/c$ (pressure $2I/c$); on a black surface each photon is absorbed, the change is $p$, and the pressure is $I/c$. Two things follow, and both are examined: the pressure depends on the intensity, not on the colour, because blue light of the same intensity has fewer photons per second but each carries proportionally more momentum; and even sunlight ($I \approx 1\ \text{kW m}^{-2}$) exerts only a few micropascals.
Worked example. Red light of intensity $160\ \text{W m}^{-2}$ falls normally on a plane mirror; each photon has momentum $9.5 \times 10^{-28}\ \text{N s}$. Find the number of photons hitting $1.0\ \text{m}^{2}$ of the mirror per second, and the pressure on it.
Photon energy $E = pc = (9.5 \times 10^{-28})(3.00 \times 10^{8}) = 2.85 \times 10^{-19}\ \text{J}$. Photons per second on $1.0\ \text{m}^{2}$: $160/2.85 \times 10^{-19} = 5.6 \times 10^{20}\ \text{s}^{-1}$. Each is reflected, so the force is $F = 5.6 \times 10^{20} \times 2 \times 9.5 \times 10^{-28} = 1.1 \times 10^{-6}\ \text{N}$ on $1.0\ \text{m}^{2}$: a pressure of $1.1 \times 10^{-6}\ \text{Pa}$ (check: $2I/c = 320/3.00 \times 10^{8} = 1.1 \times 10^{-6}\ \text{Pa}$). Replace the beam with blue light of the same intensity and the pressure is unchanged.
Worked example. A laser emits $2.0\ \text{mW}$ of light of wavelength $650\ \text{nm}$. Find the number of photons it emits per second, and the force on a surface that absorbs the beam completely.
$E = hc/\lambda = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(650 \times 10^{-9}) = 3.06 \times 10^{-19}\ \text{J}$, so the rate is $P/E = 2.0 \times 10^{-3}/3.06 \times 10^{-19} = 6.5 \times 10^{15}\ \text{s}^{-1}$. The force is the momentum delivered per second: $F = P/c = 2.0 \times 10^{-3}/3.00 \times 10^{8} = 6.7 \times 10^{-12}\ \text{N}$ (or $6.5 \times 10^{15} \times h/\lambda$, the same thing).
Radiasi elektromagnetik berperilaku seperti partikel serta seperti gelombang. Partikel radiasi EM adalah foton — paket kecil ("kuanta") energi EM yang bergerak dengan kecepatan cahaya.
"Nyatakan arti foton" (dua nilai). Foton adalah kuantum (paket diskrit) energi radiasi elektromagnetik. Kedua bagian bernilai: kuantum atau paket (atau "jumlah diskrit"), dan radiasi elektromagnetik (atau "cahaya"). "Partikel cahaya" saja tidak cukup. Fakta bahwa radiasi datang dalam paket-paket inilah yang disebut kurikulum sebagai sifat partikulat: energivideo diberikan dalam gumpalan $hf$, tidak pernah dalam potongan yang lebih kecil.
Foton dengan frekuensi $f$ memiliki energi
di mana $h = 6.63 \times 10^{-34}\ \text{J s}$ adalah konstanta Planck. Menggunakan $c = f\lambda$:
Contoh terpecahkan. Temukan energi foton cahaya hijau dengan panjang gelombang $500\ \text{nm}$. ($h = 6.63 \times 10^{-34}\ \text{J s}$, $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$.)
Foton berfrekuensi lebih tinggi (panjang-gelombang lebih pendek) membawa energi lebih banyak: satu foton sinar-$\gamma$ membawa jauh lebih banyak daripada satu foton radio.
Elektronvolt (eV) adalah satuan energi yang berguna pada skala atom:
Ini adalah energi kinetik yang diperoleh elektron bergerak melalui beda potensial sebesar 1 V. Sebagai contoh, foton terlihat ($\lambda \approx 500\ \text{nm}$) memiliki energi $\approx 2.5\ \text{eV}$. Untuk mengubah eV → J kalikan dengan $1.60 \times 10^{-19}$; J → eV bagi.
Menggunakan elektronvolt. Energi foton, fungsi kerja, dan tingkat energi semuanya beberapa eV, sehingga ujian mengutipnya dengan cara tersebut dan mengharapkan Anda berpindah antar satuan tanpa ribet: fungsi kerja $2.0\ \text{eV}$ adalah $2.0 \times 1.60 \times 10^{-19} = 3.2 \times 10^{-19}\ \text{J}$; foton $4.0 \times 10^{-19}\ \text{J}$ adalah $2.5\ \text{eV}$. Singkatan berguna untuk panjang gelombang: $hc = 1.99 \times 10^{-25}\ \text{J m} = 1240\ \text{eV nm}$, sehingga foton $2.0\ \text{eV}$ memiliki $\lambda = 1240/2.0 = 620\ \text{nm}$ (merah) dan foton $400\ \text{nm}$ membawa $3.1\ \text{eV}$.
Foton juga membawa momentum:
Memiliki massa diam nol namun momentum tak-nol $E/c$. Tekanan radiasi (foton mendorong permukaan) berasal dari hal ini.
"Buktikan bahwa $p = h/\lambda$." Mulai dari dua hubungan foton: $E = hf$ dan $p = E/c$. Kemudian $p = hf/c$, dan karena $c = f\lambda$, $f/c = 1/\lambda$, sehingga $p = h/\lambda$. Berikan kedua persamaan awal dan persamaan gelombang; poin diberikan untuk rantai penalaran, bukan hasilnya.
Contoh terpecahkan. Foton di ruang bebas memiliki momentum $9.5 \times 10^{-28}\ \text{N s}$. Tunjukkan bahwa itu adalah foton cahaya merah.
$\lambda = h/p = 6.63 \times 10^{-34}/9.5 \times 10^{-28} = 7.0 \times 10^{-7}\ \text{m} = 700\ \text{nm}$, yang terletak di ujung merah spektrum tampak ($400$–$700\ \text{nm}$). Energinya adalah $pc = 2.9 \times 10^{-19}\ \text{J} = 1.8\ \text{eV}$.
Tekanan radiasi. Sinar dengan intensitas $I$ (daya per satuan luas) jatuh pada area $A$ mengirimkan $IA/(hf)$ foton per detik, masing-masing dengan momentum $h/\lambda$. Gaya adalah laju perubahan momentum. Pada cermin, setiap foton memantul balik, sehingga perubahan momennya adalah $2p$ dan gayanya adalah $F = 2IA/c$ (tekanan $2I/c$); pada permukaan hitam, setiap foton diserap, perubahannya $p$, dan tekanannya $I/c$. Dua hal mengikuti, dan keduanya diujikan: tekanan bergantung pada intensitas, bukan pada warna, karena cahaya biru dengan intensitas yang sama memiliki lebih sedikit foton per detik tetapi masing-masing membawa momentum proporsional lebih besar; dan bahkan sinar matahari ($I \approx 1\ \text{kW m}^{-2}$) hanya memberikan beberapa mikropascal.
Contoh terpecahkan. Cahaya merah dengan intensitas $160\ \text{W m}^{-2}$ jatuh tegak lurus pada cermin datar; setiap foton memiliki momentum $9.5 \times 10^{-28}\ \text{N s}$. Tentukan jumlah foton yang mengenai $1.0\ \text{m}^{2}$ cermin per detik, dan tekanannya.
Energi foton $E = pc = (9.5 \times 10^{-28})(3.00 \times 10^{8}) = 2.85 \times 10^{-19}\ \text{J}$. Foton per detik pada $1.0\ \text{m}^{2}$: $160/2.85 \times 10^{-19} = 5.6 \times 10^{20}\ \text{s}^{-1}$. Masing-masing dipantulkan, sehingga gayanya $F = 5.6 \times 10^{20} \times 2 \times 9.5 \times 10^{-28} = 1.1 \times 10^{-6}\ \text{N}$ pada $1.0\ \text{m}^{2}$: tekanan $1.1 \times 10^{-6}\ \text{Pa}$ (cek: $2I/c = 320/3.00 \times 10^{8} = 1.1 \times 10^{-6}\ \text{Pa}$). Ganti sinar dengan cahaya biru dengan intensitas yang sama dan tekanannya tidak berubah.
Contoh terpecahkan. Laser memancarkan $2.0\ \text{mW}$ cahaya dengan panjang gelombang $650\ \text{nm}$. Tentukan jumlah foton yang dipancarkan per detik, dan gaya pada permukaan yang menyerap sinar sepenuhnya.
$E = hc/\lambda = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(650 \times 10^{-9}) = 3.06 \times 10^{-19}\ \text{J}$, sehingga lajunya $P/E = 2.0 \times 10^{-3}/3.06 \times 10^{-19} = 6.5 \times 10^{15}\ \text{s}^{-1}$. Gayanya adalah momentum yang disalurkan per detik: $F = P/c = 2.0 \times 10^{-3}/3.00 \times 10^{8} = 6.7 \times 10^{-12}\ \text{N}$ (atau $6.5 \times 10^{15} \times h/\lambda$, hal yang sama).
E = h·f
Photon energy is proportional to frequency — the gradient is Planck's constant h. · Energi foton berbanding lurus dengan frekuensi — gradiennya adalah konstanta Planck h.
| English | Bahasa Indonesia |
|---|---|
| photon/ˈfəʊtɒn/ | foton |
| quanta/ˈkwɒntə/ | kuanta |
| discrete/dɪˈskriːt/ | diskrit |
| particulate nature/pəˈtɪkjʊleɪt ˈneɪtʃə/ | sifat partikulat |
| momentum/məʊˈmentəm/ | momentum |
| Radiation pressure/ˌreɪdɪˈeɪʃn ˈpreʃə/ | tekanan radiasi |
| photoelectric effect/ˌfəʊtəʊɪˈlektrɪk ɪˈfekt/ | efek fotolistrik |
| photoelectron/ˌfəʊtəʊɪˈlektrɒn/ | fotoelektron |
| threshold frequency/ˈθreʃəʊld ˈfriːkwənsi/ | frekuensi ambang |
| threshold wavelength/ˈθreʃəʊld ˈweɪvleŋθ/ | panjang gelombang ambang |
| stopping potential/ˈstɒpɪŋ pəˈtenʃl/ | potensial penghenti |
| wave–particle duality/weɪv ˈpɑːtɪkl djuːˈælɪti/ | Dualisme gelombang-partikel |
| interference/ˌɪntəˈfɪərəns/ | intervensi |
| diffraction grating/dɪˈfrækʃn ˈɡreɪtɪŋ/ | kisi difraksi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
When EM radiation of high enough frequency hits a metal, electrons are emitted. These are photoelectrons 光电子, and the effect is the photoelectric effect 光电效应.
"State what is meant by the photoelectric effect." The emission of electrons from (the surface of) a metal when electromagnetic radiation of high enough frequency is incident on it. Two marks: emission of electrons and from a metal surface illuminated by electromagnetic radiation (or "when light is shone on it"). It is the syllabus's evidence for the particulate nature of radiation, because a wave could not explain what follows.
Each metal has a lowest photon frequency, the threshold frequency 极限频率 $f_{0}$, below which no electrons come out, however bright the light. The work function 逸出功 $\Phi$ is the least energy needed to free an electron from the surface:
Different metals have different work functions (about $2$–$5\ \text{eV}$).
The two-mark definitions. The work function energy of a metal is the minimum energy needed to remove an electron from the surface of the metal. Both "minimum" and "from the surface" carry marks: an electron deeper in the metal needs more, which is why the equation gives a maximum kinetic energy. The threshold frequency is the minimum frequency of radiation for which photoelectrons are emitted, and the threshold wavelength 极限波长 $\lambda_{0} = c/f_{0} = hc/\Phi$ is the corresponding maximum wavelength: longer wavelengths do nothing.
Worked example. Light of wavelength $400\ \text{nm}$ falls on four metals whose work functions are: caesium $2.1\ \text{eV}$, sodium $2.3\ \text{eV}$, zinc $4.3\ \text{eV}$, platinum $5.6\ \text{eV}$. Which emit photoelectrons, and with what maximum kinetic energy?
The photon energy is $hc/\lambda = 1240/400 = 3.1\ \text{eV}$. Emission needs $hf \geq \Phi$, so caesium ($3.1 - 2.1 = 1.0\ \text{eV}$) and sodium ($0.8\ \text{eV}$) emit; zinc and platinum do not, however intense the light. To make zinc emit, the wavelength must fall below $\lambda_{0} = 1240/4.3 = 290\ \text{nm}$, in the ultraviolet, which is why the electroscope demonstration needs a UV lamp on zinc.
Worked example. A polished magnesium sheet in a vacuum emits electrons only when the ultraviolet frequency is at least $8.8 \times 10^{14}\ \text{Hz}$. It is illuminated at $1.2 \times 10^{15}\ \text{Hz}$. Find the work function and the maximum speed of the photoelectrons.
$\Phi = hf_{0} = (6.63 \times 10^{-34})(8.8 \times 10^{14}) = 5.8 \times 10^{-19}\ \text{J}$ ($3.6\ \text{eV}$). Then $\tfrac{1}{2}mv_{\text{max}}^{2} = h(f - f_{0}) = (6.63 \times 10^{-34})(1.2 \times 10^{15} - 8.8 \times 10^{14}) = 2.1 \times 10^{-19}\ \text{J}$, so $v_{\text{max}} = \sqrt{2 \times 2.1 \times 10^{-19}/9.11 \times 10^{-31}} = 6.8 \times 10^{5}\ \text{m s}^{-1}$. Subtract the frequencies before multiplying by $h$; rounding $hf$ and $\Phi$ separately loses a significant figure in the small difference.
One photon gives all its energy to one electron. If the photon energy $hf$ is more than the work function, the electron escapes with kinetic energy up to a maximum:
Worked example. A metal has a work function of $2.0\ \text{eV}$. Light made of photons of energy $3.5\ \text{eV}$ shines on it. Find the maximum kinetic energy of the photoelectrons.
So the maximum KE of photoelectrons depends linearly on frequency, not on brightness.
Reading the graph. Write the equation as $E_{\text{K,max}} = hf - \Phi$: a straight line of gradient $h$, intercept $-\Phi$ on the energy axis and $f_{0} = \Phi/h$ on the frequency axis. So a graph for two metals shows two parallel lines (same gradient $h$ for every metal), the metal with the larger work function cutting the frequency axis further to the right; the intensity of the light moves neither line. This is how the Planck constant is measured, and "sketch the line for metal Y" is marked on exactly those two features.
Measuring the maximum kinetic energy. In a photocell the photoelectrons cross a vacuum to a collector and the current is a count of electrons per second. Make the collector negative and the electrons must climb a potential hill; raise the reverse p.d. until the current just reaches zero, the stopping potential 遏止电势 $V_{\text{s}}$, and then $eV_{\text{s}} = E_{\text{K,max}}$. Two results, both examined: at fixed frequency, doubling the intensity doubles the current at low reverse p.d. but leaves $V_{\text{s}}$ unchanged; raising the frequency raises $V_{\text{s}}$ but, at fixed intensity, does not raise the current.
A wave model predicts that brightness should set the electrons' kinetic energy, and that emission should happen at any frequency given enough time. But experiments show:
The photon model explains this: light arrives as photons each of energy $hf$. One photon–electron interaction either has enough energy to free the electron ($hf \geq \Phi$) or it does not.
A brighter beam of the same frequency has more photons per second, but each still carries $hf$. So the maximum KE of any electron is $hf - \Phi$ (set by $f$ only), while the rate of emission (the current) grows with the number of photons, i.e. with brightness. Doubling the brightness doubles the current but does not change the maximum KE.
Writing the explanation (a standard three-marker). (1) Each photon interacts with, and gives all its energy to, one electron. (2) The photon energy $hf$ depends only on the frequency, so the maximum energy an electron can leave with, $hf - \Phi$, is fixed by the frequency. (3) Increasing the intensity at the same frequency increases the number of photons per second, so more electrons are emitted per second (a larger current), but each still receives the same energy. A wave, by contrast, would spread its energy over the surface, so a brighter wave should have given faster electrons and a dim one should have needed a delay to accumulate energy; neither happens.
Ketika radiasi EM dengan frekuensi cukup tinggi mengenai logam, elektron dipancarkan. Ini disebut fotoelektron, dan fenomena ini adalah efek fotolistrik.
"Jelaskan makna efek fotolistrik." Pemancaran elektron dari (permukaan) logam ketika radiasi elektromagnetik dengan frekuensi cukup tinggi menyinari permukaannya. Dua nilai: pemancaran elektron dan dari permukaan logam yang diterangi radiasi elektromagnetik (atau "ketika cahaya diarahkan ke sana"). Ini merupakan bukti kurikulum tentang sifat partikel radiasi, karena gelombang tidak dapat menjelaskan apa yang terjadi selanjutnya.
Setiap logam memiliki frekuensi foton terendah, yaitu frekuensi ambang $f_{0}$, di bawahnya tidak ada elektron yang keluar, secerah apa pun cahayanya. Fungsi kerja $\Phi$ adalah energi terkecil yang diperlukan untuk membebaskan elektron dari permukaan:
Logam berbeda memiliki fungsi kerja yang berbeda (sekitar $2$–$5\ \text{eV}$).
Definisi dua nilai. Fungsi kerja energi suatu logam adalah energi minimum yang diperlukan untuk melepaskan elektron dari permukaan logam tersebut. Kata "minimum" dan "dari permukaan" masing-masing bernilai poin: elektron yang berada lebih dalam dalam logam membutuhkan energi lebih banyak, itulah sebabnya persamaan tersebut menghasilkan energi kinetik maksimum. Frekuensi ambang adalah frekuensi minimum radiasi yang memancarkan fotoelektron, dan panjang gelombang ambang $\lambda_{0} = c/f_{0} = hc/\Phi$ adalah panjang gelombang maksimum yang sesuai: panjang gelombang lebih panjang tidak akan menghasilkan apa-apa.
Contoh terpecahkan. Cahaya dengan panjang gelombang $400\ \text{nm}$ mengenai empat logam dengan fungsi kerja sebagai berikut: caesium $2.1\ \text{eV}$, natrium $2.3\ \text{eV}$, seng $4.3\ \text{eV}$, platinum $5.6\ \text{eV}$. Manakah yang memancarkan fotoelektron, dan dengan energi kinetik maksimum berapa?
Energi foton adalah $hc/\lambda = 1240/400 = 3.1\ \text{eV}$. Emisi memerlukan $hf \geq \Phi$, sehingga sesium ($3.1 - 2.1 = 1.0\ \text{eV}$) dan natrium ($0.8\ \text{eV}$) memancarkan; seng dan platinum tidak, bagaimanapun intensitas cahayanya. Agar seng memancarkan, panjang gelombang harus turun di bawah $\lambda_{0} = 1240/4.3 = 290\ \text{nm}$, di daerah ultraviolet, itulah sebabnya demonstrasi elektroskop memerlukan lampu UV pada seng.
Contoh terpecahkan. Lembar magnesium dipoles dalam vakum memancarkan elektron hanya jika frekuensi ultraviolet setidaknya $8.8 \times 10^{14}\ \text{Hz}$. Lembar tersebut diterangi pada $1.2 \times 10^{15}\ \text{Hz}$. Tentukan fungsi kerja dan kecepatan maksimum fotoelektron.
$\Phi = hf_{0} = (6.63 \times 10^{-34})(8.8 \times 10^{14}) = 5.8 \times 10^{-19}\ \text{J}$ ($3.6\ \text{eV}$). Kemudian $\tfrac{1}{2}mv_{\text{max}}^{2} = h(f - f_{0}) = (6.63 \times 10^{-34})(1.2 \times 10^{15} - 8.8 \times 10^{14}) = 2.1 \times 10^{-19}\ \text{J}$, sehingga $v_{\text{max}} = \sqrt{2 \times 2.1 \times 10^{-19}/9.11 \times 10^{-31}} = 6.8 \times 10^{5}\ \text{m s}^{-1}$. Kurangkan frekuensinya sebelum mengalikannya dengan $h$; pembulatan $hf$ dan $\Phi$ secara terpisah akan menghilangkan satu angka penting dalam selisih kecil.
Satu foton memberikan seluruh energinya kepada satu elektron. Jika energi foton $hf$ lebih besar dari fungsi kerja, elektron akan melarikan diri dengan energi kinetik hingga mencapai nilai maksimum:
Contoh terpecahkan. Sebuah logam memiliki fungsi kerja $2.0\ \text{eV}$. Cahaya yang terdiri dari foton dengan energi $3.5\ \text{eV}$ menyinari logam tersebut. Tentukan energi kinetik maksimum fotoelektron.
