Energy in simple harmonic motion · 简谐运动中的能量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | 平衡 | píng héng |
| damping/ˈdæmpɪŋ/ | 阻尼 | zǔ ní |
Where the energy goes at the top of a swing
- At the highest point of a swing you are, for an instant, completely still. Your kinetic energy is exactly zero.
- A moment later you are moving faster than at any other point in the arc. Nothing pushed you; the energy was never gone.
- It was in the potential store, and it came straight back out. An oscillation is that exchange, repeated for as long as friction allows.
- This lesson is the energy of an oscillator: how it swaps, why the total is constant, and why the total depends on the square of the amplitude.
秋千最高处的能量去了哪里
- 在秋千的最高点,你有一瞬间完全静止。你的动能恰好是零。
- 片刻之后你比弧线上任何一点都快。没有什么推了你;能量从未消失。
- 它在势能的储存里,又原样回来了。振动就是这种交换,只要摩擦允许就一直重复。
- 这一课讲振子的能量:它怎样交换、总量为什么恒定,以及总量为什么取决于振幅的平方。
The swap
- At the equilibrium 平衡 position the speed is greatest, so the kinetic energy is maximum and the potential energy is at its minimum.
- At the extremes the oscillator is momentarily at rest, so the kinetic energy is zero and the potential energy is at its maximum.
- In between, the two trade continuously. The exchange happens twice per full oscillation, since the oscillator passes through the centre twice.
Two curves in opposite phase, and a flat line above them
交换
- 在平衡(equilibrium)位置速率最大,所以动能最大,势能最小。
- 在两端振子有一瞬间静止,所以动能为零,势能最大。
- 在两者之间,两种能量不断互换。这种交换在一个完整振动中发生两次,因为振子两次经过中心。
两条相位相反的曲线,以及它们上方的一条水平线
The kinetic energy of an oscillator is greatest: · 振子的动能在哪里最大:
Speed is greatest at the middle, so KE peaks there while PE is at its lowest. · 速率在中间最大,所以动能在那里达到峰值,而势能最低。
At the extreme positions of the motion, all the energy is potential. · 在运动的两端,所有能量都是势能。
The oscillator is momentarily at rest there, so KE = 0 and all the energy is potential. · 振子在那里瞬间静止,所以动能 = 0,所有能量都是势能。
The total is constant
- With no damping 阻尼, no resistive force removing energy, the total energy stays constant. This is conservation of energy applied to the oscillator.
- The easiest place to evaluate it is at the centre, where all of it is kinetic and the speed is $\omega x_0$:
- On a graph against displacement, the kinetic energy is a downward parabola, the potential energy an upward one, and their sum is a horizontal line.
The two parabolas add to a constant at every displacement
总量恒定
- 没有阻尼(damping)——即没有耗散力带走能量——时,总能量保持恒定。这是应用于振子的能量守恒。
- 最方便计算它的地方是中心,那里全部是动能,速率为 $\omega x_0$:
- 在对位移的图上,动能是一条开口向下的抛物线,势能是开口向上的一条,它们的和是一条水平线。
两条抛物线在每个位移处相加都得到同一个常量
Energy in SHM · 简谐运动中的能量
KE + PE = constant · 动能 + 势能 = 常数
Energy trades between kinetic (fastest at the centre) and potential · 势 (max at the ends). · 能量在 动能(中心处最快)和 势能(两端最大)之间交换。
With no damping, the total energy of an oscillator stays constant. · 没有阻尼时,振子的总能量保持恒定。
Energy just swaps between KE and PE; their sum is conserved. · 能量只是在动能和势能之间互换;它们的和守恒。
Energy and amplitude
- Because $E = \tfrac12 m\omega^2 x_0^2$, the total energy is proportional to the square of the amplitude.
- Double the amplitude and the energy is four times as large. Triple it and the energy is nine times.
- This is the relationship the exam tests most often, and answering "twice" for a doubled amplitude is the commonest error in the subtopic.
能量与振幅
- 因为 $E = \tfrac12 m\omega^2 x_0^2$,总能量与振幅的平方成正比。
- **振幅加倍,能量变为四倍。**振幅变三倍,能量变九倍。
- 这是考试最常考的关系,而对振幅加倍答"两倍"是本子专题最常见的错误。
If the amplitude doubles, the total energy of an oscillator becomes: · 如果振幅加倍,振子的总能量变为:
$E \propto x_0^{2}$, so doubling $x_0$ gives $2^{2} = 4$ times the energy. · $E \propto x_0^{2}$,所以 $x_0$ 加倍给出 $2^{2} = 4$ 倍的能量。
The amplitude of an undamped oscillator is doubled. What happens to its total energy? · 无阻尼振子的振幅加倍。它的总能量会怎样?
