Representing and Analyzing SHM · 简谐运动的表示与分析
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| amplitude/ˈæmplɪtjuːd/ | 振幅 | zhèn fú |
| phase/feɪz/ | 相位 | xiàng wèi |
Drawing the wobble
- Trace a bobbing mass over time and you get a smooth wave on the graph.
- The same S-curve describes springs, pendulums, and sound.
- Its height, speed, and steepness all follow from one equation.
- A single cosine captures the whole back-and-forth.
画出这份晃动
- 把起伏的物块随时间描出来,图上就得到一条平滑的波。
- 同一条 S 形曲线描述弹簧、单摆和声音。
- 它的高度、速度和陡度都来自同一个方程。
- 一个余弦函数就抓住了整个来回摆动。
The position over time
- The displacement in SHM traces a cosine:
- $A$ is the amplitude 振幅, the maximum displacement.
- $\omega$ is the angular frequency and $\varphi$ the phase 相位, the starting point.
位置随时间的变化
- 简谐运动的位移描出一条余弦曲线:
- $A$ 是振幅,即最大位移。
- $\omega$ 是角频率,$\varphi$ 是相位,即起始点。
In $x = A\cos(\omega t + \varphi)$, the symbol A stands for the . · 在$x = A\cos(\omega t + \varphi)$中，符号A代表。
$A$ is the amplitude -- the largest displacement from equilibrium. · $A$是振幅——偏离平衡位置的最大位移。
Speed and acceleration follow
- Differentiate to get the velocity and acceleration:
- Maximum speed $v_{max} = A\omega$; maximum acceleration $a_{max} = A\omega^2$.
速度和加速度随之而来
- 求导得到速度和加速度:
- 最大速度 $v_{max} = A\omega$;最大加速度 $a_{max} = A\omega^2$。
An oscillator has $A = 0.2\ \text{m}$ and $\omega = 4\ \text{rad/s}$. Its maximum speed (in m/s)? · 振荡器具有$A = 0.2\ \text{m}$和$\omega = 4\ \text{rad/s}$。其最大速度（单位：m/s）是多少？
$v_{max} = A\omega = 0.2 \times 4 = 0.8\ \text{m/s}$.
For the same oscillator ($A = 0.2\ \text{m}$, $\omega = 4\ \text{rad/s}$), the maximum acceleration (in m/s²)? · 对于同一振荡器（$A = 0.2\ \text{m}$, $\omega = 4\ \text{rad/s}$），最大加速度（单位：m/s²）是多少？
$a_{max} = A\omega^2 = 0.2 \times 16 = 3.2\ \text{m/s}^2$.
Opposite extremes
- Speed is greatest at equilibrium (where $x = 0$) and zero at the ends.
- Acceleration is greatest at the ends and zero at equilibrium.
- So $x$ and $a$ are perfectly out of step, while $v$ runs a quarter-cycle ahead.
相反的两端
- 速度在平衡位置(即 $x = 0$)最大,在两端为零。
- 加速度在两端最大,在平衡位置为零。
- 所以 $x$ 和 $a$ 完全不同步,而 $v$ 领先四分之一周期。
SHM as a sine wave · 简谐运动作为正弦波
The displacement of a simple harmonic oscillator traces a cosine curve in time. · 简谐振荡器的位移随时间呈余弦曲线变化。
Where in the motion is the speed greatest? · 在运动的哪个位置速度最大？
Speed peaks where $x = 0$ (the middle) and is zero at the ends. · 速度在$x = 0$（中间位置）达到峰值，在两端为零。
In SHM, displacement and acceleration are exactly out of step (opposite phase). · 在简谐运动中，位移和加速度完全反相（相位相反）。
$a = -\omega^2 x$, so $a$ is always opposite to $x$. · $a = -\omega^2 x$，因此$a$始终与$x$相反。
An oscillator has $A = 0.1\ \text{m}$ and $\omega = 5\ \text{rad/s}$.
- Max speed: $v_{max} = A\omega = 0.1 \times 5 = 0.5\ \text{m/s}$, reached at the middle.
- Max acceleration: $a_{max} = A\omega^2 = 0.1 \times 25 = 2.5\ \text{m/s}^2$, reached at the ends.
一个振子 $A = 0.1\ \text{m}$,$\omega = 5\ \text{rad/s}$。
- 最大速度:$v_{max} = A\omega = 0.1 \times 5 = 0.5\ \text{m/s}$,在中间达到。
- 最大加速度:$a_{max} = A\omega^2 = 0.1 \times 25 = 2.5\ \text{m/s}^2$,在两端达到。
A mass released from rest at the far end is best described using... · 从最远端静止释放的质量最适合用...描述
At $t = 0$ the displacement is maximum, which cosine gives with $\varphi = 0$. · 在$t = 0$处位移最大，这由带有$\varphi = 0$的余弦函数给出。
Do not swap where speed and acceleration peak -- they sit at opposite places: speed is largest in the middle, acceleration largest at the extremes. If the motion starts at the far end, use cosine ($\varphi = 0$); if it starts at equilibrium moving outward, use sine. And $\omega$ here is in rad/s, not the frequency $f$.
不要弄反速度和加速度的峰值位置——它们在相反的地方:速度在中间最大,加速度在两端最大。若运动从远端开始,用余弦($\varphi = 0$);若从平衡位置向外开始,用正弦。而且这里的 $\omega$ 用 rad/s,不是频率 $f$。
SHM is one cosine: $x = A\cos(\omega t + \varphi)$, with amplitude $A$ and phase $\varphi$. Its derivatives give $v_{max} = A\omega$ (at equilibrium) and $a_{max} = A\omega^2$ (at the extremes). Displacement and acceleration are exactly out of step, and velocity leads position by a quarter cycle.
简谐运动就是一个余弦:$x = A\cos(\omega t + \varphi)$,其中振幅为 $A$、相位为 $\varphi$。求导给出 $v_{max} = A\omega$(在平衡位置)和 $a_{max} = A\omega^2$(在两端)。位移与加速度恰好不同步,而速度领先位置四分之一周期。