Oscillations
AP Physics C: Mechanics Topic 7 8:14 English narration · English + 中文 subtitles burned in
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Transcript
In a science museum, a brass ball hangs from a very long wire.
在一座科学博物馆里，一只铜球挂在很长的钢丝上。
It swings all day above a marked floor, untouched.
它整天在带刻度的地面上方摆动，无人触碰。
Every swing takes the same time as the last.
每一次摆动的时间都和上一次相同。
Pull it further out: the swing gets wider, but it still takes the same time.
把它拉得更远：摆幅变大，但用时依然不变。
That steadiness comes from one equation.
这种稳定来自一个方程。
This is Unit Seven: oscillations.
这是第七单元：振动。
The whole unit rests on one differential equation.
整个单元都建立在一个微分方程之上。
Learn to recognise it and write down its solution, and you can read off position, speed and acceleration at any moment.
学会识别它、写出它的解，你就能读出任意时刻的位置、速度和加速度。
An oscillation is periodic motion — motion that repeats — about one central point, the equilibrium position, where the net force is zero.
振动是围绕一个中心点重复的运动，这个点叫平衡位置，物体在那里所受的合力为零。
Representing and analyzing S H M starts with a picture — here the same motion is drawn three ways: a block on a spring, a dot going round a circle, and a smooth curve.
这里用三种方式画出同一种运动：弹簧上的物块、绕圆周运动的点，以及一条光滑曲线。
They stay in step because they are one motion.
它们始终同步，因为它们本来就是同一种运动。
Now the force.
再看力。
Pull the block below the middle and the spring pulls it up; push it above and the spring pulls it down.
把物块拉到中心以下，弹簧把它拉回上去；推到中心以上，弹簧把它拉回下来。
A force that always points back to equilibrium is a restoring force.
总是指回平衡位置的力叫回复力。
Simple harmonic motion needs a LINEAR restoring force: its size is also proportional to the displacement.
在简谐运动中，它的大小还与位移成正比。
The acceleration arrow always points opposite the displacement.
加速度箭头总是与位移方向相反。
Write that restoring force as minus k times the displacement.
把这个回复力写成负的 k 乘以位移。
The minus sign means: back toward the centre.
负号的意思是：方向指回中心。
Put it into Newton's second law, and the acceleration becomes the second derivative of position with time.
把它代入牛顿第二定律，加速度就成了位置对时间的二阶导数。
Divide by the mass: the constants collect into one symbol, omega, the angular frequency.
两边除以质量：常数合并成一个符号——欧米伽，也就是角频率。
This is the test — a second derivative equal to a negative constant times the quantity itself.
这就是简谐运动的判据——二阶导数等于一个负常数乘以这个量本身。
Once you have omega, the frequency and period of the S H M are free.
有了欧米伽，时间就都出来了。
The period is two pi divided by omega — or one over the frequency.
周期等于二派除以欧米伽，也就是频率的倒数。
For a block on a spring it is two pi times the square root of the mass over the spring constant.
对弹簧上的物块，它等于二派乘以质量除以劲度系数的平方根。
Half a kilogram on a fifty newton per metre spring gives about six tenths of a second.
半千克挂在五十牛顿每米的弹簧上，周期约零点六秒。
Two warnings.
有两点要注意。
The period does not depend on the amplitude: a bigger swing just moves faster.
周期与振幅无关：摆得更大，只是运动更快。
And notice what the formula leaves out — there is no little g in it, so a spring oscillator would keep perfect time in orbit.
再看公式里少了什么——里面没有重力加速度，所以弹簧振子在轨道上也能准确计时。
Now the calculus.
现在用微积分。
The solution is a cosine: amplitude times the cosine of omega t plus a phase constant.
它的解是一个余弦：振幅乘以欧米伽 t 加相位常数的余弦。
Differentiate once for the velocity — a sine, with omega in front.
求一次导数得到速度——是正弦，前面带一个欧米伽。
Differentiate again for the acceleration: minus omega squared times the position.
再求一次导数得到加速度：负的欧米伽平方乘以位置。
Our equation falls straight back out.
我们最初的方程又原样出现了。
So the top speed is amplitude times omega; the top acceleration, amplitude times omega squared.
所以最大速度是振幅乘以欧米伽；最大加速度是振幅乘以欧米伽的平方。
Three curves, one above the other.
三条曲线上下排好。
Start with the position: it swings between plus and minus the amplitude.
先看位置：它在正振幅和负振幅之间往返。
The velocity curve is the slope of the position curve, a quarter of a cycle ahead.
速度曲线是位置曲线的斜率，超前四分之一个周期。
The acceleration curve is the slope of the velocity curve — the position curve upside down.
加速度曲线是速度曲线的斜率——就是位置曲线上下翻转。
At the two ends the object has stopped: speed zero, acceleration largest.
在两端物体已经停下：速度为零，加速度最大。
At the middle the acceleration is zero and the object is fastest.
在中间加速度为零，物体最快。
A half kilogram block on a spring of stiffness two hundred newtons per metre is pulled ten centimetres from the centre and released.
半千克的物块放在劲度系数为二百牛顿每米的弹簧上，被拉离中心十厘米后释放。
Pause here and try it.
先暂停，自己试一试。
First, the angular frequency: two hundred over zero point five is four hundred; its square root is twenty radians per second.
首先求角频率：二百除以零点五等于四百，它的平方根是二十弧度每秒。
The period is two pi over twenty, about zero point three one seconds.
周期等于二派除以二十，约零点三一秒。
The greatest speed is amplitude times omega — two metres per second, at the middle.
最大速度等于振幅乘以欧米伽——二米每秒，在中间位置。
Halfway out, the speed–position formula gives one point seven metres per second.