E is proportional to the square of the amplitude, from E = m omega^2 x0^2 / 2. Answering "twice" is the commonest error in this subtopic. · 由 E = m omega^2 x0^2 / 2,E 与振幅的平方成正比。答"两倍"是本子专题最常见的错误。
Worked example: find the total energy
- An oscillator of mass $0.20\ \text{kg}$ has angular frequency $10\ \text{rad/s}$ and amplitude $0.050\ \text{m}$. Find its total energy and its kinetic energy at a displacement of $0.030\ \text{m}$.
- Total: $E = \tfrac12 m \omega^2 x_0^2 = \tfrac12 \times 0.20 \times 10^2 \times 0.050^2 = 0.025\ \text{J}$.
- At $x = 0.030\ \text{m}$: $v = \omega\sqrt{x_0^2 - x^2} = 10\sqrt{0.050^2 - 0.030^2} = 0.40\ \text{m/s}$, so $E_{\text{k}} = \tfrac12 \times 0.20 \times 0.40^2 = 0.016\ \text{J}$.
- The potential energy there is the difference, $0.025 - 0.016 = 0.009\ \text{J}$. Using the constant total to get the second energy is quicker and safer than a separate formula.
例题:求总能量
- 一个质量 $0.20\ \text{kg}$ 的振子,角频率 $10\ \text{rad/s}$,振幅 $0.050\ \text{m}$。求它的总能量,以及位移为 $0.030\ \text{m}$ 处的动能。
- 总能量:$E = \tfrac12 m \omega^2 x_0^2 = \tfrac12 \times 0.20 \times 10^2 \times 0.050^2 = 0.025\ \text{J}$。
- 在 $x = 0.030\ \text{m}$ 处:$v = \omega\sqrt{x_0^2 - x^2} = 10\sqrt{0.050^2 - 0.030^2} = 0.40\ \text{m/s}$,所以 $E_{\text{k}} = \tfrac12 \times 0.20 \times 0.40^2 = 0.016\ \text{J}$。
- 那里的势能是差值,$0.025 - 0.016 = 0.009\ \text{J}$。用恒定的总量去求第二个能量,比另用一个公式更快也更稳妥。
An oscillator has $m = 0.20\ \text{kg}$, $\omega = 10\ \dfrac{\text{rad}}{\text{s}}$ and amplitude $0.10\ \text{m}$. What is its total energy? · 一个振子的 $m = 0.20\ \text{kg}$,$\omega = 10\ \dfrac{\text{rad}}{\text{s}}$,振幅 $0.10\ \text{m}$。它的总能量是多少?
$E = \tfrac{1}{2}m\omega^{2}x_0^{2} = \tfrac{1}{2} \times 0.20 \times 10^{2} \times 0.10^{2} = 0.10\ \text{J}$. · $E = \tfrac{1}{2}m\omega^{2}x_0^{2} = \tfrac{1}{2} \times 0.20 \times 10^{2} \times 0.10^{2} = 0.10\ \text{J}$。
An oscillator of mass 0.20 kg has omega = 10 rad/s and amplitude 0.050 m. What is its total energy, in joules? · 一个质量 0.20 kg 的振子,omega = 10 rad/s,振幅 0.050 m。它的总能量是多少焦耳?
E = m omega^2 x0^2 / 2 = 0.20 x 100 x 0.0025 / 2 = 0.025 J. Evaluating at the centre, where all the energy is kinetic, is the quickest route. · E = m omega^2 x0^2 / 2 = 0.20 x 100 x 0.0025 / 2 = 0.025 J。在全部为动能的中心处计算是最快的路径。
If the total energy is 0.025 J and the kinetic energy at some displacement is 0.016 J, the potential energy there is ____ J. · 若总能量是 0.025 J,某位移处的动能是 0.016 J,则那里的势能是 ____ J。
The total is constant, so the potential energy is simply the total minus the kinetic. That subtraction is quicker and safer than a separate formula. · 总量恒定,所以势能就是总量减动能。这个减法比另用一个公式更快也更稳妥。
Worked example: reading the energy graphs
- Sketch how the kinetic energy, the potential energy and the total energy of an undamped oscillator vary with displacement.
- Kinetic energy: maximum at $x = 0$, falling to zero at $x = \pm x_0$, a downward parabola.
- Potential energy: zero at $x = 0$, rising to the maximum at $x = \pm x_0$, an upward parabola, and the mirror image of the first.
- Total: a horizontal line at the value of either maximum, since the two curves always add to the same amount.