Sehingga KE maksimum fotoelektron bergantung secara linear pada frekuensi, bukan pada kecerahan.
Membaca grafik. Tulis persamaan sebagai $E_{\text{K,max}} = hf - \Phi$: sebuah garis lurus dengan gradien $h$, intersep $-\Phi$ pada sumbu energi dan $f_{0} = \Phi/h$ pada sumbu frekuensi. Jadi grafik untuk dua logam menunjukkan dua garis sejajar (gradien sama $h$ untuk setiap logam), logam dengan fungsi kerja lebih besar memotong sumbu frekuensi lebih jauh ke kanan; intensitas cahaya tidak menggeser salah satu garis pun. Inilah cara konstanta Planck diukur, dan "sketsa garis untuk logam Y" dinilai tepat pada dua fitur tersebut.
*Menyelidiki efek: arus mengukur laju emisi, dan tegangan balik yang tepat menghentikan elektron tercepat mengukur energi kinetik maksimum mereka
Mengukur energi kinetik maksimum. Dalam sel foto, fotoelektron melintasi vakum menuju pengumpul dan arus adalah jumlah elektron per detik. Buatlah pengumpul negatif dan elektron harus menaiki bukit potensial; naikkan tegangan balik hingga arus tepat mencapai nol, potensial penghentian $V_{\text{s}}$, dan kemudian $eV_{\text{s}} = E_{\text{K,max}}$. Dua hasil, keduanya ditinjau: pada frekuensi tetap, menggandakan intensitas menggandakan arus pada tegangan balik rendah tetapi tidak mengubah $V_{\text{s}}$; menaikkan frekuensi menaikkan $V_{\text{s}}$ tetapi, pada intensitas tetap, tidak menaikkan arus.
Model gelombang memprediksi bahwa kecerahan seharusnya menentukan energi kinetik elektron, dan emisi harus terjadi pada frekuensi apa pun jika waktunya cukup. Namun eksperimen menunjukkan:
Model foton menjelaskan hal ini: cahaya datang sebagai foton-foton, masing-masing berenergi $hf$. Satu interaksi foton–elektron memiliki cukup energi untuk melepaskan elektron ($hf \geq \Phi$) atau tidak.
Sinar yang lebih terang dari frekuensi yang sama memiliki lebih banyak foton per detik, tetapi masing-masing tetap membawa $hf$. Jadi energi kinetik maksimum setiap elektron adalah $hf - \Phi$ (ditentukan hanya oleh $f$), sementara laju emisi (arus) bertambah seiring jumlah foton, yaitu sebanding dengan kecerahan. Menggandakan kecerahan menggandakan arus tetapi tidak mengubah KE maksimum.
Menulis penjelasan (standar tiga penanda). (1) Setiap foton berinteraksi dengan, dan memberikan seluruh energinya kepada, satu elektron. (2) Energi foton $hf$ hanya bergantung pada frekuensi, sehingga energi maksimum yang dapat dibawa elektron keluar, $hf - \Phi$, ditetapkan oleh frekuensi. (3) Meningkatkan intensitas pada frekuensi yang sama meningkatkan jumlah foton per detik, sehingga lebih banyak elektron teremisikan per detik (arus lebih besar), tetapi masing-masing masih menerima energi yang sama. Sebaliknya, gelombang akan menyebarkan energinya di permukaan, sehingga gelombang yang lebih terang seharusnya menghasilkan elektron lebih cepat dan gelombang redup seharusnya membutuhkan penundaan untuk mengumpulkan energi; kedua hal itu tidak terjadi.
KEmax = h·f − φ
Max KE is a straight line in frequency, with intercept −φ (the work function). · KE maksimum adalah garis lurus terhadap frekuensi, dengan intersep −φ (fungsi kerja).
| English | Bahasa Indonesia |
|---|---|
| frequency/ˈfriːkwənsi/ | frekuensi |
| Planck constant/plæŋk ˈkɒnstənt/ | konstanta Planck |
| wavelength/ˈweɪvleŋθ/ | panjang gelombang |
| electronvolt/ɪˈlektrɒnvəʊlt/ | elektronvolt |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The photoelectric effect is strong evidence for the particle nature of light. But interference 干涉 (Young's double slit, the diffraction grating 衍射光栅) and diffraction 衍射 show its wave nature. So light has both wave and particle sides — this is wave–particle duality 波粒二象性.
"Describe what is meant by wave–particle duality" (two marks). Electromagnetic radiation (and matter) can exhibit both wave properties, such as interference and diffraction, and particle properties, such as the photoelectric effect (or, for matter, discrete collisions). Asked for one piece of evidence for each nature of radiation, give: particulate, the photoelectric effect (the threshold frequency and the immediate emission); wave, diffraction or interference (Young's slits, a diffraction grating). For matter, the wave evidence is electron diffraction.
If a wave can act like particles, perhaps particles can act like waves. De Broglie proposed that any moving particle has a de Broglie wavelength 德布罗意波长:
where $p = mv$. Example: an electron at $v = 4.9 \times 10^{7}\ \text{m s}^{-1}$ has $p = 4.46 \times 10^{-23}\ \text{kg m s}^{-1}$, so $\lambda = 1.49 \times 10^{-11}\ \text{m} \approx 0.015\ \text{nm}$ — close to atomic spacings.
"State what is meant by the de Broglie wavelength." The wavelength associated with a moving particle, or the wavelength of the wave associated with a particle of momentum $p$, given by $\lambda = h/p$ where $h$ is the Planck constant. In "state the formula and the meaning of any other symbol", name $h$ as the Planck constant and $p$ as the momentum of the particle. The wavelength is small because $h$ is small: a $0.10\ \text{kg}$ ball at $10\ \text{m s}^{-1}$ has $\lambda = 6.6 \times 10^{-34}\ \text{m}$, far below any slit or lattice spacing, which is why everyday objects show no diffraction.
When electrons are fired at a crystal lattice 晶格 (e.g. thin graphite), they make a diffraction pattern of bright rings on a screen — exactly what waves of wavelength $\lambda = h/p$ would do. This is direct evidence for the wave nature of particles (electron diffraction 电子衍射): only waves diffract, yet electrons do.
A faster electron has more momentum, so a shorter de Broglie wavelength, which diffracts less — the rings move closer together. Slowing the electrons spreads the rings apart. To calculate: $p = \sqrt{2 m E_{\text{K}}}$, and for an electron accelerated through p.d. $V$, $E_{\text{K}} = eV$, so $\lambda = h/\sqrt{2m_{e} e V}$.
Worked example. An electron is accelerated from rest through a p.d. of $2500\ \text{V}$. Find its de Broglie wavelength. ($m_{e} = 9.11 \times 10^{-31}\ \text{kg}$, $e = 1.6 \times 10^{-19}\ \text{C}$, $h = 6.63 \times 10^{-34}\ \text{J s}$.)
Its kinetic energy is $E_{\text{K}} = eV$, so $\lambda = \dfrac{h}{\sqrt{2 m_{e} e V}}$:
This is close to the spacing between atoms in a crystal, which is why the electrons diffract off the graphite.
Deriving $\lambda$ for an accelerated electron. An electron of mass $m$ and charge $q$ accelerated from rest through a p.d. $V$ gains kinetic energy $qV = \tfrac{1}{2}mv^{2}$, so $v = \sqrt{2qV/m}$ and $p = mv = \sqrt{2mqV}$; hence $\lambda = h/\sqrt{2mqV}$. Two consequences the exam asks for: increasing $V$ increases the momentum and shortens the wavelength, so the diffraction rings shrink towards the centre; halving the wavelength needs four times the p.d.
Describing electron diffraction (four marks). (1) Electrons from a heated filament are accelerated through a high p.d. into a beam. (2) The beam meets a thin polycrystalline 多晶的 graphite film; the regular spacing of the carbon atoms, about $10^{-10}\ \text{m}$, acts as a diffraction grating. (3) On a fluorescent screen the electrons produce a bright central spot surrounded by concentric rings. (4) Rings are a diffraction pattern, and diffraction is a wave property, so the electrons are behaving as waves; the ring radii match a wavelength $h/p$, which confirms de Broglie's relation. Sketch the pattern as rings, not spots or a fringe pattern. A faster beam gives rings of smaller radius.
Efek fotolistrik adalah bukti kuat sifat partikel cahaya. Tetapi interferensi (celah ganda Young, kisi difraksi) dan difraksi menunjukkan sifat gelombang-nya. Jadi cahaya memiliki kedua sisi gelombang dan partikel — ini disebut dualisme gelombang–partikel.
"Jelaskan yang dimaksud dengan dualisme gelombang–partikel" (dua nilai). Radiasi elektromagnetik (dan materi) dapat menunjukkan sifat-sifat gelombang, seperti interferensi dan difraksi, serta sifat partikel, seperti efek fotolistrik (atau, untuk materi, tumbukan diskrit). Diminta satu bukti untuk setiap sifat radiasi, berikan: partikel, efek fotolistrik (frekuensi ambang dan emisi segera); gelombang, difraksi atau interferensi (celah Young, kisi difraksi). Untuk materi, bukti gelombangnya adalah difraksi elektron.
Jika gelombang bisa bertindak seperti partikel, mungkin partikel bisa bertindak seperti gelombang. De Broglie mengusulkan bahwa setiap partikel bergerak memiliki panjang gelombang de Broglie:
di mana $p = mv$. Contoh: elektron pada $v = 4.9 \times 10^{7}\ \text{m s}^{-1}$ memiliki $p = 4.46 \times 10^{-23}\ \text{kg m s}^{-1}$, sehingga $\lambda = 1.49 \times 10^{-11}\ \text{m} \approx 0.015\ \text{nm}$ — mendekati jarak antaratom.
"Nyatakan yang dimaksud dengan panjang gelombang de Broglie." Panjang gelombang yang terkait dengan partikel bergerak, atau panjang gelombang dari gelombang yang terkait dengan partikel bermomentum $p$, diberikan oleh $\lambda = h/p$ di mana $h$ adalah konstanta Planck. Dalam "nyatakan rumus dan arti simbol lainnya", sebut $h$ sebagai konstanta Planck dan $p$ sebagai momentum partikel. Panjang gelombangnya kecil karena $h$ kecil: bola $0.10\ \text{kg}$ pada $10\ \text{m s}^{-1}$ memiliki $\lambda = 6.6 \times 10^{-34}\ \text{m}$, jauh di bawah celah atau jarak kisi manapun, itulah sebabnya benda sehari-hari tidak menunjukkan difraksi.
Ketika elektron ditembakkan ke sebuah kisi kristal (misalnya grafit tipis), mereka menghasilkan pola difraksi berupa cincin-cincin terang pada layar — persis seperti yang akan dilakukan gelombang dengan panjang gelombang $\lambda = h/p$. Ini adalah bukti langsung dari sifat gelombang partikel (difraksi elektron): hanya gelombang yang mengalami difraksi, namun elektron juga melakukannya.
Elektron yang lebih cepat memiliki momentum lebih besar, sehingga panjang gelombang de Broglie yang lebih pendek, yang menyebabkan difraksi kurang — cincin-cincin tersebut bergerak lebih dekat satu sama lain. Memperlambat elektron menyebarkan cincin-cincin tersebut menjauh. Untuk menghitung: $p = \sqrt{2 m E_{\text{K}}}$, dan untuk elektron yang dipercepat melalui beda potensial p.d. $V$, $E_{\text{K}} = eV$, sehingga $\lambda = h/\sqrt{2m_{e} e V}$.
Contoh terpecahkan. Sebuah elektron dipercepat dari keadaan diam melalui beda potensial sebesar $2500\ \text{V}$. Tentukan panjang gelombang de Broglinya. ($m_{e} = 9.11 \times 10^{-31}\ \text{kg}$, $e = 1.6 \times 10^{-19}\ \text{C}$, $h = 6.63 \times 10^{-34}\ \text{J s}$.)
Energi kinetiknya adalah $E_{\text{K}} = eV$, sehingga $\lambda = \dfrac{h}{\sqrt{2 m_{e} e V}}$:
Nilai ini mendekati jarak antar atom dalam sebuah kristal, itulah sebabnya elektron mengalami difraksi oleh grafit.
Menurunkan $\lambda$ untuk elektron yang dipercepat. Elektron bermassa $m$ dan muatan $q$ yang dipercepat dari keadaan diam melalui beda potensial $V$ memperoleh energi kinetik $qV = \tfrac{1}{2}mv^{2}$, sehingga $v = \sqrt{2qV/m}$ dan $p = mv = \sqrt{2mqV}$; oleh karena itu $\lambda = h/\sqrt{2mqV}$. Dua konsekuensi yang diminta dalam ujian: meningkatkan $V$ meningkatkan momentum dan memperpendek panjang gelombang, sehingga cincin difraksi menyusut menuju pusat; memperparah panjang gelombang membutuhkan empat kali beda potensial.
Mendeskripsikan difraksi elektron (empat nilai). (1) Elektron dari filamen yang dipanaskan dipercepat melalui beda potensial tinggi menjadi sebuah sinar. (2) Sinar tersebut mengenai film grafit polikristalin yang tipis; jarak teratur atom karbon, sekitar $10^{-10}\ \text{m}$, bertindak sebagai kisi difraksi. (3) Pada layar fosfor, elektron menghasilkan titik terang di tengah yang dikelilingi oleh cincin konsentris. (4) Cincin merupakan pola difraksi, dan difraksi adalah sifat gelombang, sehingga elektron berperilaku sebagai gelombang; jari-jari cincin sesuai dengan panjang gelombang $h/p$, yang mengonfirmasi hubungan de Broglie. Gambarlah pola tersebut sebagai cincin, bukan titik atau pola interferensi. Sinar yang lebih cepat menghasilkan cincin dengan jari-jari yang lebih kecil.
λ = h/p
A particle's wavelength is inversely proportional to its momentum — faster, heavier particles have shorter waves. · Panjang gelombang partikel berbanding terbalik dengan momentumnya—partikel yang lebih cepat dan berat memiliki gelombang yang lebih pendek.
| English | Bahasa Indonesia |
|---|---|
| electron/ɪˈlektrɒn/ | elektron |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | beda potensial |
| work function/wɜːk ˈfʌŋkʃn/ | fungsi kerja |
| diffraction/dɪˈfrækʃn/ | difraksi |
| electron diffraction/ɪˈlektrɒn dɪˈfrækʃn/ | difraksi elektron |
| de Broglie wavelength/də ˈbrəʊli ˈweɪvleŋθ/ | panjang gelombang de Broglie |
| crystal lattice/ˈkrɪstl ˈlætɪs/ | kisi kristal |
| polycrystalline/ˌpɒlɪˈkrɪstəlaɪn/ | polikristalin |
| ground state/ɡraʊnd steɪt/ | keadaan dasar |
| excited state/ekˈsaɪtɪd steɪt/ | keadaan tereksitasi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
In an isolated atom, electrons can only sit at certain discrete 分立 energy levels 能级 — never in between. The lowest is the ground state 基态; the others are excited states 激发态.
By convention, energies are written negative, with $E = 0$ for an electron just free of the atom. For hydrogen the ground state is $E_{1} = -13.6\ \text{eV}$; higher states approach zero.
When an electron drops from a higher level $E_{2}$ to a lower level $E_{1}$, it emits one photon of energy
(Both energies are negative; their difference is positive.) Because the levels are discrete, only certain photon energies — and so certain wavelengths — come out. The emission spectrum 发射光谱 is a set of sharp bright lines on a dark background, one line per transition 跃迁. The pattern is a "fingerprint" of the element.
Explaining a line spectrum (the standard four-marker). (1) The electrons in an isolated atom can only occupy discrete energy levels. (2) An electron in an excited state falls to a lower level and the energy it loses is emitted as one photon. (3) The photon energy equals the difference between the two levels, $hf = E_{2} - E_{1}$, so only certain frequencies (wavelengths) are emitted. (4) Each possible transition gives one line; the same set of levels gives the same lines every time, which is why a spectrum identifies the element. Asked to match lines to transitions: the largest energy gap gives the line of highest frequency and shortest wavelength.
When white light passes through a cool gas, photons whose energy exactly matches an upward transition are absorbed. The light then shows dark lines on a bright background — the absorption spectrum 吸收光谱. The dark lines sit at the same wavelengths as the emission lines of the same gas.
Explaining the dark lines. Photons whose energy equals the difference between two levels are absorbed, raising an electron to the higher level; the excited electron soon falls back and re-emits a photon of the same energy, but in a random direction, so almost none of that light continues along the original path. The rest of the white light, whose photons match no gap, passes through unchanged. Hence dark lines at exactly the wavelengths the same gas would emit: the Sun's spectrum shows the absorption lines of the cooler gases in its outer layers.
For a transition between two known levels:
Work in consistent units — convert eV to joules (× $1.60 \times 10^{-19}$) before finding $\lambda$ in metres, or use $hc \approx 1240\ \text{eV nm}$ for a quick estimate.
Worked example. The lowest four energy levels of hydrogen are $-13.6$, $-3.40$, $-1.51$ and $-0.85\ \text{eV}$. Find the wavelengths of the three lines produced by transitions to the ground state, and the number of lines these four levels can produce altogether.
Transition $2 \to 1$: $\Delta E = 13.6 - 3.40 = 10.2\ \text{eV}$, so $\lambda = 1240/10.2 = 122\ \text{nm}$. $3 \to 1$: $12.09\ \text{eV}$, $\lambda = 103\ \text{nm}$. $4 \to 1$: $12.75\ \text{eV}$, $\lambda = 97.3\ \text{nm}$: all ultraviolet, the largest jump giving the shortest wavelength. Four levels allow $4 \to 3$, $4 \to 2$, $4 \to 1$, $3 \to 2$, $3 \to 1$ and $2 \to 1$: six lines. The visible red line of hydrogen is $3 \to 2$: $1.89\ \text{eV}$, $656\ \text{nm}$. (Working in joules: $10.2\ \text{eV} = 1.63 \times 10^{-18}\ \text{J}$, $\lambda = hc/E = 1.99 \times 10^{-25}/1.63 \times 10^{-18} = 1.22 \times 10^{-7}\ \text{m}$.)
Worked example. A laser emits red light of wavelength $650\ \text{nm}$ when electrons drop from one level to another. Find the energy gap between the two levels.
$\Delta E = hc/\lambda = 1240/650 = 1.91\ \text{eV} = 3.1 \times 10^{-19}\ \text{J}$. The gap, not either level, fixes the colour: two atoms with different levels but the same gap emit the same line.
Worked example (annihilation). An electron and a positron, each moving slowly, meet and annihilate 湮灭, producing two identical photons. Find the wavelength of each photon. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$.)
The rest energy of each particle is $E = mc^{2} = (9.11 \times 10^{-31})(3.00 \times 10^{8})^{2} = 8.2 \times 10^{-14}\ \text{J}$ ($0.51\ \text{MeV}$), and momentum conservation shares the energy between two photons moving in opposite directions, so each carries $8.2 \times 10^{-14}\ \text{J}$: $\lambda = hc/E = 1.99 \times 10^{-25}/8.2 \times 10^{-14} = 2.4 \times 10^{-12}\ \text{m}$, a gamma ray. (Any kinetic energy the pair had is added to the photon energies.)
Dalam atom terisolasi, elektron hanya dapat berada pada tingkat energi diskrit tertentu — tidak pernah di antaranya. Tingkat terendah disebut keadaan dasar; yang lainnya adalah keadaan tereksitasi.
Berdasarkan konvensi, energi ditulis negatif, dengan $E = 0$ untuk elektron yang tepat bebas dari atom. Untuk hidrogen, keadaan dasarnya adalah $E_{1} = -13.6\ \text{eV}$; tingkat-tingkat yang lebih tinggi mendekati nol.
Ketika elektron turun dari tingkat yang lebih tinggi $E_{2}$ ke tingkat yang lebih rendah $E_{1}$, ia memancarkan satu foton dengan energi
(Kedua energi bernilai negatif; selisihnya positif.) Karena tingkat-tingkatnya diskrit, hanya energi foton tertentu — dan karena itu panjang gelombang tertentu — yang keluar. Spektrum emisi adalah serangkaian garis terang tajam pada latar belakang gelap, satu garis per transisi. Pola ini adalah "sidik jari" dari elemen tersebut.