在一半位移处，用速度与位置的公式，得到一点七米每秒。
Notice the order: find omega first, and every other number follows from it.
注意顺序：先求出欧米伽，其他每个数都随之而来。
Energy tells the same story.
能量讲的是同一个故事。
With no friction the total never changes; it only changes form.
没有摩擦时，总能量不变，只改变形式。
At the two ends the object is momentarily still, so all the energy is potential. At the middle it moves fastest, so all the energy is kinetic.
在两端物体瞬间静止，能量全是势能；在中间它运动最快，能量全是动能。
Watch the bars trade: one rises exactly as the other falls.
看能量条的交换：一个升高多少，另一个就降低多少。
Now plot energy against position, not time.
现在把能量对位置作图，而不是对时间。
Kinetic energy is largest at the centre and zero at the ends.
动能在中心最大，在两端为零。
Potential energy is the mirror image.
势能正好相反。
Both are curves, because both depend on a square.
两者都是曲线，因为都与平方有关。
Their sum is a flat line — the same everywhere.
它们的和是一条水平线——处处相同。
For a spring it is one half k times the amplitude squared: double the amplitude and you get four times the energy.
对弹簧来说它等于二分之一 k 乘以振幅的平方：振幅翻倍，能量变四倍。
A favourite exam question.
一道常考题。
Where in the swing is the energy split evenly, half potential and half kinetic?
在振动的哪个位置，能量正好一半势能、一半动能？
Start with the total: one half, times two hundred, times zero point one squared — one joule.
先算总能量：二分之一乘以二百，再乘以零点一的平方——等于一焦耳。
Set the potential energy equal to half of that; the spring constant cancels.
让势能等于总能量的一半；劲度系数约掉。
So the answer is the amplitude over the square root of two: seven point one centimetres.
所以答案是振幅除以根号二：七点一厘米。
Most people guess halfway.
多数人会猜正中间。
They are wrong — potential energy grows with the square of the position.
他们错了——势能随位置的平方增长。
One more idea.
还有一个概念。
Left alone, every oscillator swings at its own natural frequency, set by the stiffness and the mass.
不受干扰时，每个振子都以自己的固有频率振动，由劲度系数和质量决定。
Drive it with a repeating force.
现在用一个反复施加的力去驱动它。
Far from that frequency, little happens.
频率相差很远时，几乎没什么变化。
Match it, and every push arrives at the right moment, adding energy faster than friction removes it.
让两者一致，每一次推动都恰好在正确的时刻到来，注入能量比摩擦消耗更快。
The amplitude climbs to a sharp peak.
振幅升到一个尖锐的峰。
That is resonance.
这就是共振。
Now the last family.
现在看最后一类。
A physical pendulum is any rigid body swinging about a fixed pivot — a rule, a hoop, a leg.
物理摆是绕固定转轴摆动的任意刚体——一把尺、一个圆环、一条腿。
Gravity acts at the centre of mass, a distance d from the pivot.
重力作用在质心，质心到转轴的距离是 d。
Tip the body through an angle and gravity supplies a restoring torque: minus m g d, times the sine of that angle.
把物体转过一个角度，重力提供一个回复力矩：负的 m g d 乘以这个角度的正弦。
The same minus sign as before.
还是之前那个负号。
But that sine spoils it.
但那个正弦破坏了一切。
Sine is not proportional to the angle, so this is not yet simple harmonic.
正弦与角度不成正比，所以这还不是简谐运动。
For small angles, though, the small-angle approximation says the sine is almost the angle itself, in radians.
不过对小角度来说，正弦非常接近角度本身，用弧度表示。
Make that swap and the torque becomes minus m g d times the angle.
做了这个替换，力矩就变成负的 m g d 乘以角度。
Now put it into Newton's second law for rotation and our pattern is back.
现在把它代入转动形式的牛顿第二定律，我们的模式又回来了。
Read omega off the constant: the period is two pi times the square root of I over m g d.
从常数中读出欧米伽：周期是二派乘以转动惯量除以 m g d 的平方根。
The simple pendulum is just that formula with a point mass on a light string; and a torsion pendulum — a disc hanging on a twisting wire — does it once more.
单摆就是质点挂在轻绳上的特例；而挂在扭丝上的圆盘，又把这件事重演了一遍。
Try the classic.
来做经典题。
A uniform rod of mass M and length L hangs from one end and swings gently.
一根质量为 M、长度为 L 的均匀细杆从一端悬挂，做小幅摆动。
Its rotational inertia about the end is one third M L squared.
它绕端点的转动惯量是三分之一 M L 平方。
The centre of mass is at the middle, so d is L over two.
质心在中点，所以 d 等于二分之一 L。
Put both into the formula: the mass cancels, leaving two pi times the square root of two L over three g.
把两者代入公式：质量约掉，剩下二派乘以三 g 分之二 L 的平方根。
Compare that with a simple pendulum of the same length — the rod swings faster, because its mass sits closer to the pivot.
把它和同样长度的单摆比较——细杆摆得更快，因为它的质量离转轴更近。
Three habits that pick up marks.
三个能拿分的习惯。
First, to show it is simple harmonic motion you must reach the standard form — second derivative equals a negative constant times the quantity — then read omega off that constant.
第一，要证明是简谐运动，必须化到标准形式—— 二阶导数等于一个负常数乘以该量——再从这个常数里读出欧米伽。
Second, match the phase to the start: released from rest at full displacement means a cosine.
第二，让相位与起始状态对应：从最大位移处由静止释放，就用余弦。
Third, on any pendulum, state the small-angle step out loud — a scored point, and the reason a big swing is not simple harmonic.
第三，凡是摆的题，都要明确写出小角度近似这一步—— 它本身是得分点，也是大幅摆动不是简谐运动的原因。