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AP Physics C: Mechanics · Topic 7

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7.1

Defining Simple Harmonic Motion
Syllabus
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.

Source: College Board AP Course and Exam Description

Simple harmonic motion

Simple harmonic motion 简谐运动 (SHM) is a special case of periodic motion 周期运动. It appears whenever two things are true: there is an equilibrium 平衡位置 position where the net force is zero, and displacing the object produces a restoring force 回复力 – a force pointing back toward equilibrium – whose magnitude is proportional to the displacement:

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

Newton's second law then gives the defining differential equation 微分方程:

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

Its solution is sinusoidal, $x(t)=A\cos(\omega t+\phi)$. You do not need to prove this solution – you need to recognise the equation: any system whose motion obeys "$\ddot{x}=-\omega^2x$" is a simple harmonic oscillator, whatever it is made of. (A mass hanging on a vertical spring works the same way: gravity only shifts the equilibrium point; the oscillation about it is unchanged.)

In SHM the acceleration always points back towards equilibrium, opposite the displacement In SHM the acceleration always points back towards equilibrium, opposite the displacement

Vocabulary Train
English Chinese Pinyin
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
Exercise sheet
7.2

Frequency and Period of SHM
Syllabus
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: $T_p = 2\pi\sqrt{\dfrac{l}{g}}.$

Source: College Board AP Course and Exam Description

The angular frequency 角频率 $\omega$ sets the period 周期 and frequency 频率:

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

For the two standard systems:

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

Two classic conceptual traps: the period of SHM never depends on the amplitude 振幅, and each system ignores one obvious variable – the pendulum's period does not involve its mass, and the spring's period does not involve $g$ (a spring–block oscillator keeps perfect time in orbit; a pendulum clock does not).

Explore

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.

Vocabulary Train
English Chinese Pinyin
angular frequency 角频率 jiǎo pín lǜ
period 周期 zhōu qī
frequency 频率 pín lǜ
amplitude 振幅 zhèn fú
Exercise sheet
7.3

Representing and Analyzing SHM
Syllabus
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.

Source: College Board AP Course and Exam Description

Differentiate $x(t)=A\cos(\omega t+\phi)$ twice:

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

So at the ends of the swing ($x=\pm A$) the speed is zero and the acceleration is largest; passing through equilibrium ($x=0$) the acceleration is zero and the speed is largest, $v_{\max}=A\omega$. The amplitude $A$ and the phase constant 相位常数 $\phi$ come from the initial conditions: where the object starts and how fast it is moving. Eliminating $t$ (or using energy, below) gives speed as a function of position:

$$v=\pm\,\omega\sqrt{A^2-x^2}.$$

Displacement varies sinusoidally with time in simple harmonic motion Displacement varies sinusoidally with time in simple harmonic motion

Worked example. A $0.50\ \text{kg}$ mass on a $k=200\ \text{N/m}$ spring: $\omega=\sqrt{k/m}=20\ \text{rad/s}$, $T=2\pi/\omega=0.31\ \text{s}$. With $A=0.10\ \text{m}$: $v_{\max}=A\omega=2.0\ \text{m/s}$ at equilibrium, and at $x=0.050\ \text{m}$ the speed is $v=\omega\sqrt{A^2-x^2}=20\sqrt{0.10^2-0.050^2}=1.7\ \text{m/s}$.

Vocabulary Train
English Chinese Pinyin
phase constant 相位常数 xiàng wèi cháng shù
Exercise sheet
7.4

Energy of Simple Harmonic Oscillators
Syllabus
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: $E_{\text{total}} = \dfrac{1}{2}kA^2$

Source: College Board AP Course and Exam Description

The total energy of the oscillator is the sum $E=U+K$, and with no friction it is constant – energy just trades back and forth between the spring's potential energy and the mass's kinetic energy:

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

All potential at the ends (where $K=0$), all kinetic at equilibrium (where $U$ is minimum). Because $E\propto A^2$, doubling the amplitude quadruples the energy.