Menjelaskan spektrum garis (penanda standar empat). (1) Elektron dalam atom terisolasi hanya dapat menempati tingkat energi diskrit. (2) Elektron dalam keadaan tereksitasi jatuh ke tingkat yang lebih rendah dan energi yang hilangnya dipancarkan sebagai satu foton. (3) Energi foton sama dengan selisih antara kedua tingkat tersebut, $hf = E_{2} - E_{1}$, sehingga hanya frekuensi tertentu (panjang gelombang tertentu) yang dipancarkan. (4) Setiap transisi yang mungkin menghasilkan satu garis; himpunan tingkat yang sama menghasilkan garis yang sama setiap saat, itulah sebabnya spektrum mengidentifikasi elemen. Ditanya mencocokkan garis dengan transisi: celah energi terbesar menghasilkan garis dengan frekuensi tertinggi dan panjang gelombang terpendek.
Ketika cahaya putih melewati gas dingin, foton-foton yang energinya sesuai persis dengan transisi naik diserap. Cahaya kemudian menampilkan garis gelap pada latar terang — spektrum absorpsi. Garis-garis gelap tersebut berada pada panjang gelombang yang sama dengan garis-garis emisi dari gas yang sama.
Menjelaskan garis-garis gelap. Foton-foton yang energinya sama dengan selisih antara dua tingkat diserap, mengangkat elektron ke tingkat yang lebih tinggi; elektron yang tereksitasi segera jatuh kembali dan memancarkan kembali foton dengan energi yang sama, tetapi dalam arah acak, sehingga hampir tidak ada cahaya itu yang melanjutkan perjalanan sepanjang jalur asli. Sisa cahaya putih, yang foton-fotonnya tidak sesuai dengan celah energi, tembus tanpa berubah. Oleh karena itu terdapat garis-garis gelap pada panjang gelombang yang sama persis dengan yang akan dipancarkan oleh gas yang sama: spektrum Matahari menunjukkan garis-garis absorpsi dari gas-gas dingin di lapisan luarnya.
Untuk transisi antara dua tahap yang diketahui:
Kerjakan dengan unit yang konsisten — ubah eV menjadi joule (× $1.60 \times 10^{-19}$) sebelum mencari $\lambda$ dalam meter, atau gunakan $hc \approx 1240\ \text{eV nm}$ untuk anggaran cepat.
Contoh terpecahkan. Empat tingkat energi terendah hidrogen adalah $-13.6$, $-3.40$, $-1.51$ dan $-0.85\ \text{eV}$. Temukan panjang gelombang dari ketiga garis yang dihasilkan oleh transisi ke keadaan dasar, dan jumlah garis yang dapat dihasilkan keempat tingkat ini secara keseluruhan.
Transisi $2 \to 1$: $\Delta E = 13.6 - 3.40 = 10.2\ \text{eV}$, sehingga $\lambda = 1240/10.2 = 122\ \text{nm}$. $3 \to 1$: $12.09\ \text{eV}$, $\lambda = 103\ \text{nm}$. $4 \to 1$: $12.75\ \text{eV}$, $\lambda = 97.3\ \text{nm}$: semua ultraviolet, lonjakan terbesar memberikan panjang gelombang terpendek. Empat tingkat memungkinkan $4 \to 3$, $4 \to 2$, $4 \to 1$, $3 \to 2$, $3 \to 1$ dan $2 \to 1$: enam garis. Garis merah terlihat pada hidrogen adalah $3 \to 2$: $1.89\ \text{eV}$, $656\ \text{nm}$. (Bekerja dalam joule: $10.2\ \text{eV} = 1.63 \times 10^{-18}\ \text{J}$, $\lambda = hc/E = 1.99 \times 10^{-25}/1.63 \times 10^{-18} = 1.22 \times 10^{-7}\ \text{m}$.)
Contoh terpecahkan. Laser memancarkan cahaya merah dengan panjang gelombang $650\ \text{nm}$ ketika elektron turun dari satu tingkat ke tingkat lain. Temukan celah energi antara kedua tingkat tersebut.
$\Delta E = hc/\lambda = 1240/650 = 1.91\ \text{eV} = 3.1 \times 10^{-19}\ \text{J}$. Celah, bukan salah satu tingkat, menentukan warna: dua atom dengan tingkat berbeda tetapi celah yang sama memancarkan garis yang sama.
Contoh terpecahkan (pembasmian). Sebuah elektron dan sebuah positron, masing-masing bergerak lambat, bertemu dan membasmi, menghasilkan dua foton identik. Temukan panjang gelombang setiap foton. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$.)
Energi diam setiap partikel adalah $E = mc^{2} = (9.11 \times 10^{-31})(3.00 \times 10^{8})^{2} = 8.2 \times 10^{-14}\ \text{J}$ ($0.51\ \text{MeV}$), dan kekekalan momentum membagi energi antara dua foton yang bergerak berlawanan arah, sehingga masing-masing membawa $8.2 \times 10^{-14}\ \text{J}$: $\lambda = hc/E = 1.99 \times 10^{-25}/8.2 \times 10^{-14} = 2.4 \times 10^{-12}\ \text{m}$, sebuah sinar gamma. (Energi kinetik apa pun yang dimiliki pasangan ditambahkan ke energi foton.)
An electron dropping between fixed energy levels emits a photon of exactly the gap's energy — a fixed wavelength and colour. Each jump is one line of the element's barcode. · Elektron yang turun antara tingkat energi tetap memancarkan foton dengan energi tepat sebesar selisihnya—panjang gelombang dan warna yang tetap. Setiap lompatan adalah satu garis dari kode batang unsur tersebut.
| English | Bahasa Indonesia |
|---|---|
| energy level/ˈenədʒi ˈlevl/ | tingkat energi |
| transition/trænˈsɪʃn/ | transisi |
| emission spectrum/ɪˈmɪʃn ˈspektrəm/ | spektrum emisi |
| absorption spectrum/əbˈsɔːpʃn ˈspektrəm/ | spektrum penyerapan |
| annihilate/əˈnaɪəleɪt/ | annihilasi |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| photon | a quantum (discrete packet) of energy of electromagnetic radiation |
| electronvolt | the energy gained by an electron accelerated through a potential difference of one volt; $1.60 \times 10^{-19}\ \text{J}$ |
| photoelectric effect | the emission of electrons from a metal surface when electromagnetic radiation of high enough frequency is incident on it |
| work function energy | the minimum energy needed to remove an electron from the surface of the metal |
| threshold frequency | the minimum frequency of radiation that causes photoelectric emission from a metal |
| threshold wavelength | the maximum wavelength of radiation that causes photoelectric emission, $\lambda_{0} = hc/\Phi$ |
| wave–particle duality | radiation and matter show both wave properties (diffraction, interference) and particle properties (photoelectric effect, discrete collisions) |
| de Broglie wavelength | the wavelength associated with a moving particle, $\lambda = h/p$ |
| energy level | one of the discrete energies an electron in an isolated atom may have |
| emission line spectrum | a set of bright lines of definite wavelengths, each from a transition between two energy levels |
| absorption line spectrum | dark lines on a continuous spectrum at the wavelengths absorbed by transitions to higher levels |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| foton | kuantum (paket diskrit) energi radiasi elektromagnetik |
| elektronvolt | energi yang diperoleh elektron yang dipercepat melalui beda potensial satu volt; $1.60 \times 10^{-19}\ \text{J}$ |
| efek fotoelektrik | pembebasan elektron dari permukaan logam ketika radiasi elektromagnetik dengan frekuensi cukup tinggi menyinari permukaannya |
| energi fungsi kerja | energi minimum yang diperlukan untuk melepaskan elektron dari permukaan logam |
| frekuensi ambang | frekuensi minimum radiasi yang menyebabkan pembebasan fotoelektrik dari suatu logam |
| panjang gelombang ambang | panjang gelombang maksimum radiasi yang menyebabkan pembebasan fotoelektrik, $\lambda_{0} = hc/\Phi$ |
| dualitas gelombang-partikel | radiasi dan materi menunjukkan sifat gelombang (difraksi, interferensi) dan sifat partikel (efek fotoelektrik, tumbukan diskrit) |
| panjang gelombang de Broglie | panjang gelombang yang terkait dengan partikel yang bergerak, $\lambda = h/p$ |
| tingkat energi | salah satu energi diskrit yang mungkin dimiliki elektron dalam atom terisolasi |
| spektrum garis emisi | serangkaian garis terang dengan panjang gelombang tertentu, masing-masing berasal dari transisi antara dua tingkat energi |
| spektrum garis absorpsi | garis-garis gelap pada spektrum kontinu pada panjang gelombang yang diserap oleh transisi ke tingkat yang lebih tinggi |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Einstein's special relativity gives the famous link (mass-energy equivalence 质能等价):
where $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$. A mass $m$ matches an energy 能量 $E$ — the two can change into each other. For a mass change $\Delta m$:
Worked example. The star Sirius loses mass through nuclear fusion at $1.09 \times 10^{11}\ \text{kg s}^{-1}$. Find the power it radiates.
Every kilogram that disappears leaves as energy: $P = c^{2} \times (\text{mass lost per second}) = (3.00 \times 10^{8})^{2}(1.09 \times 10^{11}) = 9.8 \times 10^{27}\ \text{W}$. This is the star's luminosity 光度 (Topic 25); the Sun's is $3.8 \times 10^{26}\ \text{W}$, so it loses about $4$ million tonnes a second. The same idea in reverse: a power station producing $1\ \text{GW}$ for a year converts $E/c^{2} = 3.2 \times 10^{16}/9.0 \times 10^{16} = 0.35\ \text{kg}$ of mass, which is why the fuel weighs almost the same afterwards.
In nuclear physics the masses are tiny but $c^{2}$ is huge, so a small mass change means a large energy. A mass change of $1\ \text{u}$ ($1.661 \times 10^{-27}\ \text{kg}$) matches $\Delta E \approx 1.49 \times 10^{-10}\ \text{J}$. This gives a conversion you will use again and again:
Relativitas khusus Einstein memberikan hubungan terkenal (kesetaraan massa-energi):
di mana $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$. Satu massa $m$ setara dengan energi $E$ — keduanya dapat berubah menjadi yang lain. Untuk perubahan massa $\Delta m$:
Contoh terpecahkan. Bintang Sirius kehilangan massa melalui fusi nuklir pada $1.09 \times 10^{11}\ \text{kg s}^{-1}$. Temukan daya yang dipancarkannya.
Setiap kilogram yang hilang meninggalkan sebagai energi: $P = c^{2} \times (\text{mass lost per second}) = (3.00 \times 10^{8})^{2}(1.09 \times 10^{11}) = 9.8 \times 10^{27}\ \text{W}$. Ini adalah luminositas bintang (Topik 25); luminositas Matahari adalah $3.8 \times 10^{26}\ \text{W}$, sehingga ia kehilangan sekitar $4$ juta ton per detik. Ide yang sama secara terbalik: pembangkit listrik yang menghasilkan $1\ \text{GW}$ selama satu tahun mengonversi $E/c^{2} = 3.2 \times 10^{16}/9.0 \times 10^{16} = 0.35\ \text{kg}$ massa, itulah sebabnya bahan bakar beratnya hampir sama setelah itu.
Dalam fisika nuklir, massanya sangat kecil tetapi $c^{2}$ sangat besar, sehingga perubahan massa kecil berarti energi besar. Perubahan massa $1\ \text{u}$ ($1.661 \times 10^{-27}\ \text{kg}$) setara dengan $\Delta E \approx 1.49 \times 10^{-10}\ \text{J}$. Ini memberikan konversi yang akan Anda gunakan lagi dan lagi:
A nuclear reaction 核反应 is written like
with nucleon number 核子数 conserved (top numbers: $14 + 4 = 17 + 1$) and charge conserved (bottom numbers: $7 + 2 = 8 + 1$) — this is conservation of charge 电荷守恒. Use these to fill in an unknown: identify the species, then balance the top and bottom numbers.
The decay equations. Alpha decay removes $^{4}_{2}\text{He}$, so $A$ falls by 4 and $Z$ by 2: $^{211}_{84}\text{Po} \to {}^{207}_{82}\text{Pb} + {}^{4}_{2}\text{He}$. Beta-minus decay turns a neutron into a proton, emitting an electron and an antineutrino: $^{15}_{6}\text{C} \to {}^{15}_{7}\text{N} + {}^{0}_{-1}\text{e} + \bar{\nu}$ ($A$ unchanged, $Z$ up by 1). Beta-plus decay turns a proton into a neutron, emitting a positron 正电子 and a neutrino 中微子: $^{18}_{9}\text{F} \to {}^{18}_{8}\text{O} + {}^{0}_{+1}\text{e} + \nu$ ($Z$ down by 1). Gamma emission changes neither number. In a chain of decays just keep the books: a nucleus W that emits $\beta^{-}$, then $\alpha$, then $\beta^{-}$ ends with $A - 4$ and $Z + 1 - 2 + 1 = Z$, an isotope of W. Check both lines of every equation you complete, and remember the neutrino: the mark scheme includes it.
Reaksi nuklir ditulis seperti
dengan nomor nukleon terjaga (angka atas: $14 + 4 = 17 + 1$) dan muatan terjaga (angka bawah: $7 + 2 = 8 + 1$) — ini adalah kekekalan muatan. Gunakan ini untuk mengisi yang tidak diketahui: identifikasi spesiesnya, lalu seimbangkan angka atas dan bawah.
Persamaan peluruhan. Peluruhan alpha melepaskan $^{4}_{2}\text{He}$, sehingga $A$ turun sebesar 4 dan $Z$ turun sebesar 2: $^{211}_{84}\text{Po} \to {}^{207}_{82}\text{Pb} + {}^{4}_{2}\text{He}$. Peluruhan beta-minus mengubah neutron menjadi proton, memancarkan elektron dan antineutrino: $^{15}_{6}\text{C} \to {}^{15}_{7}\text{N} + {}^{0}_{-1}\text{e} + \bar{\nu}$ ($A$ tidak berubah, $Z$ naik 1). Peluruhan beta-plus mengubah proton menjadi neutron, memancarkan positron dan neutrino: $^{18}_{9}\text{F} \to {}^{18}_{8}\text{O} + {}^{0}_{+1}\text{e} + \nu$ ($Z$ turun 1). Emisi gamma tidak mengubah kedua angka tersebut. Dalam rantai peluruhan, cukup catat: inti W yang memancarkan $\beta^{-}$, kemudian $\alpha$, lalu $\beta^{-}$ berakhir dengan $A - 4$ dan $Z + 1 - 2 + 1 = Z$, sebuah isotop dari W. Periksa kedua baris setiap persamaan yang Anda lengkapkan, dan ingat neutrino: kunci jawaban menyertakannya.
| English | Bahasa Indonesia |
|---|---|
| neutron/ˈnjuːtrɒn/ | neutron |
| proton/ˈprəʊtɒn/ | proton |
| positron/ˈpɒzɪtrɒn/ | positron |
| neutrino/njuːˈtriːnəʊ/ | neutrino |
The mass of a nucleus 原子核 is less than the total mass of its separate protons 质子 and neutrons 中子. The difference is the mass defect 质量亏损 $\Delta m$:
By $E = mc^{2}$, this "missing" mass was released as energy when the nucleus formed. To pull the nucleus fully apart you must put that energy back — the binding energy 结合能 $B$:
Worked example. A helium-4 nucleus has a mass defect of $\Delta m = 0.0304\ \text{u}$. Find its binding energy. ($1\ \text{u}$ corresponds to $931\ \text{MeV}$.)
A more tightly bound nucleus has a larger mass defect and larger binding energy. The binding energy per nucleon 比结合能 is $B/A$ (usually in MeV per nucleon) — a measure of how tightly each nucleon is held, useful for comparing nuclides.
The two-mark definitions. The mass defect of a nucleus is the difference between the total mass of its separate nucleons and the mass of the nucleus. The binding energy is the minimum energy required to separate the nucleus into its individual nucleons (equivalently, the energy released when the nucleus is formed from separate nucleons). Say "separate nucleons" or "individual protons and neutrons"; "the energy holding the nucleus together" scores nothing. Binding energy is released, not stored: the nucleus has less energy than its parts.
Worked example. The masses are: proton $1.007\,276\ \text{u}$, neutron $1.008\,665\ \text{u}$, polonium-212 nucleus $211.945\,4\ \text{u}$. Find the mass defect and the binding energy per nucleon of $^{212}_{84}\text{Po}$.
$Z = 84$ protons and $N = 212 - 84 = 128$ neutrons: total $84 \times 1.007\,276 + 128 \times 1.008\,665 = 84.611\,2 + 129.109\,1 = 213.720\,3\ \text{u}$. Mass defect $\Delta m = 213.720\,3 - 211.945\,4 = 1.774\,9\ \text{u}$. Binding energy $= 1.774\,9 \times 931.5 = 1653\ \text{MeV}$, so per nucleon $1653/212 = 7.80\ \text{MeV}$, on the falling part of the curve. Keep every decimal place until the subtraction: the defect is a small difference of two large numbers.
A graph of $B/A$ against $A$ has a typical shape:
So the curve is dome-shaped, rising to iron then falling.
Sketching the curve. The exam gives blank axes ($A$ from 1 to 250, $B/A$ up to about $9\ \text{MeV}$) and marks: a steep rise from near zero at $A = 1$, a maximum near $A = 56$ at about $8.8\ \text{MeV}$, then a slow, gentle fall to about $7.5\ \text{MeV}$ at $A = 238$. Do not start the curve at the origin exactly (hydrogen-1 has no binding energy but is a single point), do not make the fall as steep as the rise, and do not let the curve reach zero on the right. Asked to mark a nucleus that undergoes alpha decay, put X on the far right, $A > 200$; a nucleus that undergoes fusion goes at the far left, $A < 10$; both are at low $B/A$, moving up the curve when they react.
Massa inti lebih kecil daripada total massa proton dan neutron terpisahnya. Selisihnya adalah defek massa $\Delta m$:
Berdasarkan $E = mc^{2}$, massa "hilang" ini dilepaskan sebagai energi saat inti terbentuk. Untuk memisahkan inti sepenuhnya, Anda harus memasukkan kembali energi itu — energi ikat $B$:
*Inti yang berkumpul memiliki massa lebih kecil daripada nukleon-nukleon terpisahnya; massa yang hilang dilepaskan sebagai energi ikat
Contoh terpecahkan. Inti helium-4 memiliki defek massa $\Delta m = 0.0304\ \text{u}$. Temukan energi ikatnya. ($1\ \text{u}$ setara dengan $931\ \text{MeV}$.)
Inti yang terikat lebih kuat memiliki defek massa dan energi ikat yang lebih besar. Energi ikat per nukleon adalah $B/A$ (biasanya dalam MeV per nukleon) — ukuran seberapa kuat setiap nukleon diikat, berguna untuk membandingkan nuklida.
Definisi dua nilai. Defek massa suatu inti adalah selisih antara total massa nukleon-nukleon terpisahnya dan massa intinya. Energi ikat adalah energi minimum yang diperlukan untuk memisahkan inti menjadi nukleon-nukleon individualnya (setara dengan energi yang dilepaskan ketika inti terbentuk dari nukleon-nukleon terpisah). Katakan "nukleon-nukleon terpisah" atau "proton dan neutron individual"; "energi yang mengikat inti bersama" tidak mendapat nilai. Energi ikat dilepaskan, bukan disimpan: inti memiliki energi lebih sedikit daripada bagian-bagiannya.
Contoh terpecahkan. Massanya adalah: proton $1.007\,276\ \text{u}$, neutron $1.008\,665\ \text{u}$, inti polonium-212 $211.945\,4\ \text{u}$. Temukan defek massa dan energi ikat per nukleon dari $^{212}_{84}\text{Po}$.
$Z = 84$ proton dan $N = 212 - 84 = 128$ neutron: total $84 \times 1.007\,276 + 128 \times 1.008\,665 = 84.611\,2 + 129.109\,1 = 213.720\,3\ \text{u}$. Defek massa $\Delta m = 213.720\,3 - 211.945\,4 = 1.774\,9\ \text{u}$. Energi ikat $= 1.774\,9 \times 931.5 = 1653\ \text{MeV}$, sehingga per nukleon $1653/212 = 7.80\ \text{MeV}$, pada bagian menurun kurva. Pertahankan setiap desimal hingga pengurangan: defeknya adalah selisih kecil dari dua bilangan besar.
