振动

AP 物理 1 · 第 7 主题

训练

讲义词汇表

7.1

简谐运动（SHM）的定义

大纲

Learning Objective

Essential Knowledge

7.1.A
Describe simple harmonic motion.

7.1.A.1 Simple harmonic motion is a special case of periodic motion.
7.1.A.2 SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position.
- Equation: $ma_x = -k\Delta x$
- 7.1.A.2.i A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position.
- 7.1.A.2.ii An equilibrium position is a location at which the net force exerted on an object or system is zero.
- 7.1.A.2.iii The motion of a pendulum with a small angular displacement can be modeled as simple harmonic motion because the restoring torque is proportional to the angular displacement.

来源:美国大学理事会 AP 课程与考试说明

简谐运动

简谐运动(simple harmonic motion)(SHM)是由一个回复力(restoring force)造成的一个来回振动(oscillation),该回复力与从平衡的位移成比例并总是指回向它:$F=-kx$。一个弹簧上的质量和(对小角度)一个摆是标准的例子。因为力随位移增长,运动是平滑而重复的,在时间上描出一条正弦曲线。

在 SHM 里加速度总是指回向平衡,与位移相反

SHM 的判据正是这个:加速度与位移成比例且方向相反,$a=-\dfrac{k}{m}x$。一个摆只对小摆动遵循它,那里 $\sin\theta\approx\theta$;大的摆动不完全是 SHM。

A spring under load: Hooke's law F = −kx is the restoring force that produces SHM

词汇表训练

英文	中文	拼音
Simple harmonic motion	简谐运动	jiǎn xié yùn dòng
oscillation	振动	zhèn dòng
restoring force	回复力	huí fù lì

练习卷

7.2

简谐运动的频率与周期

大纲

Learning Objective

Essential Knowledge

7.2.A
Describe the frequency and period of an object exhibiting SHM.

7.2.A.1 The period of SHM is related to the frequency $f$ of the object's motion by the following equation:
- Equation: $T = \dfrac{1}{f}$
- 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation $T_s = 2\pi\sqrt{\dfrac{m}{k}}$.
- 7.2.A.1.ii The period of a simple pendulum displaced by a small angle is given by the equation $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$.

来源:美国大学理事会 AP 课程与考试说明

周期(period)$T$ 是一个完整循环的时间。
频率(frequency)$f=\dfrac{1}{T}$ 是每秒的循环数(赫兹)。

对于 SHM 这些只取决于系统,不取决于振幅:

$$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{L}{g}}.$$

所以一个更硬的弹簧或更小的质量振动更快;一个更长的摆摆动更慢。

Worked example. 一个 $0.25\ \text{kg}$ 的质量挂在一个刚度 $k=100\ \text{N/m}$ 的弹簧上。它的周期是

$$T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{0.25}{100}}=0.31\ \text{s},\qquad f=\frac{1}{T}=3.2\ \text{Hz}.$$

Worked example. 一个摆钟以恰好 $2.0\ \text{s}$ 的周期滴答。它多长?重新排列 $T=2\pi\sqrt{L/g}$,

$$L=g\left(\frac{T}{2\pi}\right)^2=9.8\times\left(\frac{2.0}{2\pi}\right)^2=0.99\ \text{m}.$$

注意振幅从未进入——一个宽的或窄的摆动保持相同的时间,这就是使摆成为好钟的东西。

探索

Time a pendulum's swing

A pendulum's period depends on its length and gravity, not its mass or (small) amplitude: $T=2\pi\sqrt{L/g}$. Lengthen it and each swing takes longer.

词汇表训练

英文	中文	拼音
period	周期	zhōu qī
frequency	频率	pín lǜ

练习卷

7.3

简谐运动的表示与分析

大纲

Learning Objective

Essential Knowledge

7.3.A
Describe the displacement, velocity, and acceleration of an object exhibiting SHM.

7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$.
- 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
- 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
7.3.A.2 Changing the amplitude of a system exhibiting SHM will not change the period of that system.
7.3.A.3 Properties of SHM can be determined and analyzed using graphical representations.

来源:美国大学理事会 AP 课程与考试说明

位移正弦地变化:$x(t)=A\cos(\omega t)$(或正弦),其中 $A$ 是振幅(amplitude)(最大位移)而 $\omega=2\pi f$ 是角频率(angular frequency)。读运动:

位移在简谐运动里随时间正弦地变化

在极端($x=\pm A$):位移和回复力最大,所以加速度最大,但速度是零。
在平衡(equilibrium)($x=0$):力和加速度是零,但速率最大。

速度和加速度也是正弦的,与位移在相位(phase)上偏移——速度比位移领先四分之一循环,而加速度与位移恰好相反。

SHM 里的位移、速度和加速度,每个相隔四分之一循环

把图当作一个故事读:$x$ 最大的地方(转折点),$v$ 刚落到零而 $a$ 处于它最负的,把质量拖回;四分之一循环后质量以顶速冲过中间,加速度为零。

词汇表训练

英文	中文	拼音
amplitude	振幅	zhèn fú
angular frequency	角频率	jiǎo pín lǜ
phase	相位	xiàng wèi
equilibrium	平衡	píng héng

练习卷

7.4

简谐振子的能量

大纲

Learning Objective

Essential Knowledge

7.4.A
Describe the mechanical energy of a system exhibiting SHM.

7.4.A.1 The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies.
- Equation: $E_{\text{total}} = U + K$
7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
- 7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
- 7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system.
- Relevant equation for a spring–object system: $E_{\text{total}} = \dfrac{1}{2}kA^2$

来源:美国大学理事会 AP 课程与考试说明

能量守恒：动能↔势能

能量在动能和势能之间晃动,而总量保持恒定(没有摩擦):

$$E=\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2+\tfrac{1}{2}mv^2.$$

在极端它是全势能;在平衡它是全动能(最大速率)。因为 $E\propto A^2$,把振幅加倍使能量变为四倍。

动能和势能在一个循环上交换而总能量保持恒定

Worked example. 一个刚度 $k=200\ \text{N/m}$ 的弹簧上的 $0.50\ \text{kg}$ 质量以振幅 $A=0.10\ \text{m}$ 振动。求它的最大速率。所有的能量在平衡点是动能,所以 $\tfrac12 kA^2=\tfrac12 mv_{\max}^2$:

$$v_{\max}=A\sqrt{\frac{k}{m}}=0.10\times\sqrt{\frac{200}{0.50}}=0.10\times20=2.0\ \text{m/s}.$$

当你只需要最大速率时,能量方法比追踪正弦函数更快。

探索

Trade kinetic and potential energy in SHM

In simple harmonic motion, energy sloshes between kinetic (fastest at the centre) and potential (greatest at the extremes) while the total stays constant.

练习卷

7.4

考试技巧

判据 SHM:加速度必须与位移成比例并指回向中间($a=-\tfrac{k}{m}x$)。
周期不取决于振幅——用 $T=2\pi\sqrt{m/k}$(弹簧)或 $T=2\pi\sqrt{L/g}$(摆,小角度)。
速率在中间最大(全动能)而在极端为零(全势能);加速度在极端最大。
用能量($\tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2$)快速求最大速率:$v_{\max}=A\sqrt{k/m}$。
总能量 $\propto A^2$,所以把振幅加倍使能量变为四倍。

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