跳到主要内容

振动

AP 物理 C:力学 · 第 7 主题

训练
讲义 词汇表
7.1

简谐运动的定义

大纲
Learning ObjectiveEssential 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.
    • Derived 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.

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)是周期运动(periodic motion)的一个特殊情况。它每当两件事为真时出现:有一个合力为零的平衡位置(equilibrium)位置,而移开物体产生一个回复力(restoring force)——一个指回向平衡的力——它的大小与位移成比例:

$$F=-k\,\Delta x.$$

牛顿第二定律然后给出定义性的微分方程(differential equation):

$$\frac{d^2x}{dt^2}=-\frac{k}{m}\,x=-\omega^2 x,\qquad \omega=\sqrt{\frac{k}{m}}.$$

它的解是正弦的,$x(t)=A\cos(\omega t+\phi)$。你不需要证明这个解——你需要认出这个方程:任何运动遵循"$\ddot{x}=-\omega^2x$"的系统都是一个简谐振子,无论它由什么制成。(一个挂在一个竖直弹簧上的质量以相同的方式工作:重力只移动平衡点;绕它的振动不变。)

在 SHM 里加速度总是指回向平衡,与位移相反
在 SHM 里加速度总是指回向平衡,与位移相反
词汇表 训练
英文 中文 拼音
Simple harmonic motion 简谐运动 jiǎn xié yùn dòng
periodic motion 周期运动 zhōu qī yùn dòng
equilibrium 平衡位置 píng héng wèi zhì
restoring force 回复力 huí fù lì
differential equation 微分方程 wēi fēn fāng chéng
period 周期 zhōu qī
练习卷
7.2

简谐运动的频率与周期

大纲
Learning ObjectiveEssential 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 angular frequency, $\omega$, of the object's motion by the following equation:
    • Equation: $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$
    • 7.2.A.1.i The period of an object–ideal-spring oscillator is given by the equation
      • 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
      • Equation: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$

来源:美国大学理事会 AP 课程与考试说明

角频率(angular frequency)$\omega$ 设定周期(period)和频率(frequency):

$$T=\frac{2\pi}{\omega}=\frac{1}{f}.$$

对于两个标准系统:

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

两个经典的概念陷阱:SHM 的周期从不取决于振幅(amplitude),而每个系统忽略一个明显的变量——摆的周期不涉及它的质量,而弹簧的周期不涉及 $g$(一个弹簧-木块振子在轨道里保持完美的时间;一个摆钟不)。

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
angular frequency 角频率 jiǎo pín lǜ
frequency 频率 pín lǜ
amplitude 振幅 zhèn fú
7.3

简谐运动的表示与分析

大纲
Learning ObjectiveEssential 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 The position as a function of time for an object exhibiting SHM is a solution of the second-order differential equation derived from the application of Newton's second law.
    • Derived equation: $\dfrac{d^2 x}{dt^2} = -\omega^2 x$
  • 7.3.A.3 Characteristics of SHM, such as velocity and acceleration, can be determined by or derived from the equation $x = A\cos(\omega t + \phi).$
    • 7.3.A.3.i The acceleration of an object exhibiting SHM is related to the object's angular frequency and position.
      • Derived equation: $a = -\omega^2 x$
    • 7.3.A.3.ii It can be shown that the maximum velocity and acceleration of an object exhibiting SHM are related to the angular frequency of the object's motion.
      • Derived equations: $v_{\max} = A\omega$
      • $a_{\max} = A\omega^2$
  • 7.3.A.4 In the presence of a sinusoidal external force, a system may exhibit resonance.
    • 7.3.A.4.i Resonance occurs when an external force is exerted at the natural frequency of an oscillating system.
    • 7.3.A.4.ii Resonance increases the amplitude of oscillating motion.
    • 7.3.A.4.iii The natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position.
  • 7.3.A.5 Changing the amplitude of a system exhibiting SHM will not change its period.
  • 7.3.A.6 Properties of SHM can be determined and analyzed using graphical representations.

Boundary statement: AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct.

来源:美国大学理事会 AP 课程与考试说明

$x(t)=A\cos(\omega t+\phi)$ 求导两次:

$$v=-A\omega\sin(\omega t+\phi),\qquad a=-A\omega^2\cos(\omega t+\phi)=-\omega^2x.$$

所以在摆动的两端($x=\pm A$)速率是零而加速度最大;通过平衡($x=0$)加速度是零而速率最大,$v_{\max}=A\omega$。振幅 $A$相位常数(phase constant)$\phi$ 来自初始条件:物体从哪里开始以及它移动多快。消去 $t$(或用能量,下面)给出速率作为位置的一个函数:

$$v=\pm\,\omega\sqrt{A^2-x^2}.$$
位移在简谐运动里随时间正弦地变化
位移在简谐运动里随时间正弦地变化

Worked example. 一个 $k=200\ \text{N/m}$ 弹簧上的 $0.50\ \text{kg}$ 质量:$\omega=\sqrt{k/m}=20\ \text{rad/s}$$T=2\pi/\omega=0.31\ \text{s}$。以 $A=0.10\ \text{m}$:在平衡 $v_{\max}=A\omega=2.0\ \text{m/s}$,而在 $x=0.050\ \text{m}$ 速率是 $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$

词汇表 训练
英文 中文 拼音
phase constant 相位常数 xiàng wèi cháng shù
7.4

简谐振子的能量

大纲
Learning ObjectiveEssential 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.
    • Relevant 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

振子的总能量是和 $E=U+K$,而没有摩擦它恒定——能量只是在弹簧的势能和质量的动能之间来回交换:

$$E=\tfrac12kA^2=\tfrac12kx^2+\tfrac12mv^2.$$

在两端全是势能(那里 $K=0$),在平衡全是动能(那里 $U$ 最小)。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
动能和势能在一个循环上交换而总能量保持恒定

Worked example. 能量在哪里均匀分割?令 $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$,所以 $x=A/\sqrt2\approx0.71A$ ——比到中间更接近端。对于上面的系统($A=0.10\ \text{m}$):$x=0.071\ \text{m}$

不受打扰时,一个系统以它自己的固有频率(natural frequency)振荡(由 $k$$m$ 决定:$f_0=\frac{1}{2\pi}\sqrt{k/m}$)。用一个周期性的力在那个相同频率驱动它,振幅急剧增大——共振(resonance)。每一次恰好定时的推动加入能量的速度快于阻尼移除的速度,所以当驱动频率匹配 $f_0$ 时驱动振幅尖锐地达到峰值(荡秋千按节奏推、桥被风摇晃)。

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.