Grafik $B/A$ terhadap $A$ memiliki bentuk khas:
Jadi kurvanya berbentuk kubah, naik menuju besi lalu turun.
Menggambar sketsa kurva. Ujian menyediakan sumbu kosong ($A$ dari 1 hingga 250, $B/A$ hingga sekitar $9\ \text{MeV}$) dan menandai: kenaikan curam dari dekat nol pada $A = 1$, maksimum di sekitar $A = 56$ pada sekitar $8.8\ \text{MeV}$, kemudian penurunan lambat dan landai hingga sekitar $7.5\ \text{MeV}$ pada $A = 238$. Jangan mulai kurva tepat di titik asal (hidrogen-1 tidak memiliki energi ikat tetapi merupakan titik tunggal), jangan buat penurunan securam kenaikan, dan jangan biarkan kurva mencapai nol di sisi kanan. Diminta untuk menandai inti yang mengalami peluruhan alpha, letakkan X di ujung kanan, $A > 200$; inti yang mengalami fusi diletakkan di ujung kiri, $A < 10$; keduanya berada pada $B/A$ rendah, bergerak naik sepanjang kurva saat bereaksi.
E = delta m c^2
Change mass defect and see binding energy rise with E = mc^2. · Ubah kehilangan massa dan lihat energi ikat meningkat dengan E = mc^2.
| English | Bahasa Indonesia |
|---|---|
| mass-energy equivalence/mæs ˈenədʒi ɪˈkwɪvələns/ | ekuivalensi massa-energi |
| energy/ˈenədʒi/ | tenaga |
| nuclear fusion/ˈnjuːklɪə ˈfjuːʒn/ | fusi nuklir |
| luminosity/ˌluːmɪˈnɒsɪti/ | luminositas |
| nuclear reaction/ˈnjuːklɪə rɪˈækʃn/ | reaksi nuklir |
| nucleon number/ˈnjuːklɪən ˈnʌmbə/ | nomor nukleon |
| conservation of charge/ˌkɒnsəˈveɪʃn ɒv tʃɑːdʒ/ | kekekalan muatan |
| nucleus/ˈnjuːklɪəs/ | inti sel |
| mass defect/mæs ˈdiːfekt/ | kehilangan massa |
| binding energy/ˈbaɪndɪŋ ˈenədʒi/ | energi ikat |
| binding energy per nucleon/ˈbaɪndɪŋ ˈenədʒi pɜː ˈnjuːklɪən/ | energi ikat per nukleon |
| nuclear fission/ˈnjuːklɪə ˈfɪʃn/ | fisi nuklir |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| electrostatic/ɪˌlektrəʊˈstætɪk/ | elektrostatik |
| strong nuclear force/strɒŋ ˈnjuːklɪə fɔːs/ | gaya nuklir kuat |
Energy is released when nuclei move towards the iron peak — by joining light nuclei or splitting heavy ones.
Nuclear fusion 核聚变 joins two light nuclei into one heavier nucleus:
The product has greater binding energy per nucleon than the reactants, so energy is released. Fusion powers stars. It needs very high temperatures (millions of kelvin) so the nuclei have enough kinetic energy 动能 to beat their electrostatic 静电 repulsion and get close enough for the strong nuclear force 强核力 to take over.
Nuclear fission 核裂变 splits a heavy nucleus into two lighter ones:
The products have higher binding energy per nucleon than $^{235}$U, so energy is released. The extra neutrons can cause more fissions — a chain reaction 链式反应 in a large enough mass of fuel (the critical mass 临界质量). This is the basis of nuclear power and weapons.
"Describe the differences between fission and fusion." Fission: a heavy nucleus (large $A$) splits into two lighter nuclei of roughly similar mass, usually after absorbing a neutron, releasing further neutrons. Fusion: two light nuclei (small $A$) join to form one heavier nucleus; it needs very high temperature and pressure to overcome the electrostatic repulsion between the nuclei. Both release energy, but per kilogram of fuel fusion releases more.
"Explain, with reference to the curve, why energy is released." In both processes the products lie higher on the binding-energy-per-nucleon curve than the reactants: each nucleon ends up more tightly bound, so the total binding energy increases. The extra binding energy is released (as the kinetic energy of the products and as photons), and the total mass of the products is less than that of the reactants by $\Delta E/c^{2}$. A nucleus near the peak (iron) can release energy by neither process, which is why the stars' fusion stops at iron.
In kg this gives joules; in atomic mass units use $\Delta E\ (\text{MeV}) = \Delta m\ (\text{u}) \times 931$.
Worked example. In a nuclear reaction the total mass decreases by $0.020\ \text{u}$. Find the energy released.
Worked example (fusion). The mass defect of deuterium $^{2}_{1}\text{H}$ is $0.002\,388\ \text{u}$ and that of helium-4 is $0.030\,377\ \text{u}$. Find the energy released when two deuterium nuclei fuse to form one helium-4 nucleus.
Energy released $=$ (binding energy of the products) $-$ (binding energy of the reactants) $= [0.030\,377 - 2 \times 0.002\,388] \times 931.5 = 0.025\,601 \times 931.5 = 23.8\ \text{MeV}$ ($3.82 \times 10^{-12}\ \text{J}$). Mass defects can be used directly like this because the number of nucleons is the same on both sides; the difference in the defects is the mass converted. Per kilogram of deuterium this is $5.7 \times 10^{14}\ \text{J}$, about a million times a chemical fuel.
Worked example (fission). $^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n}$. Masses: U-235 $235.043\,9\ \text{u}$, n $1.008\,665\ \text{u}$, Ba-141 $140.914\,4\ \text{u}$, Kr-92 $91.926\,2\ \text{u}$. Find the energy released.
Reactants: $235.043\,9 + 1.008\,665 = 236.052\,6\ \text{u}$. Products: $140.914\,4 + 91.926\,2 + 3 \times 1.008\,665 = 235.866\,6\ \text{u}$. $\Delta m = 0.186\,0\ \text{u}$, so $\Delta E = 0.186\,0 \times 931.5 = 173\ \text{MeV} = 2.8 \times 10^{-11}\ \text{J}$ per fission. Count the three neutrons on the right and the one on the left: forgetting one changes the answer by a whole nucleon mass.
Worked example (alpha decay and momentum). A stationary $^{238}_{92}\text{U}$ nucleus decays to $^{234}_{90}\text{Th}$ by emitting an $\alpha$-particle; the total kinetic energy released is $4.27\ \text{MeV}$. Find the kinetic energy of the $\alpha$-particle.
Momentum is conserved and the parent was at rest, so the $\alpha$ and the thorium nucleus have equal and opposite momenta $p$. With $E_{\text{K}} = p^{2}/2m$, the energies are in the inverse ratio of the masses: $E_{\alpha}/E_{\text{Th}} = m_{\text{Th}}/m_{\alpha} = 234/4$. So $E_{\alpha} = 4.27 \times 234/238 = 4.20\ \text{MeV}$ and the thorium recoil 反冲 takes only $0.07\ \text{MeV}$. The $\alpha$-particles from a given decay are all emitted with this one energy, which is the sign that the energy is shared between just two bodies.
Worked example (a radioactive power source). A space probe is powered by $0.874\ \text{kg}$ of plutonium-238, half-life $87.7$ years, each decay releasing $5.59\ \text{MeV}$. Find the power available at launch.
Number of nuclei: $N = 0.874/(238 \times 1.661 \times 10^{-27}) = 2.21 \times 10^{24}$. Decay constant: $\lambda = 0.693/(87.7 \times 3.156 \times 10^{7}) = 2.50 \times 10^{-10}\ \text{s}^{-1}$. Activity: $A = \lambda N = 5.53 \times 10^{14}\ \text{Bq}$. Power $= A \times E = 5.53 \times 10^{14} \times 5.59 \times 1.60 \times 10^{-13} = 490\ \text{W}$, falling to half after $87.7$ years. A nuclide with a shorter half-life would give more power per kilogram but would not last the mission; polonium-210 ($138$ days) would be far more powerful at first and useless within two years.
Energi dilepaskan ketika inti bergerak mendekati puncak besi — dengan menggabungkan inti ringan atau memecah inti berat.
Fusi nuklir menggabungkan dua inti ringan menjadi satu inti lebih berat:
Produk memiliki energi ikat per nukleon lebih besar daripada reaktan, sehingga energi dilepaskan. Fusi menghidupkan bintang-bintang. Proses ini membutuhkan suhu sangat tinggi (jutaan kelvin) agar inti memiliki energi kinetik yang cukup untuk mengatasi elektrostatik tolakan dan mendekat cukup dekat sehingga gaya nuklir kuat mengambil alih.
Fisi nuklir memecah inti berat menjadi dua yang lebih ringan:
Produk memiliki energi ikat per nukleon lebih tinggi daripada $^{235}$U, sehingga energi dilepaskan. Neutron tambahan dapat menyebabkan lebih banyak fisi — sebuah reaksi berantai dalam massa bahan bakar yang cukup besar (massa kritis). Ini adalah dasar dari tenaga nuklir dan senjata.
"Jelaskan perbedaan antara fisi dan fusi." Fisi: inti berat (massa $A$ besar) terpecah menjadi dua inti lebih ringan dengan massa kira-kira sama, biasanya setelah menyerap neutron, melepaskan neutron lebih lanjut. Fusi: dua inti ringan (massa $A$ kecil) bergabung membentuk satu inti lebih berat; proses ini memerlukan suhu dan tekanan sangat tinggi untuk mengatasi tolakan elektrostatik antar inti. Keduanya membebaskan energi, tetapi per kilogram bahan bakar fusi menghasilkan lebih banyak.
"Jelaskan, merujuk pada kurva, mengapa energi dilepaskan." Pada kedua proses, produk terletak lebih tinggi pada kurva energi ikat per nukleon daripada reaktan: setiap nukleon berakhir terikat lebih erat, sehingga total energi ikat meningkat. Energi ikat ekstra tersebut dilepaskan (sebagai energi kinetik produk dan sebagai foton), dan total massa produk lebih sedikit daripada massa reaktan sebesar $\Delta E/c^{2}$. Inti di dekat puncak (besi) tidak dapat membebaskan energi melalui salah satu proses, itulah sebabnya fusi bintang berhenti pada besi.
Dalam kg hasilnya joule; dalam satuan massa atom gunakan $\Delta E\ (\text{MeV}) = \Delta m\ (\text{u}) \times 931$.
Contoh terkerjakan. Dalam reaksi nuklir, total massa berkurang sebesar $0.020\ \text{u}$. Temukan energi yang dilepaskan.
Contoh terkerjakan (fusi). Cacat massa deuterium $^{2}_{1}\text{H}$ adalah $0.002\,388\ \text{u}$ dan cacat massa helium-4 adalah $0.030\,377\ \text{u}$. Temukan energi yang dilepaskan ketika dua inti deuterium menyatu membentuk satu inti helium-4.
Energi yang dilepaskan $=$ (energi ikat produk) $-$ (energi ikat reaktan) $= [0.030\,377 - 2 \times 0.002\,388] \times 931.5 = 0.025\,601 \times 931.5 = 23.8\ \text{MeV}$ ($3.82 \times 10^{-12}\ \text{J}$). Cacat massa dapat digunakan langsung seperti ini karena jumlah nukleon sama di kedua sisi; perbedaan cacatnya adalah massa yang terkonversi. Per kilogram deuterium ini adalah $5.7 \times 10^{14}\ \text{J}$, sekitar satu juta kali bahan bakar kimia.
Contoh terpecah (fisi). $^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n}$. Massa: U-235 $235.043\,9\ \text{u}$, n $1.008\,665\ \text{u}$, Ba-141 $140.914\,4\ \text{u}$, Kr-92 $91.926\,2\ \text{u}$. Hitung energi yang dilepaskan.
Reaktan: $235.043\,9 + 1.008\,665 = 236.052\,6\ \text{u}$. Produk: $140.914\,4 + 91.926\,2 + 3 \times 1.008\,665 = 235.866\,6\ \text{u}$. $\Delta m = 0.186\,0\ \text{u}$, sehingga $\Delta E = 0.186\,0 \times 931.5 = 173\ \text{MeV} = 2.8 \times 10^{-11}\ \text{J}$ per fisi. Hitung tiga neutron di kanan dan satu di kiri: melupakan satu akan mengubah jawaban sebesar massa satu nukleon utuh.
Contoh terkerjakan (peluruhan alfa dan momentum). Inti $^{238}_{92}\text{U}$ yang diam meluruh menjadi $^{234}_{90}\text{Th}$ dengan memancarkan partikel $\alpha$; total energi kinetik yang dilepaskan adalah $4.27\ \text{MeV}$. Temukan energi kinetik dari partikel $\alpha$.
Momentum kekal dan induknya berada dalam keadaan diam, sehingga $\alpha$ dan inti thorium memiliki momentum sama besar dan berlawanan arah $p$. Dengan $E_{\text{K}} = p^{2}/2m$, energinya berbanding terbalik dengan massa: $E_{\alpha}/E_{\text{Th}} = m_{\text{Th}}/m_{\alpha} = 234/4$. Jadi $E_{\alpha} = 4.27 \times 234/238 = 4.20\ \text{MeV}$ dan recoil thorium hanya mengambil $0.07\ \text{MeV}$. Partikel-$\alpha$ dari peluruhan tertentu dipancarkan dengan satu energi ini, yang merupakan tanda bahwa energi dibagi antara hanya dua benda.
Contoh terkerjakan (sumber daya radioaktif). Probe luar angkasa digerakkan oleh $0.874\ \text{kg}$ plutonium-238, waktu paruh $87.7$ tahun, setiap peluruhan membebaskan $5.59\ \text{MeV}$. Temukan daya yang tersedia saat peluncuran.
Jumlah inti: $N = 0.874/(238 \times 1.661 \times 10^{-27}) = 2.21 \times 10^{24}$. Konstanta peluruhan: $\lambda = 0.693/(87.7 \times 3.156 \times 10^{7}) = 2.50 \times 10^{-10}\ \text{s}^{-1}$. Aktivitas: $A = \lambda N = 5.53 \times 10^{14}\ \text{Bq}$. Daya $= A \times E = 5.53 \times 10^{14} \times 5.59 \times 1.60 \times 10^{-13} = 490\ \text{W}$, turun menjadi setengah setelah $87.7$ tahun. Nuclida dengan waktu paruh lebih pendek akan menghasilkan daya lebih besar per kilogram tetapi tidak akan bertahan selama misi; polonium-210 ($138$ hari) akan jauh lebih kuat pada awalnya dan tidak berguna dalam dua tahun.
A neutron splits a heavy nucleus, releasing energy and more neutrons — which split more nuclei. · Sebuah neutron memecah inti berat, melepaskan energi dan neutron lebih lanjut — yang kemudian memecah inti lain.
| English | Bahasa Indonesia |
|---|---|
| chain reaction/tʃeɪn rɪˈækʃn/ | reaksi berantai |
| critical mass/ˈkrɪtɪkl mæs/ | massa kritis |
| recoil/rɪˈkɔɪl/ | recoil |
| half-life/hɑːf laɪf/ | waktu paruh |
| decay constant/dɪˈkeɪ ˈkɒnstənt/ | konstanta peluruhan |
| activity/ækˈtɪvɪti/ | aktivitas |
| random/ˈrændəm/ | acak |
| spontaneous/spɒnˈteɪnɪəs/ | spontan |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
Radioactive decay is:
Evidence for randomness: the count rate fluctuates. A Geiger counter 盖革计数器 next to a source clicks at uneven intervals — never a steady stream — although the long-run mean rate is well-defined.
The two-mark definitions. Radioactive decay is the spontaneous and random emission of a particle ($\alpha$ or $\beta$) or a photon ($\gamma$) from an unstable nucleus. Spontaneous means the decay is not affected by external factors: temperature, pressure, chemical state, or the presence of other nuclei. Random means it is impossible to predict which nucleus will decay next, or when a given nucleus will decay; only a probability can be given. Evidence for randomness: the count rate fluctuates from one interval to the next, even for a source whose activity is not changing over the experiment.
For $N$ undecayed nuclei of a radionuclide 放射性核素, the rate of decay is
$\lambda$ is fixed for a nuclide; a larger sample (larger $N$) has proportionally larger activity.
The one-mark definitions. Activity is the number of decays (of nuclei) per unit time, or the rate of decay. The decay constant is the probability per unit time that a (given) nucleus will decay. Not "the rate of decay": that is the activity. Note the units are the same, $\text{s}^{-1}$, but $A$ counts events and $\lambda$ is a probability per second.
Worked example. Fluorine-18 has a half-life of $110$ minutes. Show that its decay constant is $1.05 \times 10^{-4}\ \text{s}^{-1}$, and find the activity of $2.1 \times 10^{-12}\ \text{kg}$ of fluorine-18.
$\lambda = 0.693/(110 \times 60) = 1.05 \times 10^{-4}\ \text{s}^{-1}$. The number of nuclei is the mass divided by the mass of one nucleus: $N = 2.1 \times 10^{-12}/(18 \times 1.661 \times 10^{-27}) = 7.0 \times 10^{13}$. So $A = \lambda N = 1.05 \times 10^{-4} \times 7.0 \times 10^{13} = 7.4 \times 10^{9}\ \text{Bq}$. A tiny mass gives a huge activity because the half-life is short; the same mass of uranium-238 (half-life $4.5 \times 10^{9}$ years) would give about $10^{-5}\ \text{Bq}$.
Since $\lambda$ is the fractional decay rate, $\dfrac{dN}{dt} = -\lambda N$, whose solution is an exponential decay 指数衰减:
Because $A = \lambda N$, the activity (and any count rate 计数率 proportional to it) decays the same way:
Why exponential? For each nucleus, $\lambda$ is a fixed probability per unit time, independent of the others and of the nucleus's age. So the same fraction decays in each time interval, which gives exponential decay.
Writing the explanation (three marks). (1) The decay constant is the probability per unit time of decay and is the same for every nucleus of the nuclide, whatever its age. (2) So the rate of decay, the activity, is proportional to the number of undecayed nuclei present: $A = \lambda N$. (3) A rate of change proportional to the quantity itself gives an exponential change; equivalently, the same fraction of the remaining nuclei decays in every equal time interval, so the number never reaches zero but halves in every half-life.
The half-life 半衰期 $t_{1/2}$ is the time for the number of undecayed nuclei (or the activity, or the count rate) to fall to half. From $N = N_{0} e^{-\lambda t}$ with $N = N_{0}/2$:
Worked example. A radioactive isotope has a half-life of $6.0$ hours. Find its decay constant.
Convert the half-life to seconds: $6.0\ \text{h} = 21\,600\ \text{s}$. Then
A larger decay constant means a shorter half-life. After $n$ half-lives the surviving fraction is $(1/2)^{n}$; after 5 half-lives only about 3% remains.
The definition of half-life. The time taken for the number of undecayed nuclei (or the activity) of a sample to fall to half its initial value. "Half the atoms decay" is accepted; "half the sample disappears" is not, since the decayed nuclei are still there as the daughter product.
Reading a two-isotope graph. When X decays to a stable Y, $N_{\text{Y}} = N_{0} - N_{\text{X}} = N_{0}(1 - e^{-\lambda t})$. The half-life is the time at which the curves cross ($N_{\text{X}} = N_{\text{Y}} = N_{0}/2$), or the time for $N_{\text{X}}$ to halve. If the graph gives the initial number $N_{0}$ and the initial mass $m$, the nucleon number follows from $m = N_{0} A u$: for $m = 7.3 \times 10^{-4}\ \text{kg}$ and $N_{0} = 2.0 \times 10^{21}$, $A = m/(N_{0} u) = 7.3 \times 10^{-4}/(2.0 \times 10^{21} \times 1.661 \times 10^{-27}) = 220$.
Worked example. A sample of a single radioactive isotope has an activity of $180\ \text{Bq}$ at $t = 0$ and $45\ \text{Bq}$ at $t = 8.4$ minutes. Find the half-life and the decay constant, and the activity after a further $8.4$ minutes.