Kinetic and potential energy swap over a cycle while the total energy stays constant Kinetic and potential energy swap over a cycle while the total energy stays constant

Worked example. Where is the energy split evenly? Set $\tfrac12kx^2=\tfrac12E=\tfrac14kA^2$, so $x=A/\sqrt2\approx0.71A$ – much closer to the end than to the middle. For the system above ($A=0.10\ \text{m}$): $x=0.071\ \text{m}$.

Left alone, a system oscillates at its own natural frequency 固有频率 (set by $k$ and $m$: $f_0=\frac{1}{2\pi}\sqrt{k/m}$). Drive it with a periodic force at that same frequency and the amplitude grows dramatically – resonance 共振. Each well-timed push adds energy faster than damping removes it, so the driven amplitude peaks sharply when the driving frequency matches $f_0$ (a swing pumped in time, a bridge shaken by wind).

Explore

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.

Vocabulary Train
English Chinese Pinyin
natural frequency 固有频率 gù yǒu pín lǜ
resonance 共振 gòng zhèn
Exercise sheet
7.5

Simple and Physical Pendulums
Syllabus
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: $I\alpha = -k\Delta\theta$

Source: College Board AP Course and Exam Description

Energy conservation: KE ⇄ PE

A physical pendulum 物理摆 is any rigid body swinging about a fixed pivot. Displace it by an angle $\theta$ and gravity, acting at the center of mass a distance $d$ from the pivot, supplies a restoring torque

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

For small angles, apply the small-angle approximation 小角度近似 $\sin\theta\approx\theta$ and Newton's second law in rotational form ($\tau=I\alpha$):

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

This is the same "$\ddot\theta=-\omega^2\theta$" pattern as before – recognising it is the derivation. A simple pendulum 单摆 is the special case of a point mass on a light string: $I=m\ell^2$ and $d=\ell$ give $T=2\pi\sqrt{\ell/g}$.

Gravity acting at the center of mass provides a physical pendulum's restoring torque Gravity acting at the center of mass provides a physical pendulum's restoring torque

Worked example. A uniform rod (mass $M$, length $L$) swings from one end: $I=\tfrac13ML^2$, $d=\tfrac{L}{2}$, so

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

– shorter than a simple pendulum of length $L$, because the rod's mass sits nearer the pivot.

A torsion pendulum 扭摆 – a disk hanging from a wire that twists – is SHM one more time: the wire's restoring torque is proportional to the twist angle, $I\alpha=-\kappa\,\Delta\theta$, giving $T=2\pi\sqrt{I/\kappa}$.

Exam skill. Every pendulum FRQ wants the same three steps: write the restoring torque about the pivot, apply the small-angle approximation, and match the result to $\ddot\theta=-\omega^2\theta$ to read off $\omega$. State the small-angle step explicitly – it is a scored point, and it is why large-amplitude swings are not simple harmonic.

A large Foucault pendulum swinging over a marked floor A Foucault pendulum: a long, heavy pendulum whose slow swing reveals the Earth turning beneath it

Vocabulary Train
English Chinese Pinyin
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
Exercise sheet
7.5

Exam tips
  • Identify SHM from a linear restoring force $F=-kx$, which gives $\tfrac{d^2x}{dt^2}=-\omega^2 x$ with $\omega=\sqrt{k/m}$.
  • Write the solution $x=A\cos(\omega t+\phi)$ and get $v,a$ by differentiating; period $T=\tfrac{2\pi}{\omega}$ is amplitude-independent.
  • Energy trades between $\tfrac12 kx^2$ and $\tfrac12 mv^2$, with total $\tfrac12 kA^2$.
  • For a pendulum, use the small-angle approximation $\sin\theta\approx\theta$ to reach SHM.
  • Match phase to the start: released from rest at maximum displacement uses cosine.

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