词汇表 训练
英文 中文 拼音
natural frequency 固有频率 gù yǒu pín lǜ
resonance 共振 gòng zhèn
7.5

单摆与物理摆

大纲
Learning ObjectiveEssential Knowledge

7.5.A
Describe the properties of a physical pendulum.

  • 7.5.A.1 A physical pendulum is a rigid body that undergoes oscillation about a fixed axis.
  • 7.5.A.2 For small amplitudes of motion, the period of a physical pendulum is derived from the application of Newton's second law in rotational form.
    • Relevant equation: $T_{\text{phys}} = 2\pi\sqrt{\dfrac{I}{mgd}}$
    • 7.5.A.2.i When displaced from equilibrium, the gravitational force exerted on a physical pendulum's center of mass provides a restoring torque.
      • Derived equation: $\tau = -mgd\sin\theta$
    • 7.5.A.2.ii For small amplitudes of motion, the small-angle approximation can be applied to the restoring torque.
      • Derived equation: $\sin\theta \approx \theta$
      • $\tau = -mgd\theta = I\alpha$
    • 7.5.A.2.iii The small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM:
      • Equation: $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$
  • 7.5.A.3 A simple pendulum is a special case of physical pendulums in which the hanging object can be modeled as a point mass at a distance, $l$, from the pivot point.
    • Relevant equation: $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
  • 7.5.A.4 A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system. For example, a horizontal disk that is suspended from a wire attached to its center of mass may undergo rotational oscillations about the wire in the horizontal plane.
    • Derived equation: $I\alpha = -k\Delta\theta$

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

一个物理摆(physical pendulum)是任何绕一个固定枢轴摆动的刚体。把它移开一个角 $\theta$,而重力,作用在距枢轴一个距离 $d$ 的质心处,供应一个回复力矩

$$\tau=-mgd\sin\theta.$$

对于小角度,应用小角度近似(small-angle approximation)$\sin\theta\approx\theta$ 和旋转形式的牛顿第二定律($\tau=I\alpha$):

$$\frac{d^2\theta}{dt^2}=-\frac{mgd}{I}\,\theta=-\omega^2\theta\quad\Rightarrow\quad T_{\text{phys}}=2\pi\sqrt{\frac{I}{mgd}}.$$

这是与之前相同的"$\ddot\theta=-\omega^2\theta$"模式——认出它就是这个推导。一个单摆(simple pendulum)是一个轻绳上的点质量的特殊情况:$I=m\ell^2$$d=\ell$ 给出 $T=2\pi\sqrt{\ell/g}$

作用在质心处的重力提供一个物理摆的回复力矩
作用在质心处的重力提供一个物理摆的回复力矩

Worked example. 一根均匀的杆(质量 $M$、长度 $L$)从一端摆动:$I=\tfrac13ML^2$$d=\tfrac{L}{2}$,所以

$$T=2\pi\sqrt{\frac{\tfrac13ML^2}{Mg\,\tfrac{L}{2}}}=2\pi\sqrt{\frac{2L}{3g}}$$

——比一个长度 $L$ 的单摆更短,因为杆的质量坐得离枢轴更近。

一个扭摆(torsion pendulum)——一个从一根扭转的金属丝悬挂的圆盘——又是 SHM 一次:金属丝的回复力矩与扭转角成比例,$I\alpha=-\kappa\,\Delta\theta$,给出 $T=2\pi\sqrt{I/\kappa}$

Exam skill. 每个摆 FRQ 想要相同的三个步骤:写绕枢轴的回复力矩、应用小角度近似,并把结果匹配到 $\ddot\theta=-\omega^2\theta$ 以读出 $\omega$。明确地陈述小角度步骤——它是一个评分点,而它是为什么大振幅摆动是简谐的。

词汇表 训练
英文 中文 拼音
physical pendulum 物理摆 wù lǐ bǎi
small-angle approximation 小角度近似 xiǎo jiǎo dù jìn sì
simple pendulum 单摆 dān bǎi
torsion pendulum 扭摆 niǔ bǎi
练习卷
7.5

考试技巧

  • 从一个线性回复力 $F=-kx$ 辨别 SHM,它给出 $\tfrac{d^2x}{dt^2}=-\omega^2 x$,$\omega=\sqrt{k/m}$
  • 写解 $x=A\cos(\omega t+\phi)$ 并通过求导得到 $v,a$;周期 $T=\tfrac{2\pi}{\omega}$ 与振幅无关。
  • 能量在 $\tfrac12 kx^2$$\tfrac12 mv^2$ 之间交换,总量 $\tfrac12 kA^2$
  • 对于一个摆,用小角度近似 $\sin\theta\approx\theta$ 达到 SHM。
  • 把相位匹配到起点:从最大位移静止释放用余弦。

本主题的互动课程

逐步学习,并即时检测练习。

AP 物理 C:力学历年真题

AP 物理 C:力学的更多主题

登录或创建账号

IGCSE, A-Level & AP