$45/180 = 1/4 = (1/2)^{2}$: two half-lives in $8.4$ minutes, so $t_{1/2} = 4.2\ \text{min} = 252\ \text{s}$ and $\lambda = 0.693/252 = 2.8 \times 10^{-3}\ \text{s}^{-1}$. After another two half-lives the activity is $45/4 = 11\ \text{Bq}$. When the ratio is not a neat power of two, use $t = \ln(A_{0}/A)/\lambda$: for the activity to fall from $180$ to $50\ \text{Bq}$ takes $\ln(3.6)/2.75 \times 10^{-3} = 466\ \text{s}$. Always convert the half-life to seconds before finding $\lambda$ if the activity is in becquerels.
Given $A_{0}$ and $A$ at time $t$:
Taking logs of $A = A_{0} e^{-\lambda t}$ gives $\ln A = \ln A_{0} - \lambda t$, so a plot of $\ln A$ against $t$ is a straight line with gradient $-\lambda$. Use this with several data points.
Count rate is not activity. A detector records only the radiation that reaches it and is absorbed in it: a fraction set by the solid angle it covers, by absorption in the air and in the source itself, and by its efficiency. So the measured count rate is smaller than the activity, but proportional to it, and the half-life obtained from a count-rate graph is correct. Subtract the background radiation 本底辐射 (measured with the source removed) from every reading before taking ratios or logarithms; an unsubtracted background makes the curve flatten and the half-life appear too long.
Tracers in medicine. Fluorine-18 and oxygen-15 are $\beta^{+}$ emitters used as tracers 示踪剂 in PET scanning (Topic 24): the positron annihilates with an electron, giving two gamma photons that leave in opposite directions and reveal where the tracer is. A short half-life ($110$ minutes, $2$ minutes) is chosen so that the activity falls quickly after the scan and the dose to the patient stays small, at the price of having to make the isotope close to the hospital and use it at once.
Peluruhan radioaktif adalah:
Bukti kenacakan: tingkat hitung berfluktuasi. Sebuah pembilang Geiger di samping sumber berklik pada interval yang tidak teratur — tidak pernah sebagai aliran yang stabil — meskipun rata-rata jangka panjangnya terdefinisi dengan jelas.
Definisi bernilai dua poin. Peluruhan radioaktif adalah emisi spontan dan acak dari sebuah partikel ($\alpha$ atau $\beta$) atau foton ($\gamma$) dari inti yang tidak stabil. Spontan berarti peluruhan tidak dipengaruhi oleh faktor eksternal: suhu, tekanan, keadaan kimia, atau keberadaan inti lainnya. Acak berarti mustahil untuk memprediksi inti mana yang akan meluruh selanjutnya, atau kapan inti tertentu akan meluruh; hanya probabilitas yang dapat diberikan. Bukti kenacakan: tingkat hitung berfluktuasi dari satu interval ke interval berikutnya, bahkan untuk sumber yang aktivitasnya tidak berubah selama percobaan.
Untuk $N$ inti yang belum meluruh dari radionuklida, laju peluruhannya adalah
$\lambda$ tetap untuk suatu nuclida; sampel yang lebih besar ($N$ yang lebih besar) memiliki aktivitas yang sebanding lebih besar.
Definisi bernilai satu poin. Aktivitas adalah jumlah peluruhan (inti) per satuan waktu, atau laju peluruhan. Konstanta peluruhan adalah probabilitas per satuan waktu bahwa (suatu) inti akan meluruh. Bukan "laju peluruhan": itu adalah aktivitas. Perhatikan bahwa satuannya sama, $\text{s}^{-1}$, tetapi $A$ menghitung peristiwa dan $\lambda$ adalah probabilitas per detik.
Contoh terpecahkan. Fluorin-18 memiliki waktu paruh $110$ menit. Tunjukkan bahwa konstanta peluruhannya adalah $1.05 \times 10^{-4}\ \text{s}^{-1}$, dan temukan aktivitas dari $2.1 \times 10^{-12}\ \text{kg}$ fluorin-18.
$\lambda = 0.693/(110 \times 60) = 1.05 \times 10^{-4}\ \text{s}^{-1}$. Jumlah inti adalah massa dibagi dengan massa satu inti: $N = 2.1 \times 10^{-12}/(18 \times 1.661 \times 10^{-27}) = 7.0 \times 10^{13}$. Jadi $A = \lambda N = 1.05 \times 10^{-4} \times 7.0 \times 10^{13} = 7.4 \times 10^{9}\ \text{Bq}$. Massa yang sangat kecil menghasilkan aktivitas yang sangat besar karena waktu paruhnya pendek; massa yang sama dari uranium-238 (waktu paruh $4.5 \times 10^{9}$ tahun) akan menghasilkan sekitar $10^{-5}\ \text{Bq}$.
Karena $\lambda$ adalah tingkat peluruhan fraksional, $\dfrac{dN}{dt} = -\lambda N$, yang solusinya adalah peluruhan eksponensial:
Karena $A = \lambda N$, aktivitas (dan setiap tingkat hitung yang sebanding dengannya) meluruh dengan cara yang sama:
Mengapa eksponensial? Untuk setiap inti, $\lambda$ adalah probabilitas tetap per satuan waktu, independen dari yang lain dan dari usia inti. Jadi fraksi yang sama meluruh dalam setiap interval waktu, yang memberikan peluruhan eksponensial.
Menulis penjelasan (tiga poin). (1) Konstanta peluruhan adalah probabilitas per satuan waktu untuk peluruhan dan sama untuk setiap inti dari nuclida tersebut, terlepas dari usianya. (2) Jadi laju peluruhan, aktivitas, sebanding dengan jumlah inti yang belum meluruh yang ada: $A = \lambda N$. (3) Laju perubahan yang sebanding dengan kuantitas itu sendiri memberikan perubahan eksponensial; setara dengan, fraksi yang sama dari inti yang tersisa meluruh dalam setiap interval waktu yang sama, sehingga jumlahnya tidak pernah mencapai nol tetapi menjadi setengah dalam setiap waktu paruh.
Waktu paruh $t_{1/2}$ adalah waktu untuk jumlah inti yang belum meluruh (atau aktivitas, atau tingkat hitung) turun menjadi setengah. Dari $N = N_{0} e^{-\lambda t}$ dengan $N = N_{0}/2$:
Contoh terpecahkan. Isotop radioaktif memiliki waktu paruh $6.0$ jam. Temukan konstanta peluruhannya.
Ubah waktu paruh menjadi detik: $6.0\ \text{h} = 21\,600\ \text{s}$. Maka
Konstanta peluruhan yang lebih besar berarti waktu paruh yang lebih pendek. Setelah $n$ waktu paruh, fraksi yang tersisa adalah $(1/2)^{n}$; setelah 5 waktu paruh hanya sekitar 3% yang tersisa.
Definisi waktu paruh. Waktu yang dibutuhkan agar jumlah inti yang belum meluruh (atau aktivitas) dari suatu sampel turun menjadi setengah nilainya awal. "Setengah atom meluruh" diterima; "setengah sampel menghilang" tidak, karena inti yang telah meluruh masih ada sebagai produk anak.
Membaca grafik dua isotop. Ketika X meluruh menjadi stabil Y, $N_{\text{Y}} = N_{0} - N_{\text{X}} = N_{0}(1 - e^{-\lambda t})$. Waktu paruh adalah waktu saat kurva berpotongan ($N_{\text{X}} = N_{\text{Y}} = N_{0}/2$), atau waktu agar $N_{\text{X}}$ berkurang setengah. Jika grafik memberikan jumlah awal $N_{0}$ dan massa awal $m$, nomor nukleon mengikuti dari $m = N_{0} A u$: untuk $m = 7.3 \times 10^{-4}\ \text{kg}$ dan $N_{0} = 2.0 \times 10^{21}$, $A = m/(N_{0} u) = 7.3 \times 10^{-4}/(2.0 \times 10^{21} \times 1.661 \times 10^{-27}) = 220$.
Contoh terpecahkan. Sebuah sampel isotop radioaktif tunggal memiliki aktivitas $180\ \text{Bq}$ pada $t = 0$ dan $45\ \text{Bq}$ pada $t = 8.4$ menit. Temukan waktu paruh dan konstanta peluruhan, serta aktivitas setelah $8.4$ menit lagi.
$45/180 = 1/4 = (1/2)^{2}$: dua waktu paruh dalam $8.4$ menit, sehingga $t_{1/2} = 4.2\ \text{min} = 252\ \text{s}$ dan $\lambda = 0.693/252 = 2.8 \times 10^{-3}\ \text{s}^{-1}$. Setelah dua waktu paruh lagi, aktivitasnya menjadi $45/4 = 11\ \text{Bq}$. Ketika rasionya bukan pangkat dua yang bulat, gunakan $t = \ln(A_{0}/A)/\lambda$: untuk aktivitas turun dari $180$ ke $50\ \text{Bq}$ membutuhkan $\ln(3.6)/2.75 \times 10^{-3} = 466\ \text{s}$. Selalu konversi waktu paruh ke detik sebelum mencari $\lambda$ jika aktivitas dalam becquerel.
Diberikan $A_{0}$ dan $A$ pada waktu $t$:
Mengambil logaritma dari $A = A_{0} e^{-\lambda t}$ menghasilkan $\ln A = \ln A_{0} - \lambda t$, sehingga plot $\ln A$ terhadap $t$ adalah garis lurus dengan kemiringan $-\lambda$. Gunakan ini bersama beberapa titik data.
Tingkat hitung bukanlah aktivitas. Detektor hanya merekam radiasi yang mencapai dan diserap olehnya: sebagian ditentukan oleh sudut pejal yang diliputnya, penyerapan di udara dan sumber itu sendiri, serta efisiensinya. Jadi tingkat hitung yang terukur lebih kecil daripada aktivitas, tetapi sebanding dengannya, dan waktu paruh yang diperoleh dari grafik tingkat hitung adalah benar. Kurangi radiasi latar belakang (diukur tanpa sumber) dari setiap pembacaan sebelum mengambil rasio atau logaritma; latar belakang yang tidak dikurangkan membuat kurva mendatar dan waktu paruh tampak terlalu lama.
Penelusur dalam kedokteran. Fluorin-18 dan oksigen-15 adalah pemancar $\beta^{+}$ yang digunakan sebagai penelusur dalam pemindaian PET (Topik 24): positron annihilate dengan elektron, menghasilkan dua foton gamma yang keluar ke arah berlawanan dan mengungkap lokasi penelusur. Waktu paruh pendek ($110$ menit, $2$ menit) dipilih agar aktivitas cepat menurun setelah pemindaian dan dosis bagi pasien tetap kecil, dengan harga harus membuat isotop dekat rumah sakit dan menggunakannya segera.
Choose a decay type; the daughter nuclide is fixed so the nucleon number A and the proton number Z both balance. · Pilih jenis peluruhan; nuclid anak tetap sehingga nomor nukleon A dan nomor proton Z seimbang.
Each nucleus has a fixed chance of decaying, at random. Move time forward: about half the remaining nuclei decay every half-life — so the count halves, then halves again. · Setiap inti memiliki peluang tetap untuk meluruh secara acak. Majukan waktu: sekitar setengah inti yang tersisa meluruh setiap waktu paruh — sehingga hitungannya berkurang setengah, lalu berkurang lagi.
| English | Bahasa Indonesia |
|---|---|
| count rate/kaʊnt reɪt/ | laju hitung |
| Geiger counter/ˈɡaɪɡə ˈkaʊntə/ | pesawat Geiger |
| radionuclide/ˌreɪdiəʊˈnjuːklaɪd/ | radionuklida |
| becquerel/ˈbekwərəl/ | bekerekel |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ | peluruhan eksponensial |
| background radiation/ˈbækɡraʊnd ˌreɪdɪˈeɪʃn/ | radiasi latar belakang |
| tracers/ˈtreɪsəz/ | pelacak radioaktif |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| mass defect | the difference between the total mass of the separate nucleons and the mass of the nucleus |
| binding energy | the minimum energy needed to separate a nucleus into its individual nucleons (the energy released when it forms from them) |
| nuclear fusion | two light nuclei combine to form a single heavier nucleus |
| nuclear fission | a heavy nucleus splits into two lighter nuclei of similar mass (usually after absorbing a neutron) |
| radioactive decay | the spontaneous and random emission of α, β or γ radiation from an unstable nucleus |
| spontaneous | not affected by external factors such as temperature, pressure or chemical state |
| random | it cannot be predicted which nucleus will decay, or when a given nucleus will decay |
| activity | the number of nuclei decaying per unit time |
| decay constant | the probability per unit time that a nucleus decays |
| half-life | the time for the number of undecayed nuclei (or the activity) to fall to half its initial value |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| cacah massa | selisih antara total massa nukleon terpisah dan massa inti |
| energi ikat | energi minimum yang diperlukan untuk memisahkan inti menjadi nukleon-nukleon individualnya (energi yang dilepaskan saat terbentuk darinya) |
| fusi nuklir | dua inti ringan bergabung membentuk satu inti lebih berat |
| fisi nuklir | inti berat terbelah menjadi dua inti lebih ringan dengan massa serupa (biasanya setelah menyerap neutron) |
| peluruhan radioaktif | emisi spontan dan acak radiasi α, β, atau γ dari inti yang tidak stabil |
| spontan | tidak dipengaruhi faktor eksternal seperti suhu, tekanan, atau keadaan kimia |
| acak | tidak dapat diprediksi inti mana yang akan meluruh, atau kapan inti tertentu akan meluruh |
| aktivitas | jumlah inti yang meluruh per satuan waktu |
| konstanta peluruhan | probabilitas per satuan waktu bahwa sebuah inti meluruh |
| waktu paruh | waktu agar jumlah inti tak meluruh (atau aktivitas) turun menjadi setengah nilai awalnya |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A piezo-electric 压电 crystal changes shape a little when a p.d. is put across it, and the reverse: it makes an electromotive force 电动势 (e.m.f.) across itself when its shape is changed. Quartz and PZT are common examples. Two linked effects:
A transducer 换能器 uses this effect to both make and detect ultrasound 超声波.
Generating ultrasound (three marks). (1) An alternating p.d. is applied across the crystal. (2) The crystal changes shape, expanding and contracting at the frequency of the p.d., so it vibrates. (3) The p.d. is chosen at the crystal's resonant frequency, so the vibration is large and the crystal's faces push on the tissue, sending out a longitudinal wave of that frequency. Detecting ultrasound (two marks). The returning wave's pressure variations change the shape of the crystal; a changing shape generates an e.m.f. across it, alternating at the frequency of the wave, which is amplified and recorded. The same crystal is used for both, switched between transmitting and receiving.
So one transducer is both emitter and detector, switching between sending and listening.
To see inside the body (pulse-echo 脉冲回波 imaging):
Worked example. An ultrasound pulse returns to the transducer $60\ \mu\text{s}$ after it was sent. The speed of sound in the tissue is $1500\ \text{m s}^{-1}$. Find the depth of the reflecting boundary.
The pulse travels there and back, so $d = \dfrac{ct}{2}$:
A coupling gel 耦合剂 is put between the transducer and the skin to push out the air; without it almost all the ultrasound would reflect at the skin–air boundary and never enter the body.
"Outline the use of ultrasound to obtain diagnostic information" (four marks). (1) A pulse of ultrasound is sent into the body by the transducer, which is coupled to the skin with gel. (2) At each boundary between tissues part of the pulse is reflected; the reflected pulse returns to the transducer and is detected. (3) The time delay between emission and echo, with the speed of sound, gives the depth of the boundary ($d = ct/2$). (4) The intensity of the echo, set by the change in acoustic impedance at the boundary, shows what kind of boundary it is. Sweeping the transducer, or using an array, builds up an image.
Why pulses, and why megahertz. The transducer sends a short pulse and then listens: the echo from every boundary must arrive before the next pulse leaves, otherwise echoes from different pulses could not be told apart, and the crystal cannot transmit and receive at the same moment. The frequency is a compromise: a higher frequency means a shorter wavelength and so finer detail (resolution 分辨率), but it is attenuated more strongly, so a deep organ needs a lower frequency than a shallow one. Typical medical scanning uses $1$–$15\ \text{MHz}$, wavelengths of a fraction of a millimetre in tissue.
The specific acoustic impedance 声阻抗 of a medium is
where $\rho$ is the density 密度 and $c$ the speed of sound. Unit: $\text{kg m}^{-2}\ \text{s}^{-1}$. Bone has large $Z$; air has small $Z$; soft tissue is in between.
Worked example. Find the specific acoustic impedance of soft tissue. (Density $1060\ \text{kg m}^{-3}$, speed of sound $1540\ \text{m s}^{-1}$.)
The definition (two marks). The specific acoustic impedance of a medium is the product of its density and the speed of sound in it. Both quantities must be named; "how hard sound finds it to pass" scores nothing. Because $Z$ combines density and speed, two media with very different densities can still have similar $Z$ (water and soft tissue), and it is $Z$, not density alone, that decides how much sound reflects.
At a boundary between media of impedance $Z_{1}$ and $Z_{2}$, the intensity reflection coefficient 强度反射系数 (fraction reflected) is
Worked example (why gel). Air: density $1.29\ \text{kg m}^{-3}$, speed of sound $343\ \text{m s}^{-1}$. Gel: $Z = 1.50 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Soft tissue: $Z = 1.63 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Find the intensity reflection coefficient at an air–tissue boundary and at a gel–tissue boundary.
$Z_{\text{air}} = 1.29 \times 343 = 442\ \text{kg m}^{-2}\ \text{s}^{-1}$. Air–tissue: $\left(\dfrac{442 - 1.63 \times 10^{6}}{442 + 1.63 \times 10^{6}}\right)^{2} = 0.9989$: $99.9\%$ of the intensity reflects and only $0.1\%$ enters the body. Gel–tissue: $\left(\dfrac{1.50 - 1.63}{1.50 + 1.63}\right)^{2} = 0.0017$: only $0.2\%$ reflects. So the gel, whose impedance is close to that of skin, replaces the air layer that would otherwise reflect nearly everything, and almost all of the pulse enters the body (and the echoes can get back out again by the same route).
Worked example. Water: density $1000\ \text{kg m}^{-3}$, speed $1480\ \text{m s}^{-1}$. Glass: density $2500\ \text{kg m}^{-3}$, speed $5600\ \text{m s}^{-1}$. What fraction of the intensity is reflected at a water–glass boundary?
$Z_{\text{water}} = 1.48 \times 10^{6}$, $Z_{\text{glass}} = 1.40 \times 10^{7}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Coefficient $= \left(\dfrac{1.48 - 14.0}{1.48 + 14.0}\right)^{2} = 0.65$: two-thirds reflected, one-third transmitted. Working in units of $10^{6}$ keeps the arithmetic clean, since the ratio is what matters. A steel implant in tissue ($Z_{\text{steel}} = 4.7 \times 10^{7}$) gives $0.87$: nearly all the pulse reflects from its surface, so nothing behind it can be imaged, but the implant itself shows up very brightly.
As ultrasound goes through tissue, its intensity falls with distance:
where $\mu$ is the attenuation coefficient 衰减系数 (unit $\text{m}^{-1}$). The same form applies to X-rays.
Worked example (an echo's intensity). A pulse of intensity $I_{0}$ passes through $4.0\ \text{cm}$ of tissue with attenuation coefficient $0.50\ \text{cm}^{-1}$ to a boundary whose intensity reflection coefficient is $0.010$. Find the intensity of the echo returning to the transducer, as a fraction of $I_{0}$.
To the boundary: $e^{-\mu x} = e^{-0.50 \times 4.0} = 0.135$. Reflected: $\times 0.010$. Back through the same $4.0\ \text{cm}$: $\times 0.135$ again. Echo: $I/I_{0} = 0.135 \times 0.010 \times 0.135 = 1.8 \times 10^{-4}$, about $0.02\%$. Two things the examiner checks: the tissue is crossed twice, and the reflection coefficient multiplies the intensity at the boundary, not $I_{0}$. This is why echoes from deep boundaries are amplified more than shallow ones before display.
Kristal piezo-elektrik berubah bentuk sedikit ketika p.d. diterapkan melaluinya, dan sebaliknya: menghasilkan gaya gerak listrik (e.m.l.) di dirinya sendiri saat bentuknya diubah. Kuarsa dan PZT adalah contoh umum. Dua efek yang terkait:
Sebuah transduser menggunakan efek ini untuk baik membuat maupun mendeteksi ultrasonik.