- Against time instead of displacement, both curves are still parabola-shaped in energy but now oscillate at twice the frequency of the displacement, because energy peaks twice per cycle.
例题:读能量图
- 画出无阻尼振子的动能、势能和总能量随位移变化的图。
- 动能:在 $x = 0$ 处最大,到 $x = \pm x_0$ 降为零,是开口向下的抛物线。
- 势能:在 $x = 0$ 处为零,到 $x = \pm x_0$ 升到最大,是开口向上的抛物线,与前一条互为镜像。
- 总量:一条位于任一最大值处的水平线,因为两条曲线相加总是同一个量。
- 若横轴换成时间而不是位移,两条能量曲线仍是抛物线形状,但现在以位移两倍的频率振荡,因为能量每周期出现两次峰值。
Match each energy of an undamped oscillator to its graph against displacement. · 把无阻尼振子的每种能量与它对位移的图像配对。
The two parabolas are mirror images, so they add to the same value at every displacement. That flat line is conservation of energy drawn out. · 两条抛物线互为镜像,所以在每个位移处相加都是同一个值。那条水平线就是画出来的能量守恒。
Plotted against time, the kinetic energy of an oscillator repeats at twice the frequency of the displacement. · 以时间为横轴时,振子的动能以位移两倍的频率重复。
The oscillator passes through the centre twice per cycle, so kinetic energy peaks twice per cycle while displacement peaks once. · 振子每周期两次经过中心,所以动能每周期出现两个峰,而位移只出现一个。
Energy against time runs at double the frequency
- Plot the displacement against time and you get one cycle per period. Plot the kinetic energy against time and you get two.
- The reason is the square. Kinetic energy depends on $v^2$, and $v$ passes through its maximum magnitude twice in every cycle, once going each way.
- So both energy curves complete two maxima per oscillation, and their frequency is double that of the displacement.
- The two curves are exact mirrors of each other about the half-total line, and their sum is a horizontal line at the total energy.
- A sketch question here is marked on three things: the doubled frequency, the two curves being out of step by half their period, and the flat total.
能量对时间的图,频率是两倍
- 把位移对时间作图,一个周期一个循环。把动能对时间作图,却得到两个。
- 原因在于那个平方。动能取决于 $v^2$,而 $v$ 在每个周期里两次达到最大值,来回各一次。
- 所以两条能量曲线在每次振动中都完成两个极大,它们的频率是位移频率的两倍。
- 两条曲线关于总能量一半那条线正好互为镜像,而它们的和是一条水平线,高度是总能量。
- 这里的作图题按三点评分:频率翻倍、两条曲线相差半个它们自己的周期、以及总和是平的。
The kinetic energy of an oscillator varies at twice the frequency of its displacement. · 振子的动能变化频率是其位移变化频率的两倍。
KE depends on v squared, and the speed reaches its maximum twice per cycle, once in each direction. The potential energy does the same, and the two sum to a flat line. · 动能取决于 v 的平方,而速率每个周期两次达到最大,来回各一次。势能也是如此,两者之和是一条水平线。
Marks that slip away
- Energy is proportional to amplitude squared. Doubling the amplitude gives four times the energy, not twice.
- Kinetic energy is maximum at the centre, potential at the ends. Do not swap them.
- Against time, the energy curves repeat at twice the frequency of the displacement, because there are two energy peaks per oscillation.
- The total is constant only when there is no damping. Say so when the question mentions resistive forces.
容易丢掉的分
- 能量与振幅的平方成正比。振幅加倍给出四倍的能量,不是两倍。
- 动能在中心最大,势能在两端最大。不要互换。
- 以时间为横轴时,能量曲线以位移两倍的频率重复,因为每次振动有两个能量峰。
- 只有没有阻尼时总量才恒定。题目提到耗散力时要说出这一点。
You've got it
- kinetic energy is maximum at the centre and zero at the extremes; potential energy does the opposite, and they exchange twice per cycle
- with no damping the total is constant: $E = \tfrac12 m\omega^2 x_0^2$, most easily evaluated at the centre where all of it is kinetic
- against displacement the energies are two opposite parabolas whose sum is a horizontal line
- $E \propto x_0^2$: double the amplitude, four times the energy
你掌握了
- 动能在中心最大、在两端为零;势能恰好相反,两者每周期交换两次
- 没有阻尼时总量恒定:$E = \tfrac12 m\omega^2 x_0^2$,在全部为动能的中心处最容易计算
- 以位移为横轴时,两种能量是两条相反的抛物线,其和是一条水平线
- $E \propto x_0^2$:振幅加倍,能量四倍