Energy of Simple Harmonic Oscillators · 简谐振子的能量
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| potential energy/pəˈtenʃl ˈenədʒi/ | 势能 | shì néng |
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
The endless trade
- A swing slows to a stop at the top, then rushes fastest through the bottom.
- Nothing is lost -- the motion just keeps swapping form.
- Stretch is stored; speed is spent; back and forth forever.
- Two stores hand the same total to and fro.
永不停止的交换
- 秋千在最高点慢下来停住,又在最低点冲得最快。
- 什么也没损失——运动只是不断变换形式。
- 拉伸被储存;速度被花掉;来来回回,永不停歇。
- 两个储库把同一个总量来回传递。
Two stores in a spring
- The spring holds potential energy 势能 $U = \tfrac{1}{2}kx^2$.
- The mass carries kinetic energy 动能 $K = \tfrac{1}{2}mv^2$.
- As it oscillates, energy pours from one into the other.
弹簧里的两个储库
- 弹簧储存势能 $U = \tfrac{1}{2}kx^2$。
- 物块携带动能 $K = \tfrac{1}{2}mv^2$。
- 随着振荡,能量从一个倾注到另一个。
A spring with $k = 100\ \text{N/m}$ is displaced $x = 0.1\ \text{m}$. Its stored potential energy (in J)? · 劲度系数为$k = 100\ \text{N/m}$的弹簧被位移了$x = 0.1\ \text{m}$。其储存的势能（单位：J）是多少？
$U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(100)(0.1)^2 = 0.5\ \text{J}$.
The total never changes
- Add them and the total stays fixed:
- At the ends it is all potential (speed zero); at the middle it is all kinetic.
- The amplitude alone sets the total.
总量从不改变
- 把它们相加,总量保持固定:
- 在两端全是势能(速度为零);在中间全是动能。
- 仅由振幅就决定了总量。
A spring with $k = 100\ \text{N/m}$ oscillates with amplitude $A = 0.2\ \text{m}$. Its total energy (in J)? · 劲度系数为$k = 100\ \text{N/m}$的弹簧以振幅$A = 0.2\ \text{m}$振动。其总能量（单位：J）是多少？
$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(100)(0.2)^2 = 2\ \text{J}$.
At which point is the oscillator's energy entirely kinetic? · 在哪个点振荡器的能量完全是动能？
At equilibrium $x = 0$ so $U = 0$ and all the energy is kinetic. · 在平衡位置$x = 0$，所以$U = 0$，所有能量都是动能。
Cashing energy for speed
- At equilibrium every bit is kinetic, so:
- This recovers $v_{max} = A\omega$, since $\omega^2 = k/m$.
- Double the amplitude and you get four times the energy, twice the top speed.
用能量换速度
- 在平衡位置每一分都是动能,所以:
- 由此复原 $v_{max} = A\omega$,因为 $\omega^2 = k/m$。
- 振幅加倍,能量变四倍,最高速度变两倍。
Energy of an oscillator · 振荡器的能量
Over a cycle, energy shifts between kinetic and potential while the total stays fixed. · 在一个周期内，能量在动能和势能之间转换，而总能量保持不变。
A spring with $k = 200\ \text{N/m}$ oscillates with amplitude $A = 0.05\ \text{m}$.
- Total energy: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.05)^2 = 0.25\ \text{J}$.
- It is all kinetic at the middle and all potential at the ends.
一根 $k = 200\ \text{N/m}$ 的弹簧以振幅 $A = 0.05\ \text{m}$ 振荡。
- 总能量:$E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.05)^2 = 0.25\ \text{J}$。
- 它在中间全是动能,在两端全是势能。
If you double the amplitude of an oscillator, its total energy... · 如果你将振荡器的振幅加倍，其总能量...
$E = \tfrac{1}{2}kA^2 \propto A^2$, so $2^2 = 4$ times the energy. · $E = \tfrac{1}{2}kA^2 \propto A^2$，因此$2^2 = 4$倍的能量。
Kinetic and potential energy can both reach their maximum at the same instant. · 动能和势能不可能同时达到最大值。
Each is largest where the other is zero -- they are never both maximal. · 各自在对方为零时达到最大——它们永远不会同时最大。
In a real oscillator with friction, over time the amplitude... · 在有摩擦的真实振荡器中，随着时间的推移，振幅...
Friction drains energy, so the amplitude decays -- this is damping. · 摩擦力消耗能量，因此振幅衰减——这是阻尼。
The energy goes as amplitude squared ($E = \tfrac{1}{2}kA^2$), so doubling $A$ quadruples the energy, not doubles it. Kinetic and potential energy are each largest where the other is zero, never both at once. And "the total stays constant" assumes no friction or drag -- a real oscillator slowly loses energy and its amplitude decays (damping).
能量按振幅的平方变化($E = \tfrac{1}{2}kA^2$),所以 $A$ 加倍会使能量变成四倍,而不是两倍。动能和势能各自在对方为零处最大,绝不会同时最大。而"总量保持不变"假设没有摩擦或阻力——真实的振子会缓慢损失能量,振幅衰减(阻尼)。
An oscillator's energy trades between potential energy $U = \tfrac{1}{2}kx^2$ and kinetic energy $K = \tfrac{1}{2}mv^2$, with a fixed total $E = \tfrac{1}{2}kA^2$. It is all potential at the extremes and all kinetic at equilibrium, giving $\tfrac{1}{2}mv_{max}^2 = \tfrac{1}{2}kA^2$. Because $E \propto A^2$, doubling the amplitude quadruples the energy.
振子的能量在势能 $U = \tfrac{1}{2}kx^2$ 和动能 $K = \tfrac{1}{2}mv^2$ 之间交换,总量固定为 $E = \tfrac{1}{2}kA^2$。在两端全是势能,在平衡位置全是动能,给出 $\tfrac{1}{2}mv_{max}^2 = \tfrac{1}{2}kA^2$。因为 $E \propto A^2$,振幅加倍会使能量变四倍。