Menghasilkan ultrasonik (tiga nilai). (1) Sebuah p.d. bolak-balik diterapkan melintasi kristal. (2) Kristal berubah bentuk, mengembang dan menyusut pada frekuensi p.d., sehingga bergetar. (3) p.d. dipilih pada frekuensi resonansi kristal, sehingga getarannya besar dan wajah kristal mendorong jaringan, memancarkan gelombang longitudinal dengan frekuensi tersebut. Mendeteksi ultrasonik (dua nilai). Variasi tekanan gelombang yang kembali mengubah bentuk kristal; perubahan bentuk menghasilkan e.m.l. melintasinya, bolak-balik pada frekuensi gelombang, yang diperkuat dan direkam. Kristal yang sama digunakan untuk keduanya, beralih antara mengirim dan menerima.
Jadi satu transduser adalah pemancar sekaligus pendetektor, beralih antara mengirim dan mendengarkan.
Untuk melihat ke dalam tubuh (pencitraan pulse-echo):
Contoh terpecahkan. Sebuah pulsa ultrasonik kembali ke transduser $60\ \mu\text{s}$ setelah dikirim. Kecepatan bunyi di jaringan adalah $1500\ \text{m s}^{-1}$. Temukan kedalaman batas pemantul.
Pulsa bergerak pergi dan kembali, jadi $d = \dfrac{ct}{2}$:
Sebuah gel penghubung diletakkan antara transduser dan kulit untuk mengusir udara; tanpanya hampir seluruh ultrasonik akan dipantulkan di batas kulit-udara dan tidak pernah masuk ke dalam tubuh.
"Jelaskan penggunaan ultrasonik untuk mendapatkan informasi diagnostik" (empat nilai). (1) Sebuah pulsa ultrasonik dikirim ke dalam tubuh oleh transduser, yang terhubung ke kulit dengan gel. (2) Di setiap batas antar jaringan sebagian pulsa dipantulkan; pulsa yang dipantulkan kembali ke transduser dan dideteksi. (3) Penundaan waktu antara emisi dan gema, dengan kecepatan bunyi, memberikan kedalaman batas ($d = ct/2$). (4) Intensitas gema, ditentukan oleh perubahan impedansi akustik di batas, menunjukkan jenis batas apa itu. Menyapu transduser, atau menggunakan array, membangun gambar.
Mengapa pulsa, dan mengapa megahertz. Transduser mengirim pulsa pendek lalu mendengarkan: gema dari setiap batas harus tiba sebelum pulsa berikutnya keluar, jika tidak gema dari pulsa berbeda tidak dapat dibedakan, dan kristal tidak dapat mengirim dan menerima pada saat yang sama. Frekuensinya adalah kompromi: frekuensi lebih tinggi berarti panjang gelombang lebih pendek sehingga detail lebih halus (resolusi), tetapi mengalami peredaman lebih kuat, sehingga organ dalam membutuhkan frekuensi lebih rendah daripada yang dangkal. Pemindaian medis umum menggunakan $1$–$15\ \text{MHz}$, panjang gelombang se pecahan milimeter di jaringan.
Impedansi akustik spesifik suatu medium adalah
di mana $\rho$ adalah densitas dan $c$ kecepatan bunyi. Satuan: $\text{kg m}^{-2}\ \text{s}^{-1}$. Tulang memiliki $Z$ besar; udara memiliki $Z$ kecil; jaringan lunak berada di antaranya.
Contoh terpecahkan. Temukan impedansi akustik spesifik jaringan lunak. (Densitas $1060\ \text{kg m}^{-3}$, kecepatan bunyi $1540\ \text{m s}^{-1}$.)
Definisi (dua nilai). Impedansi akustik spesifik suatu medium adalah hasil kali dari kerapatan dan kecepatan bunyi di dalamnya. Kedua besaran harus disebutkan; "seberapa sulit suara melewatinya" tidak bernilai. Karena $Z$ menggabungkan kerapatan dan kecepatan, dua medium dengan kerapatan yang sangat berbeda masih bisa memiliki $Z$ yang mirip (air dan jaringan lunak), dan ini $Z$, bukan kerapatan saja, yang menentukan berapa banyak suara yang dipantulkan.
Pada batas antara medium dengan impedansi $Z_{1}$ dan $Z_{2}$, koefisien pantulan intensitas (fraksi yang dipantulkan) adalah
Contoh penyelesaian (mengapa menggunakan gel). Udara: kerapatan $1.29\ \text{kg m}^{-3}$, kecepatan bunyi $343\ \text{m s}^{-1}$. Gel: $Z = 1.50 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Jaringan lunak: $Z = 1.63 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Hitunglah koefisien pantulan intensitas pada batas udara-jaringan dan pada batas gel-jaringan.
$Z_{\text{air}} = 1.29 \times 343 = 442\ \text{kg m}^{-2}\ \text{s}^{-1}$. Batas udara-jaringan: $\left(\dfrac{442 - 1.63 \times 10^{6}}{442 + 1.63 \times 10^{6}}\right)^{2} = 0.9989$: $99.9\%$ dari intensitas terpantul dan hanya $0.1\%$ masuk ke dalam tubuh. Batas gel-jaringan: $\left(\dfrac{1.50 - 1.63}{1.50 + 1.63}\right)^{2} = 0.0017$: hanya $0.2\%$ terpantul. Jadi gel, yang impedansinya dekat dengan kulit, menggantikan lapisan udara yang jika tidak akan memantulkan hampir semua, sehingga hampir seluruh pulsa masuk ke dalam tubuh (dan gema dapat kembali keluar melalui jalur yang sama).
Contoh terpecah. Air: densitas $1000\ \text{kg m}^{-3}$, kecepatan $1480\ \text{m s}^{-1}$. Kaca: densitas $2500\ \text{kg m}^{-3}$, kecepatan $5600\ \text{m s}^{-1}$. Berapa bagian intensitas yang dipantulkan pada batas air–kaca?
$Z_{\text{water}} = 1.48 \times 10^{6}$, $Z_{\text{glass}} = 1.40 \times 10^{7}\ \text{kg m}^{-2}\ \text{s}^{-1}$. Koefisien $= \left(\dfrac{1.48 - 14.0}{1.48 + 14.0}\right)^{2} = 0.65$: dua pertiga terpantul, sepertiga ditransmisikan. Menggunakan satuan $10^{6}$ membuat perhitungan lebih mudah, karena yang penting adalah rasio. Implan baja dalam jaringan ($Z_{\text{steel}} = 4.7 \times 10^{7}$) menghasilkan $0.87$: hampir seluruh pulsa terpantul dari permukaannya, sehingga tidak ada yang berada di belakangnya yang dapat dicita, tetapi implant itu sendiri muncul sangat terang.
Saat gelombang ultrasonik melewati jaringan, intensitasnya menurun seiring jarak:
di mana $\mu$ adalah koefisien peredaman (satuan $\text{m}^{-1}$). Bentuk yang sama berlaku untuk sinar-X.
*Satu gema, tiga faktor: peredaman masuk, pemantulan, peredaman keluar. Jaringan dilintasi dua kali, sehingga eksponennya berlipat ganda
Contoh penyelesaian (intensitas gema). Sebuah pulsa dengan intensitas $I_{0}$ melewati $4.0\ \text{cm}$ jaringan dengan koefisien peredaman $0.50\ \text{cm}^{-1}$ menuju batas yang koefisien pantulannya adalah $0.010$. Hitunglah intensitas gema yang kembali ke transduser, sebagai pecahan dari $I_{0}$.
Menuju batas: $e^{-\mu x} = e^{-0.50 \times 4.0} = 0.135$. Terpantul: $\times 0.010$. Kembali melewati $4.0\ \text{cm}$ yang sama: $\times 0.135$ lagi. Gema: $I/I_{0} = 0.135 \times 0.010 \times 0.135 = 1.8 \times 10^{-4}$, sekitar $0.02\%$. Dua hal yang diperiksa penguji: jaringan dilintasi dua kali, dan koefisien pemantulan mengalikan intensitas pada batas, bukan $I_{0}$. Inilah mengapa gema dari batas yang dalam diperkuat lebih banyak daripada yang dangkal sebelum ditampilkan.
Follow a pulse from transducer to echo image. · Ikuti denyut dari transduser hingga gambar gema.
| English | Bahasa Indonesia |
|---|---|
| ultrasound/ˌʊltrəˈsaʊnd/ | ultrasound |
| piezo-electric/piˌeɪtsəʊ ɪˈlektrɪk/ | piezo-elektrik |
| electromotive force/ɪˌlektrəʊˈməʊtɪv fɔːs/ | gaya gerak listrik |
| transducer/trænsˈdjuːsə/ | transduser |
| longitudinal/ˌlɒŋɡɪˈtjuːdɪnl/ | longitudinal |
| pulse-echo/pʌls ˈekəʊ/ | pulsa-eco |
| coupling gel/ˈkʌplɪŋ dʒel/ | gel perpaduan |
| resolution/ˌrezəˈluːʃn/ | resolusi |
| specific acoustic impedance/spəˈsɪfɪk əˈkuːstɪk ɪmˈpiːdəns/ | impedansi akustik spesifik |
| density/ˈdensɪti/ | densitas |
| intensity reflection coefficient/ɪnˈtensɪti rɪˈflekʃn ˌkəʊɪˈfɪʃənt/ | koefisien pantulan intensitas |
| attenuation coefficient/əˌtenjuːˈeɪʃn ˌkəʊɪˈfɪʃənt/ | koefisien redaman |
| X-ray tube/eks reɪ tjuːb/ | tabung sinar-X |
| cathode/ˈkæθəʊd/ | katoda |
| electron/ɪˈlektrɒn/ | elektron |
| thermionic emission/ˌθɜːmɪˈɒnɪk ɪˈmɪʃn/ | emisi termionik |
| vacuum/ˈvækjuːm/ | vakum |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
X-rays come from an X-ray tube X射线管:
"Explain how X-rays are produced" (four marks). (1) Electrons are emitted from a heated filament (cathode) by thermionic emission. (2) They are accelerated through a high potential difference (tens of kV) across an evacuated tube towards a metal target (the anode). (3) They decelerate rapidly when they strike the target, and (4) the kinetic energy lost is emitted as X-ray photons (with most of it becoming heat in the target, which is why the anode is cooled or rotated).
Controlling the beam. The intensity (energy per unit area per second) is set by the number of electrons hitting the target per second, so it is controlled by the filament current (a hotter filament emits more electrons); the hardness 硬度, meaning the penetrating power, is set by the photon energies, so it is controlled by the accelerating p.d.: a larger p.d. gives higher-energy, shorter-wavelength, more penetrating X-rays. A metal filter removes the softest X-rays, which would be absorbed in the patient's skin without reaching the detector.
Worked example (heating of the target). In a tube run at $75\ \text{kV}$ the electron beam current is $30\ \text{mA}$ and $99\%$ of the electrons' energy becomes heat in a tungsten target of mass $15\ \text{g}$ (specific heat capacity $134\ \text{J kg}^{-1}\ \text{K}^{-1}$). Find the rate of temperature rise if no heat were lost.
Power delivered $= IV = 0.030 \times 75\,000 = 2250\ \text{W}$; heating power $= 0.99 \times 2250 = 2230\ \text{W}$. Then $mc\,\Delta\theta/\Delta t = 2230$, so $\Delta\theta/\Delta t = 2230/(0.015 \times 134) = 1.1 \times 10^{3}\ \text{K s}^{-1}$. A target would melt within seconds, which is why exposures are short and anodes rotate to spread the heat.
The most energy one X-ray photon can have is the full kinetic energy of one accelerated electron, lost in a single event. For accelerating p.d. $V$, the KE is $eV$, so
This is the short-wavelength cut-off. The continuous Bremsstrahlung spectrum tails off above $\lambda_{\text{min}}$, with sharp characteristic peaks set by the target metal.
Worked example (minimum wavelength). Electrons are accelerated through $75\ \text{kV}$. Find the maximum photon energy and the minimum wavelength of the X-rays.
Maximum energy $= eV = 75\ \text{keV} = 75\,000 \times 1.60 \times 10^{-19} = 1.2 \times 10^{-14}\ \text{J}$ (or $0.075\ \text{MeV}$). Then $\lambda_{\text{min}} = hc/(eV) = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(1.2 \times 10^{-14}) = 1.7 \times 10^{-11}\ \text{m}$. The maximum photon momentum is $p = E/c = 4.0 \times 10^{-23}\ \text{N s}$. Doubling the p.d. halves the minimum wavelength.
Explaining the shape of the spectrum. The spectrum is continuous because each electron may lose any fraction of its energy in one or several decelerations, so photons of every energy up to the maximum are produced. It has a sharp cut-off at $\lambda_{\text{min}}$ because a photon cannot carry more energy than one electron has, $eV$, and that happens only when an electron loses all its energy in a single event. The peaks are characteristic of the target metal: an incoming electron knocks an inner electron out of a target atom, and an outer electron falls into the vacancy, emitting a photon whose energy is the difference between the two levels (Topic 22).
X-rays pass through the patient onto a detector. Tissues that attenuate 衰减 more (bone, high $Z$) cast a stronger shadow and look lighter; tissues that attenuate less (soft tissue, lung) look darker.
The contrast 对比度 is the difference in attenuation between tissues. A contrast medium 造影剂 (e.g. a barium meal) can be given to make soft tissues stand out.
Sharpness and contrast. Sharpness 清晰度 is how well defined the edges of structures are in the image: a small X-ray source (a small spot on the target), a stationary patient and a detector close to the patient all improve it. Contrast is the difference in degree of blackening (or brightness) between neighbouring regions of the image, produced by a difference in attenuation. The two are independent: an image can be sharp but have poor contrast, or the reverse.
"Explain why X-ray images of internal structures have good contrast." Bone and soft tissue have very different attenuation coefficients, so equal thicknesses transmit very different intensities and the detector is exposed very differently behind each. Where two soft tissues have similar coefficients (the stomach and its surroundings), a contrast medium with a high attenuation coefficient, such as barium, is swallowed or injected to outline one of them. Using a lower p.d. (softer X-rays) also increases contrast, at the cost of a larger dose.
Worked example. X-rays pass through $3.0\ \text{cm}$ of soft tissue ($\mu = 0.20\ \text{cm}^{-1}$) in one region of the body and $3.0\ \text{cm}$ of bone ($\mu = 0.60\ \text{cm}^{-1}$) in another. Compare the transmitted intensities.
Soft tissue: $I/I_{0} = e^{-0.20 \times 3.0} = e^{-0.60} = 0.55$. Bone: $e^{-0.60 \times 3.0} = e^{-1.80} = 0.17$. The soft-tissue region receives more than three times the exposure of the bone region, so the bone appears white and the tissue dark. The ratio of exposures, $e^{(\mu_{\text{bone}} - \mu_{\text{tissue}})x}$, grows with thickness: contrast improves for thicker structures, but so does the total attenuation.
Higher-energy X-rays penetrate further (smaller $\mu$); bone has a much larger $\mu$ than soft tissue. To find the thickness for a given fraction, take logs: $x = \dfrac{1}{\mu} \ln\dfrac{I_{0}}{I}$. The half-value thickness 半值厚度 $x_{1/2} = \ln 2 / \mu$ halves the intensity (like half-life in decay).
Worked example. X-rays pass through $3.0\ \text{cm}$ of tissue with attenuation coefficient $\mu = 40\ \text{m}^{-1}$. Find the fraction of the intensity that gets through.
Worked example (percentage absorbed). X-rays pass through $2.8\ \text{cm}$ of a medium with attenuation coefficient $1.4\ \text{cm}^{-1}$. What percentage of the X-ray energy is absorbed?
Transmitted fraction $e^{-1.4 \times 2.8} = e^{-3.92} = 0.020$, so $98\%$ is absorbed. The question asks for what is absorbed, not what gets through: read it twice. The half-value thickness here is $\ln 2/\mu = 0.50\ \text{cm}$, so $2.8\ \text{cm}$ is $5.6$ half-value thicknesses and $(1/2)^{5.6} = 0.02$ checks the answer.
Worked example (two layers). A beam of intensity $I_{0}$ passes through $2.0\ \text{cm}$ of material P ($\mu = 0.35\ \text{cm}^{-1}$) and then $1.5\ \text{cm}$ of material Q ($\mu = 0.90\ \text{cm}^{-1}$). Find the transmitted intensity.
The exponentials multiply: $I = I_{0}\, e^{-0.35 \times 2.0}\, e^{-0.90 \times 1.5} = I_{0}\, e^{-(0.70 + 1.35)} = I_{0}\, e^{-2.05} = 0.13\, I_{0}$. Add the exponents ($\mu_{1}x_{1} + \mu_{2}x_{2}$); never add the thicknesses or the coefficients. The order of the layers makes no difference to the total transmitted intensity.
A computed tomography 计算机断层扫描 (CT) scan builds a 3-D image:
CT shows far more than a single X-ray, because overlapping soft tissues are separated by the many-angle reconstruction.
"Explain how CT scanning produces a three-dimensional image" (five marks). (1) The X-ray tube (and a ring of detectors) rotates around the patient, so that (2) many X-ray images of one slice (section) are taken from different angles. (3) A computer combines them to produce a two-dimensional image of that slice. (4) The patient is moved and the process is repeated for successive slices. (5) The slices are combined to build a 3-D image, which can be rotated and viewed from any angle. Purpose: to image a section of the body without the overlapping of structures that a single X-ray suffers, revealing structures that would be hidden and their positions in depth. The price is a much larger dose than a single X-ray.
Sinar-X berasal dari tabung sinar-X X:
"Jelaskan bagaimana sinar-X diproduksi (empat nilai). (1) Elektron dipancarkan dari filamen yang dipanaskan (katode) melalui emisi termionik. (2) Mereka dipercepat melalui beda potensial tinggi (puluhan kV) melintasi tabung yang dikosongkan menuju target logam (anoda). (3) Mereka melambat secara cepat ketika menabrak target, dan (4) energi kinetik yang hilang dipancarkan sebagai foton sinar-X (dengan sebagian besar berubah menjadi panas di target, itulah sebabnya anoda didinginkan atau diputar).
Mengontrol berkas. Intensitas (energi per satuan luas per detik) ditentukan oleh jumlah elektron yang menabrak target per detik, sehingga dikendalikan oleh arus filamen (filamen yang lebih panas memancarkan lebih banyak elektron); kekerasan, yang berarti daya tembusnya, ditentukan oleh energi foton, sehingga dikendalikan oleh beda potensial percepatan: beda potensial yang lebih besar menghasilkan sinar-X dengan energi lebih tinggi, panjang gelombang lebih pendek, dan daya tembus lebih kuat. Saringan logam menghilangkan sinar-X paling lunak, yang akan terserap oleh kulit pasien tanpa mencapai detektor.
Contoh terpecahkan (pemanasan target). Dalam tabung yang dioperasikan pada $75\ \text{kV}$, arus berkas elektron adalah $30\ \text{mA}$ dan $99\%$ dari energi elektron berubah menjadi panas pada target tungsten bermassa $15\ \text{g}$ (kapasitas kalor spesifik $134\ \text{J kg}^{-1}\ \text{K}^{-1}$). Hitung laju kenaikan suhu jika tidak ada panas yang hilang.
Daya yang disalurkan $= IV = 0.030 \times 75\,000 = 2250\ \text{W}$; daya pemanas $= 0.99 \times 2250 = 2230\ \text{W}$. Maka $mc\,\Delta\theta/\Delta t = 2230$, sehingga $\Delta\theta/\Delta t = 2230/(0.015 \times 134) = 1.1 \times 10^{3}\ \text{K s}^{-1}$. Target akan meleleh dalam hitungan detik, itulah mengapa paparan dilakukan singkat dan anoda berputar untuk menyebarkan panas.
Energi maksimum yang dapat dimiliki satu foton sinar-X adalah seluruh energi kinetik satu elektron dipercepat, hilang dalam satu peristiwa. Untuk beda potensial percepatan $V$, energi kinetiknya (KE) adalah $eV$, sehingga
Ini adalah batas pemotongan panjang gelombang pendek. Spektrum Bremsstrahlung kontinu mereda di atas $\lambda_{\text{min}}$, dengan puncak karakteristik tajam yang ditentukan oleh logam target.
Contoh terpecahkan (panjang gelombang minimum). Elektron dipercepat melalui $75\ \text{kV}$. Temukan energi foton maksimum dan panjang gelombang minimum sinar-X.
Energi maksimum $= eV = 75\ \text{keV} = 75\,000 \times 1.60 \times 10^{-19} = 1.2 \times 10^{-14}\ \text{J}$ (atau $0.075\ \text{MeV}$). Maka $\lambda_{\text{min}} = hc/(eV) = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(1.2 \times 10^{-14}) = 1.7 \times 10^{-11}\ \text{m}$. Momentum foton maksimum adalah $p = E/c = 4.0 \times 10^{-23}\ \text{N s}$. Menggandakan beda potensial memperparuh panjang gelombang minimum.
Menjelaskan bentuk spektrum. Spektrum ini kontinu karena setiap elektron dapat kehilangan sebagian energinya dalam satu atau beberapa perlambatan, sehingga foton dengan segala energi hingga maksimum dihasilkan. Spektrum memiliki batas pemotongan tajam pada $\lambda_{\text{min}}$ karena foton tidak dapat membawa energi lebih dari apa yang dimiliki satu elektron, $eV$, dan hal itu hanya terjadi ketika elektron kehilangan seluruh energinya dalam satu peristiwa. Puncak-puncaknya khas untuk logam target: elektron masuk menghentakkan elektron dalam dari atom target, dan elektron luar jatuh mengisi kekosongan tersebut, memancarkan foton yang energinya adalah selisih antara kedua tingkat (Topik 22).
Sinar-X menembus pasien menuju detektor. Jaringan yang melemahkan lebih banyak (tulang, $Z$ tinggi) menghasilkan bayangan lebih kuat dan terlihat lebih terang; jaringan yang melemahkan lebih sedikit (jaringan lunak, paru-paru) terlihat lebih gelap.
Kontras adalah perbedaan pelemahan antar jaringan. Media kontras (misalnya barium meal) dapat diberikan agar jaringan lunak menonjol.
*Rontgen dada nyata: tulang padat menyerap lebih banyak sinar-X dan terlihat putih; paru-paru berisi udara meneruskan sinar-X dan terlihat gelap
Ketajaman dan kontras. Ketajaman adalah seberapa jelas tepi struktur dalam gambar: sumber sinar-X kecil (titik kecil pada target), pasien diam, dan detektor dekat dengan pasien semuanya meningkatkan ketajaman. Kontras adalah perbedaan derajat penghitaman (atau kecerahan) antara wilayah tetangga dalam gambar, yang dihasilkan oleh perbedaan pelemahan. Keduanya saling independen: gambar bisa tajam tetapi memiliki kontras buruk, atau sebaliknya.
*Kontras adalah perbedaan pelemahan: tulang melemahkan sekitar tiga kali lebih kuat daripada jaringan lunak per sentimeter, sehingga ketebalan yang sama memberikan eksposur yang sangat berbeda
"Jelaskan mengapa gambar sinar-X struktur internal memiliki kontras yang baik." Tulang dan jaringan lunak memiliki koefisien pelemahan yang sangat berbeda, sehingga ketebalan yang sama mentransmisikan intensitas yang sangat berbeda dan detektor terekspos sangat berbeda di belakang masing-masing. Di mana dua jaringan lunak memiliki koefisien serupa (lambung dan sekitarnya), media kontras dengan koefisien pelemahan tinggi, seperti barium, ditelan atau disuntikkan untuk menggarisbawahi salah satunya. Menggunakan beda potensial yang lebih rendah (sinar-X lebih lunak) juga meningkatkan kontras, dengan biaya dosis yang lebih besar.
Contoh terpecahkan. Sinar-X menembus $3.0\ \text{cm}$ jaringan lunak ($\mu = 0.20\ \text{cm}^{-1}$) di satu bagian tubuh dan $3.0\ \text{cm}$ tulang ($\mu = 0.60\ \text{cm}^{-1}$) di bagian lain. Bandingkan intensitas yang ditransmisikan.
Jaringan lunak: $I/I_{0} = e^{-0.20 \times 3.0} = e^{-0.60} = 0.55$. Tulang: $e^{-0.60 \times 3.0} = e^{-1.80} = 0.17$. Wilayah jaringan lunak menerima lebih dari tiga kali eksposur wilayah tulang, sehingga tulang terlihat putih dan jaringan gelap. Rasio eksposur, $e^{(\mu_{\text{bone}} - \mu_{\text{tissue}})x}$, meningkat seiring ketebalan: kontras membaik untuk struktur yang lebih tebal, namun begitu pula total pelemahannya.
Sinar-X berenergi lebih tinggi menembus lebih jauh ($\mu$ lebih kecil); tulang memiliki $\mu$ jauh lebih besar daripada jaringan lunak. Untuk menemukan ketebalan untuk fraksi tertentu, ambil logaritma: $x = \dfrac{1}{\mu} \ln\dfrac{I_{0}}{I}$. Ketebalan nilai setengah $x_{1/2} = \ln 2 / \mu$ mengurangi intensitas menjadi setengah (seperti waktu paruh dalam peluruhan).
Contoh terpecahkan. Sinar-X menembus $3.0\ \text{cm}$ jaringan dengan koefisien redaman $\mu = 40\ \text{m}^{-1}$. Tentukan pecahan intensitas yang berhasil tembus.
Contoh terpecahkan (persentase diserap). Sinar-X menembus $2.8\ \text{cm}$ medium dengan koefisien redaman $1.4\ \text{cm}^{-1}$. Berapa persentase energi sinar-X yang diserap?
Pecahan yang ditransmisikan $e^{-1.4 \times 2.8} = e^{-3.92} = 0.020$, sehingga $98\%$ diserap. Pertanyaan menanyakan apa yang diserap, bukan apa yang tembus: bacalah dua kali. Ketebalan nilai setengah di sini adalah $\ln 2/\mu = 0.50\ \text{cm}$, sehingga $2.8\ \text{cm}$ merupakan $5.6$ ketebalan nilai setengah dan $(1/2)^{5.6} = 0.02$ memeriksa jawaban.
Contoh terpecahkan (dua lapisan). Sebuah berkas dengan intensitas $I_{0}$ menembus $2.0\ \text{cm}$ material P ($\mu = 0.35\ \text{cm}^{-1}$) dan kemudian $1.5\ \text{cm}$ material Q ($\mu = 0.90\ \text{cm}^{-1}$). Temukan intensitas yang ditransmisikan.
Eksponensialnya dikalikan: $I = I_{0}\, e^{-0.35 \times 2.0}\, e^{-0.90 \times 1.5} = I_{0}\, e^{-(0.70 + 1.35)} = I_{0}\, e^{-2.05} = 0.13\, I_{0}$. Tambahkan eksponennya ($\mu_{1}x_{1} + \mu_{2}x_{2}$); jangan pernah menambahkan ketebalan atau koefisiennya. Urutan lapisan tidak memengaruhi total intensitas yang ditransmisikan.
Pemindaian tomografi terkomputerisasi (CT) membangun gambar 3-D:
CT menampilkan jauh lebih banyak daripada sinar-X tunggal, karena jaringan lunak yang tumpang tindih dipisahkan oleh rekonstruksi sudut banyak.
*Dalam pemindaian CT, tabung sinar-X dan detektor berputar mengelilingi pasien untuk memindai irisan dari banyak sudut
"Jelaskan bagaimana pemindaian CT menghasilkan gambar tiga dimensi" (lima poin). (1) Tabung sinar-X (dan cincin detektor) berputar mengelilingi pasien, sehingga (2) banyak gambar sinar-X dari satu irisan (potongan) diambil dari sudut-sudut berbeda. (3) Komputer menggabungkan mereka untuk menghasilkan gambar dua dimensi dari irisan tersebut. (4) Pasien dipindahkan dan proses diulang untuk irisan berturut-turut. (5) Irisan-irisan tersebut digabungkan untuk membangun gambar 3-D, yang dapat diputar dan dilihat dari sudut mana pun. Tujuan: untuk memindai bagian tubuh tanpa tumpang tindih struktur yang dialami sinar-X tunggal, mengungkap struktur yang akan tersembunyi dan posisi kedalaman mereka. Harganya adalah dosis yang jauh lebih besar daripada sinar-X tunggal.
Follow electrons from cathode to X-ray photons. · Ikuti elektron dari katode hingga foton sinar-X.
| English | Bahasa Indonesia |
|---|---|
| target/ˈtɑːɡɪt/ | target |
| anode/ˈænəʊd/ | anoda |
| tungsten/ˈtʌŋstn/ | tungsten |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | energi kinetik |
| energy/ˈenədʒi/ | tenaga |
| photon/ˈfəʊtɒn/ | foton |
| Bremsstrahlung/ˈbremʃtrɑːlʊŋ/ | Bremsstrahlung |
| characteristic/ˌkærɪktəˈrɪstɪk/ | karakteristik |
| hardness/ˈhɑːdnəs/ | kekerasan |
| momentum/məʊˈmentəm/ | momentum |
| attenuate/əˈtenjuːeɪt/ | meredam |
| contrast/ˈkɒntræst/ | kontras |
| contrast medium/ˈkɒntræst ˈmiːdɪəm/ | media kontras |
| sharpness/ˈʃɑːpnəs/ | ketajaman |
| half-value thickness/hɑːf ˈvæljuː ˈθɪknəs/ | tebal nilai setengah |
| computed tomography/kəmˈpjuːtɪd təˈmɒɡrəfi/ | tomografi komputasi |
| tracer/ˈtreɪsə/ | pelacak |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A tracer 示踪剂 is a substance with radioactive nuclei put into the body. It is taken up more by the tissue being studied (e.g. a tumour takes up more glucose-tagged tracer due to its high metabolism 代谢). Its decay is detected from outside.
In positron emission tomography 正电子发射断层扫描 (PET), the tracer is a $\beta^{+}$ emitter — it gives out a positron 正电子. A common one is fluorine-18 on a glucose analogue (FDG).
"Explain what is meant by a tracer and how it is used" (three marks). A tracer is a substance containing radioactive nuclei (for PET, a $\beta^{+}$ emitter such as fluorine-18) that is introduced into the body (injected or swallowed), usually bound to a molecule such as glucose, and is absorbed by the tissue being studied; the radiation it emits leaves the body and is detected, so its distribution shows where that tissue is active. The nuclide is chosen with a short half-life, so the activity is high during the scan and the patient's dose afterwards is small, but not so short that it decays before it reaches the tissue: fluorine-18 ($110$ minutes) and oxygen-15 ($2$ minutes, made and used on the spot) are typical.
When a particle meets its antiparticle 反粒子 they annihilate 湮灭: their mass turns into electromagnetic energy. In PET:
By energy conservation, the total photon energy equals the pair's rest energy 能量:
Each photon has $h f = m_{e} c^{2} \approx 8.2 \times 10^{-14}\ \text{J} \approx 0.51\ \text{MeV}$, with $\lambda \approx 2.4 \times 10^{-12}\ \text{m}$.
Explaining the annihilation (four marks). (1) The positron emitted by the tracer travels a short distance and meets an electron in the tissue. (2) The pair annihilates: their mass is converted to energy, (3) emitted as two gamma-ray photons, since (4) momentum must be conserved: the pair had almost no momentum, so the two photons must travel in opposite directions with equal momenta. Each photon carries the rest energy of one particle, $0.51\ \text{MeV}$; a single photon could not conserve momentum.
Worked example. Find the total energy released when a positron and an electron, each moving slowly, annihilate, and the wavelength of each photon.
$E = 2m_{\text{e}}c^{2} = 2 \times 9.11 \times 10^{-31} \times (3.00 \times 10^{8})^{2} = 1.64 \times 10^{-13}\ \text{J}$ ($1.02\ \text{MeV}$), shared equally: $8.2 \times 10^{-14}\ \text{J}$ each. $\lambda = hc/E = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(8.2 \times 10^{-14}) = 2.4 \times 10^{-12}\ \text{m}$, a gamma ray. If instead the electron and positron each move at $4.9 \times 10^{7}\ \text{m s}^{-1}$ in opposite directions, each has kinetic energy $\tfrac{1}{2}mv^{2} = 1.1 \times 10^{-15}\ \text{J}$, about $1\%$ of its rest energy, and the photons carry $8.3 \times 10^{-14}\ \text{J}$ each: the rest energy dominates, which is why the annihilation photons always have very nearly the same energy.
The two photons leave the body in opposite directions and hit detector rings around the patient. Recording the two simultaneous arrivals (a "coincidence") fixes the line the annihilation happened on. Many coincidences from many angles let the computer build a 3-D map of the tracer — showing tissues with high metabolic activity. Comparing the two arrival times can refine the position along that line (time-of-flight PET).
"Explain how the gamma photons are used to form an image" (four marks). (1) The two photons leave the body in opposite directions and are detected by the ring of detectors. (2) Two photons arriving at (almost) the same time are taken to come from one annihilation, which therefore lies on the line joining the two detectors. (3) The small difference in arrival times gives the position along that line ($d = c\,\Delta t/2$ from the midpoint; a $1\ \text{ns}$ difference is $15\ \text{cm}$, so the timing must be very precise). (4) A computer collects many such events and maps the concentration of tracer in the tissue, producing an image; a higher concentration means more active tissue, such as a tumour. Photons that arrive singly (their partner absorbed in the body) are rejected.
Sebuah penanda adalah zat dengan inti radioaktif yang dimasukkan ke dalam tubuh. Zat ini lebih banyak diserap oleh jaringan yang sedang dipelajari (misalnya tumor menyerap lebih banyak penanda yang diberi label glukosa karena metabolisme tinggi). Peluruhannya dideteksi dari luar.
Dalam tomografi emisi positron (PET), penandanya adalah pemancar $\beta^{+}$ — ia memancarkan positron. Yang umum adalah fluorin-18 pada analog glukosa (FDG).
"Jelaskan apa yang dimaksud dengan penanda dan bagaimana cara menggunakannya" (tiga poin). Penanda adalah zat yang mengandung inti radioaktif (untuk PET, pemancar $\beta^{+}$ seperti fluorin-18) yang dimasukkan ke dalam tubuh (disuntikkan atau ditelan), biasanya terikat pada molekul seperti glukosa, dan diserap oleh jaringan yang dipelajari; radiasi yang dipancarkannya keluar dari tubuh dan dideteksi, sehingga distribusinya menunjukkan di mana jaringan tersebut aktif. Nuklida dipilih dengan waktu paruh pendek, sehingga aktivitasnya tinggi selama pemindaian dan dosis pasien afterwards kecil, tetapi tidak terlalu pendek sehingga meluruh sebelum mencapai jaringan: fluorin-18 ($110$ menit) dan oksigen-15 ($2$ menit, dibuat dan digunakan di tempat) adalah contoh tipikal.
Ketika partikel bertemu antipartikelnya, mereka berannihilasi: massa mereka berubah menjadi energi elektromagnetik. Dalam PET:
*PET: annihilasi menghasilkan dua foton 511 keV ke arah berlawanan; koincidensi menetapkan garis
Berdasarkan kekekalan energi, total energi foton sama dengan energi diam pasangan:
Setiap foton memiliki $h f = m_{e} c^{2} \approx 8.2 \times 10^{-14}\ \text{J} \approx 0.51\ \text{MeV}$, dengan $\lambda \approx 2.4 \times 10^{-12}\ \text{m}$.
Menjelaskan annihilasi (empat poin). (1) Positron yang dipancarkan penanda bergerak jarak pendek dan bertemu elektron dalam jaringan. (2) Pasangan itu berannihilasi: massanya diubah menjadi energi, (3) dipancarkan sebagai dua foton sinar-gamma, karena (4) momentum harus kekal: pasangan memiliki hampir tidak ada momentum, jadi dua foton harus bergerak ke arah berlawanan dengan momentum yang sama. Setiap foton membawa energi diam satu partikel, $0.51\ \text{MeV}$; satu foton saja tidak dapat mengkekalkan momentum.
Contoh terpecahkan. Temukan total energi yang dilepaskan ketika sebuah positron dan elektron, masing-masing bergerak lambat, berannihilasi, dan panjang gelombang setiap foton.
$E = 2m_{\text{e}}c^{2} = 2 \times 9.11 \times 10^{-31} \times (3.00 \times 10^{8})^{2} = 1.64 \times 10^{-13}\ \text{J}$ ($1.02\ \text{MeV}$), dibagi sama: $8.2 \times 10^{-14}\ \text{J}$ masing-masing. $\lambda = hc/E = (6.63 \times 10^{-34})(3.00 \times 10^{8})/(8.2 \times 10^{-14}) = 2.4 \times 10^{-12}\ \text{m}$, sinar gamma. Jika sebaliknya elektron dan positron masing-masing bergerak pada $4.9 \times 10^{7}\ \text{m s}^{-1}$ dalam arah berlawanan, masing-masing memiliki energi kinetik $\tfrac{1}{2}mv^{2} = 1.1 \times 10^{-15}\ \text{J}$, sekitar $1\%$ dari energi diamnya, dan foton-foton tersebut membawa $8.3 \times 10^{-14}\ \text{J}$ masing-masing: energi diam mendominasi, itulah sebabnya foton annihilasi selalu memiliki energi yang sangat hampir sama.
Dua foton keluar dari tubuh ke arah yang berlawanan dan mengenai cincin detektor di sekitar pasien. Pencatatan kedatangan dua foton secara bersamaan (sebuah "kecocokan") menentukan garis tempat annihilasi terjadi. Banyak kecocokan dari berbagai sudut memungkinkan komputer membangun peta 3-D pelacak — menunjukkan jaringan dengan aktivitas metabolik tinggi. Membandingkan dua waktu kedatangan dapat mempersempit posisi sepanjang garis tersebut (PET waktu-tempuh).
*Melokalisasi annihilasi: dua foton yang dideteksi bersama-sama menentukan garis, dan perbedaan waktu kedatangannya menentukan di mana pada garis tersebut.
"Jelaskan bagaimana foton gamma digunakan untuk membentuk gambar" (empat nilai). (1) Dua foton meninggalkan tubuh ke arah yang berlawanan dan dideteksi oleh cincin detektor. (2) Dua foton yang tiba pada (hampir) waktu yang sama dianggap berasal dari satu annihilasi, sehingga terletak pada garis yang menghubungkan kedua detektor. (3) Perbedaan waktu kedatangan yang kecil memberikan posisi sepanjang garis itu ($d = c\,\Delta t/2$ dari titik tengah; perbedaan $1\ \text{ns}$ adalah $15\ \text{cm}$, sehingga pengatur waktu harus sangat presisi). (4) Komputer mengumpulkan banyak peristiwa seperti itu dan memetakan konsentrasi pelacak dalam jaringan, menghasilkan gambar; konsentrasi lebih tinggi berarti jaringan lebih aktif, seperti tumor. Foton yang tiba tunggal (pasangannya diserap dalam tubuh) ditolak.
Follow positron emission to a ring of detected photons. · Ikuti emisi positron hingga cincin foton terdeteksi.
| English | Bahasa Indonesia |
|---|---|
| metabolism/məˈtæbəlɪzəm/ | metabolisme |
| positron emission tomography/ˈpɒzɪtrɒn ɪˈmɪʃn təˈmɒɡrəfi/ | peti pemancaran positron |
| positron/ˈpɒzɪtrɒn/ | positron |
| antiparticle/ˌæntɪˈpɑːtɪkl/ | antipartikel |
| annihilate/əˈnaɪəleɪt/ | annihilasi |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| piezo-electric effect | a crystal changes shape when a p.d. is applied across it, and generates an e.m.f. when its shape is changed |
| specific acoustic impedance | the product of the density of a medium and the speed of sound in it, $Z = \rho c$ |
| intensity reflection coefficient | the fraction of the incident intensity reflected at a boundary, $(Z_{1} - Z_{2})^{2}/(Z_{1} + Z_{2})^{2}$ |
| attenuation coefficient | the constant $\mu$ in $I = I_{0}e^{-\mu x}$; the larger it is, the faster the intensity falls with thickness |
| hardness (of X-rays) | the penetrating power of the beam, set by the photon energies (the accelerating p.d.) |
| contrast | the difference in degree of blackening between neighbouring regions of an image |
| sharpness | how well defined the edges of structures are in an image |
| CT scanning | X-ray images of one section taken from many angles are combined by computer into a 2-D image of the slice; successive slices give a 3-D image |
| tracer | a substance containing radioactive nuclei that is introduced into the body and absorbed by the tissue under study |
| annihilation | a particle and its antiparticle interact and their mass is converted to energy (photons), with mass–energy and momentum conserved |
Soal definisi dinilai berdasarkan frasa tetap. Hafalkan ini persis, dan berikan hanya satu jawaban.
| Istilah | Definisi |
|---|---|
| efek piezoelektrik | kristal berubah bentuk ketika beda potensial diterapkan melaluinya, dan menghasilkan ggl ketika bentuknya diubah |
| impedansi akustik spesifik | hasil kali densitas medium dan kecepatan bunyi di dalamnya, $Z = \rho c$ |
| koefisien pantulan intensitas | pecahan intensitas insiden yang dipantulkan pada batas, $(Z_{1} - Z_{2})^{2}/(Z_{1} + Z_{2})^{2}$ |
| koefisien peredaman | konstanta $\mu$ dalam $I = I_{0}e^{-\mu x}$; semakin besar nilainya, semakin cepat intensitas turun dengan ketebalan |
| kekerasan (sinar-X) | kemampuan tembus sinar, ditentukan oleh energi foton (beda potensial akselerasi) |
| kontras | perbedaan tingkat penghitaman antara daerah berdekatan dalam gambar |
| ketajaman | seberapa jelas didefinisikan tepi struktur dalam gambar |
| pemindaian CT | gambar sinar-X dari satu irisan yang diambil dari banyak sudut digabungkan oleh komputer menjadi gambar 2-D irisan tersebut; irisan berturut-turut memberikan gambar 3-D |
| pelacak | zat yang mengandung inti radioaktif yang dimasukkan ke dalam tubuh dan diserap oleh jaringan yang diteliti |
| annihilasi | partikel dan antipartikelnya berinteraksi dan massanya dikonversi menjadi energi (foton), dengan kekekalan massa-energi dan momentum |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The luminosity 光度 $L$ of a star is the total power 功率 of radiation it gives out — the energy 能量 radiated per second in all directions. Unit: watt (W).
The one-mark definition. The luminosity of a star is the total power of radiation emitted by the star (or the total energy emitted per unit time). It is a property of the star alone; how bright it looks from the Earth depends also on how far away it is. Asked for two reasons why some stars appear brighter than others, give exactly those two: a greater luminosity, and a smaller distance (the flux falls as $1/d^{2}$).
At distance $d$, this power has spread over a sphere of area $4\pi d^{2}$. The radiant flux intensity 辐射通量密度 $F$ (power per unit area) at distance $d$ is
Worked example. The Sun's luminosity is $L = 3.8 \times 10^{26}\ \text{W}$. Find the radiant flux intensity at the Earth, a distance $d = 1.5 \times 10^{11}\ \text{m}$ away.
Unit: $\text{W m}^{-2}$. This is the inverse-square law 平方反比定律 for flux: doubling the distance cuts the flux to a quarter. A telescope measures $F$; if $L$ is known, the distance follows:
Reading the graph. A graph of $F$ against $1/d^{2}$ for one star is a straight line through the origin with gradient $L/4\pi$, so $L = 4\pi \times \text{gradient}$; a more luminous star gives a steeper line. A star whose galaxy is receding still obeys the inverse-square law, but the light we receive is redshifted, so the flux at each wavelength is shifted along the spectrum, and a detector sensitive to one band may see a different fraction of it.
Worked example. The Sun has radius $6.96 \times 10^{8}\ \text{m}$ and surface temperature $5780\ \text{K}$. A space probe carrying a $2.0\ \text{m}^{2}$ solar panel is $4.5 \times 10^{10}\ \text{m}$ from the Sun's centre. Find the power falling on the panel when it faces the Sun.
Luminosity: $L = 4\pi\sigma r^{2}T^{4} = 4\pi(5.67 \times 10^{-8})(6.96 \times 10^{8})^{2}(5780)^{4} = 3.85 \times 10^{26}\ \text{W}$. Flux at the probe: $F = L/(4\pi d^{2}) = 3.85 \times 10^{26}/[4\pi(4.5 \times 10^{10})^{2}] = 1.5 \times 10^{4}\ \text{W m}^{-2}$, eleven times the flux at the Earth. Power on the panel: $P = FA = 1.5 \times 10^{4} \times 2.0 = 3.0 \times 10^{4}\ \text{W}$. The flux is power per unit area perpendicular to the radiation; a tilted panel receives $FA\cos\theta$.
| English | Bahasa Indonesia |
|---|---|
| luminosity/ˌluːmɪˈnɒsɪti/ | luminositas |
| power/ˈpaʊə/ | kuasa |
| energy/ˈenədʒi/ | tenaga |
| radiant flux intensity/ˈreɪdɪənt flʌks ɪnˈtensɪti/ | intensitas fluks radiatif |
| inverse-square law/ɪnˈvɜːs skweə lɔː/ | hukum kuadrat terbalik |
A standard candle 标准烛光 is an object whose luminosity is known from its type. Once you find one in a distant galaxy and measure the flux $F$ from it, you get its distance from $d = \sqrt{L/(4\pi F)}$.
Examples:
A standard candle gives $L$ without first knowing the distance, so it reaches galaxies far beyond parallax 视差.
The definition. A standard candle is an object (a star or a supernova) of known luminosity. The luminosity is known because it is fixed by a property that can be measured from any distance: the pulsation period of a Cepheid variable, or the type of a supernova.
"Explain how a standard candle is used to determine the distance of a galaxy" (three marks). (1) A standard candle in the galaxy is identified (a Cepheid variable, whose period gives its luminosity, or a Type Ia supernova) so that its luminosity $L$ is known. (2) The radiant flux intensity $F$ of its light arriving at the Earth is measured. (3) The distance follows from the inverse-square law, $d = \sqrt{L/(4\pi F)}$. The order matters: the luminosity comes from the candle's type, not from the distance, which is the thing being found.
Worked example. A Type Ia supernova, luminosity $1.0 \times 10^{36}\ \text{W}$ at its peak, is observed with a peak flux of $2.0 \times 10^{-14}\ \text{W m}^{-2}$. How far away is its galaxy?
$d = \sqrt{L/(4\pi F)} = \sqrt{1.0 \times 10^{36}/(4\pi \times 2.0 \times 10^{-14})} = 6.3 \times 10^{24}\ \text{m}$, about $670$ million light years 光年 ($1\ \text{ly} = 9.5 \times 10^{15}\ \text{m}$). A Cepheid, with a luminosity of order $10^{30}\ \text{W}$, could not be seen at that distance; supernovae are the candles for the far Universe.
brightness proportional to 1 / distance^2 · kecerahan sebanding dengan 1 / jarak^2
Move distance and see why brightness falls quickly. · Pindahkan jarak dan lihat mengapa kecerahan turun dengan cepat.
| English | Bahasa Indonesia |
|---|---|
| standard candle/ˈstændəd ˈkændl/ | lilin standar |
| Cepheid variables/ˈsefiɪd ˈveərɪəblz/ | bintang variabel Cepheid |
| Type Ia supernovae/taɪp wʌn ˈeɪ ˌsuːpəˈnəʊviː/ | supernova Tipe Ia |
| parallax/ˈpærəlæks/ | paralaks |
| light years/laɪt jɪəz/ | tahun cahaya |
| supernovae/ˌsuːpəˈnəʊviː/ | supernova |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
A hot body gives out a continuous (blackbody 黑体) spectrum with a peak at a wavelength 波长 $\lambda_{\text{max}}$ set by its temperature 温度. Wien's displacement law 维恩位移定律:
Worked example. A star's blackbody spectrum peaks at $\lambda_{\text{max}} = 500\ \text{nm}$. Find its surface temperature. ($b = 2.90 \times 10^{-3}\ \text{m K}$.)
Hotter stars peak at shorter wavelengths: a cool red star ($\sim 3000\ \text{K}$) peaks in the infrared; the Sun ($\sim 5800\ \text{K}$) peaks near $500\ \text{nm}$; a hot blue-white star ($\sim 20{,}000\ \text{K}$) peaks in the ultraviolet. Measuring $\lambda_{\text{max}}$ gives the surface temperature.
"State Wien's displacement law" (two marks). The wavelength at which the intensity of the radiation from a black body is a maximum is inversely proportional to its thermodynamic temperature: $\lambda_{\text{max}} \propto 1/T$, or $\lambda_{\text{max}} T = \text{constant}$ ($2.90 \times 10^{-3}\ \text{m K}$). Say "wavelength of maximum intensity", not just "the wavelength", and the temperature must be in kelvin. The law describes the peak of the continuous black-body curve, not the spectral lines.
A star, treated as a blackbody sphere of radius $r$ and surface temperature $T$, has luminosity
where $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ is the Stefan–Boltzmann constant 斯特藩-玻尔兹曼常量 (the Stefan–Boltzmann law 斯特藩-玻尔兹曼定律). Two strong dependences:
Worked example. A star has radius $r = 7.0 \times 10^{8}\ \text{m}$ and surface temperature $T = 5800\ \text{K}$. Find its luminosity. ($\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$.)
Combine the two laws:
This is how astronomers estimate radii of stars they cannot see as a disc.
Worked example (the radius of the Sun). The radiant flux intensity of sunlight at the Earth is $1370\ \text{W m}^{-2}$ at a distance of $1.50 \times 10^{11}\ \text{m}$, and the Sun's spectrum peaks at $500\ \text{nm}$. Estimate the radius of the Sun.
Luminosity: $L = 4\pi d^{2}F = 4\pi(1.50 \times 10^{11})^{2}(1370) = 3.87 \times 10^{26}\ \text{W}$. Temperature: $T = b/\lambda_{\text{max}} = 2.90 \times 10^{-3}/(500 \times 10^{-9}) = 5800\ \text{K}$. Radius: $r = \sqrt{L/(4\pi\sigma T^{4})} = \sqrt{3.87 \times 10^{26}/[4\pi(5.67 \times 10^{-8})(5800)^{4}]} = 6.9 \times 10^{8}\ \text{m}$. Three steps, each a one-line formula; keep the full precision of $L$ and $T$ until the end, because $T$ is raised to the fourth power.
Worked example. A star in a distant galaxy has a radiant flux intensity of $2.52 \times 10^{-8}\ \text{W m}^{-2}$ at the Earth, a distance of $4.16 \times 10^{17}\ \text{m}$ away, and a surface temperature of $9500\ \text{K}$. Find its radius.
$L = 4\pi d^{2}F = 4\pi(4.16 \times 10^{17})^{2}(2.52 \times 10^{-8}) = 5.48 \times 10^{28}\ \text{W}$. Then $r = \sqrt{5.48 \times 10^{28}/[4\pi(5.67 \times 10^{-8})(9500)^{4}]} = 3.1 \times 10^{9}\ \text{m}$, about four times the Sun's radius. Check the sense: a star $140$ times as luminous as the Sun but only $1.6$ times as hot must be considerably bigger, since $L \propto r^{2}T^{4}$.
Why $T^{4}$ matters so much. Two stars of the same radius at $3000\ \text{K}$ and $6000\ \text{K}$ differ in luminosity by $2^{4} = 16$ times; a small error in the temperature is a large error in the luminosity, which is why the peak wavelength must be read carefully and, for a receding galaxy, corrected for redshift (below).
L ∝ r²
For a given surface temperature, a star's luminosity grows with the SQUARE of its radius (Stefan's law). · Bagi suhu permukaan tertentu, kecerahan sebuah bintang bertambah dengan KUASA DUA jejari-nya (hukum Stefan).
| English | Bahasa Indonesia |
|---|---|
| blackbody/ˈblækbɒdi/ | benda hitam |
| wavelength/ˈweɪvleŋθ/ | panjang gelombang |
| temperature/ˈtemprɪtʃə/ | suhu |
| Wien's displacement law/viːnz dɪˈspleɪsmənt lɔː/ | hukum pergeseran Wien |
| Stefan–Boltzmann constant/ˈstefən ˈbɒltsmən ˈkɒnstənt/ | konstanta Stefan–Boltzmann |
| Stefan–Boltzmann law/ˈstefən ˈbɒltsmən lɔː/ | Hukum Stefan–Boltzmann |
Source: Cambridge International syllabus · Sumber: Silabus Cambridge International
The spectral lines 谱线 of light from distant galaxies are seen at longer wavelengths than their known laboratory values — the whole spectrum is stretched towards the red. This is redshift 红移.
Reading it as a Doppler shift, the galaxy is moving away. For $v \ll c$:
where $\Delta\lambda = \lambda_{\text{observed}} - \lambda_{\text{emitted}}$ and $v$ is the speed of recession 退行. Example: light emitted at $4.62 \times 10^{-7}\ \text{m}$ but seen at $4.91 \times 10^{-7}\ \text{m}$ gives $\Delta\lambda = 0.29 \times 10^{-7}\ \text{m}$ and
Almost every distant galaxy is redshifted (a few near ones are blueshifted 蓝移 by local motion). So galaxies are, on average, moving apart — not just from us but from each other. The Universe is expanding, with the space between galaxies stretching. More distant galaxies are redshifted more.
"State what is meant by redshift." The observed wavelength of the radiation (its spectral lines) from a source is longer than the wavelength emitted, because the source is moving away from the observer. It is the Doppler effect 多普勒效应 for light: the fractional change in wavelength equals the fractional change in frequency and, for $v \ll c$, the ratio $v/c$.
Worked example. A hydrogen line measured in the laboratory at $656.3\ \text{nm}$ is observed in the light from a galaxy at $660.9\ \text{nm}$. Find the galaxy's speed of recession and, using $H_{0} = 2.3 \times 10^{-18}\ \text{s}^{-1}$, its distance.
$\Delta\lambda = 660.9 - 656.3 = 4.6\ \text{nm}$, so $v = c\,\Delta\lambda/\lambda = (3.00 \times 10^{8})(4.6/656.3) = 2.1 \times 10^{6}\ \text{m s}^{-1}$. Distance: $d = v/H_{0} = 2.1 \times 10^{6}/2.3 \times 10^{-18} = 9.1 \times 10^{23}\ \text{m}$ (about $100$ million light years). Divide by the emitted (laboratory) wavelength, and keep the nanometres consistent in the ratio.
Worked example (the other way round). A galaxy in Corona Borealis recedes at $21\,400\ \text{km s}^{-1}$. At what wavelength is its $656.3\ \text{nm}$ hydrogen line observed?
$\Delta\lambda = \lambda v/c = 656.3 \times (2.14 \times 10^{7}/3.00 \times 10^{8}) = 46.8\ \text{nm}$, so the line appears at $703\ \text{nm}$, moved from the red almost into the infrared. Every line and the whole continuous spectrum are stretched by the same factor $(1 + v/c) = 1.071$.
A trap the exam sets. If the peak wavelength of a receding galaxy's spectrum is fed straight into Wien's law, the temperature comes out too low, because the observed $\lambda_{\text{max}}$ is longer than the emitted one. Correct the peak first: $\lambda_{\text{emitted}} = \lambda_{\text{observed}}/(1 + v/c)$. Asked to sketch the observed spectrum on the same axes as the emitted one, draw the same shape shifted to longer wavelengths, peak included.
"Explain how redshift leads to the idea that the Universe is expanding" (three marks). (1) The spectral lines from (almost) all distant galaxies are shifted to longer wavelengths, so (2) by the Doppler effect the galaxies are moving away from us, and (3) the further away a galaxy is, the greater its redshift and so its speed: this is what would be seen from any galaxy if the space between all galaxies were expanding, so the Universe as a whole is expanding, not just moving away from the Earth.
The link between recession speed $v$ and distance $d$ is Hubble's law 哈勃定律:
where $H_{0}$ is the Hubble constant 哈勃常数 ($\approx 2.3 \times 10^{-18}\ \text{s}^{-1}$). Always use SI units. Example: a galaxy receding at $1.9 \times 10^{7}\ \text{m s}^{-1}$ is at $d = v/H_{0} \approx 8.3 \times 10^{24}\ \text{m}$.
"State Hubble's law" (two marks). The speed of recession of a galaxy is (directly) proportional to its distance from the Earth (the observer): $v = H_{0}d$, where $v$ is the recession speed, $d$ the distance and $H_{0}$ the Hubble constant. Identify every symbol when asked. The value of $H_{0}$ is quoted in the exam in SI units ($\text{s}^{-1}$), so $v$ must be in $\text{m s}^{-1}$ and $d$ in metres, never kilometres per second per megaparsec.
Worked example. A star in a distant galaxy emits radiation whose intensity peaks at $4.62 \times 10^{-7}\ \text{m}$; the peak in the light received at the Earth is at $4.91 \times 10^{-7}\ \text{m}$. Find the star's surface temperature, the galaxy's recession speed, and its distance ($H_{0} = 2.3 \times 10^{-18}\ \text{s}^{-1}$).
Temperature from the emitted peak: $T = 2.90 \times 10^{-3}/4.62 \times 10^{-7} = 6300\ \text{K}$. Speed: $v = c\,\Delta\lambda/\lambda = (3.00 \times 10^{8})(0.29/4.62) = 1.9 \times 10^{7}\ \text{m s}^{-1}$, about $6\%$ of $c$. Distance: $d = v/H_{0} = 1.9 \times 10^{7}/2.3 \times 10^{-18} = 8.2 \times 10^{24}\ \text{m}$ (about $870$ million light years). Using the observed peak for the temperature would have given $5900\ \text{K}$, $400\ \text{K}$ too low.
Hubble's law means the galaxies were once together. Running the expansion backwards, all distances shrink to zero at $t = -1/H_{0}$ — the Universe was once a tiny, hugely dense, hot point. This is the Big Bang 大爆炸. The age of the Universe (for steady expansion) is about
The expansion, the redshift of galaxies, the cosmic microwave background 宇宙微波背景, and the hydrogen/helium abundances are the main evidence for the Big Bang.
"Explain how Hubble's law leads to the Big Bang theory" (three marks). (1) Galaxies are receding with speeds proportional to their distances, so (2) if time is run backwards, every galaxy, at whatever distance, arrives at the same point at the same time, $t = d/v = 1/H_{0}$ ago; (3) so the Universe must have begun as a single point of enormous density and temperature that has been expanding ever since. The age estimate: $1/H_{0} = 1/(2.3 \times 10^{-18}) = 4.3 \times 10^{17}\ \text{s}$, and dividing by $3.16 \times 10^{7}\ \text{s}$ per year gives $1.4 \times 10^{10}$ years. This assumes the expansion rate has not changed; it is an estimate, not a measurement.
Astronomers combine methods, each calibrated by the one below:
v = H₀·d
Recession speed is proportional to distance — the gradient is Hubble's constant. · Kelajuan surut adalah berkadar dengan jarak — kecerunan adalah pemalar Hubble.
| English | Bahasa Indonesia |
|---|---|
| spectral lines/ˈspektrəl laɪnz/ | garis spektral |
| redshift/ˈredʃɪft/ | pergeseran merah |
| recession/rɪˈseʃn/ | resesi |
| blueshifted/ˈbluːʃɪftɪd/ | bergeser ke biru |
| Doppler effect/ˈdɒplə ɪˈfekt/ | Efek Doppler |
| Hubble's law/ˈhʌblz lɔː/ | Hukum Hubble |
| Hubble constant/ˈhʌbl ˈkɒnstənt/ | konstanta Hubble |
| Big Bang/bɪɡ bæŋ/ | Big Bang |
| cosmic microwave background/ˈkɒzmɪk ˈmaɪkrəʊweɪv ˈbækɡraʊnd/ | latar belakang gelombang mikro kosmik |
A definition question is marked against fixed wording. Learn these exactly, and give one answer only.
| Term | Definition |
|---|---|
| luminosity | the total power of radiation emitted by a star |
| radiant flux intensity | the power of radiation received per unit area (at right angles to the radiation), $F = L/(4\pi d^{2})$ |
| standard candle | an object of known luminosity, from which a distance can be found by measuring the flux received |
| Wien's displacement law | the wavelength of maximum intensity of a black body is inversely proportional to its thermodynamic temperature, $\lambda_{\text{max}} T = \text{constant}$ |
| Stefan–Boltzmann law | the luminosity of a black body of radius $r$ is $L = 4\pi\sigma r^{2}T^{4}$ |
| redshift | the increase in the observed wavelength of radiation from a source moving away from the observer |
| Hubble's law | the speed of recession of a galaxy is proportional to its distance from the observer, $v = H_{0}d$ |
| Hubble constant | the constant of proportionality in Hubble's law, in $\text{s}^{-1}$; $1/H_{0}$ estimates the age of the Universe |
| Big Bang theory | the Universe began from a single point of very high density and temperature and has been expanding since |
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