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Oscillations

AP Physics 1 · 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.
    • 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.

Source: College Board AP Course and Exam Description

Simple harmonic motion

Simple harmonic motion 简谐运动 (SHM) is a back-and-forth oscillation 振动 caused by a restoring force 回复力 that is proportional to the displacement from equilibrium and always points back toward it: $F=-kx$. A mass on a spring and (for small angles) a pendulum are the standard examples. Because the force grows with displacement, the motion is smooth and repeating, tracing a sine curve in time.

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

The test for SHM is exactly this: acceleration proportional to displacement and opposite in direction, $a=-\dfrac{k}{m}x$. A pendulum only obeys it for small swings, where $\sin\theta\approx\theta$; large swings are not quite SHM.

Vocabulary Train
English Chinese Pinyin
Simple harmonic motion 简谐运动 jiǎn xié yùn dòng
oscillation 振动 zhèn dòng
restoring force 回复力 huí fù lì
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 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 $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$.

Source: College Board AP Course and Exam Description

  • The period 周期 $T$ is the time for one full cycle.
  • The frequency 频率 $f=\dfrac{1}{T}$ is cycles per second (hertz).

For SHM these depend only on the system, not on the amplitude:

$$T_{\text{spring}}=2\pi\sqrt{\frac{m}{k}},\qquad T_{\text{pendulum}}=2\pi\sqrt{\frac{L}{g}}.$$
So a stiffer spring or smaller mass oscillates faster; a longer pendulum swings slower.

Worked example. A $0.25\ \text{kg}$ mass hangs on a spring of stiffness $k=100\ \text{N/m}$. Its period is

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

Worked example. A pendulum clock ticks with a period of exactly $2.0\ \text{s}$. How long is it? Rearranging $T=2\pi\sqrt{L/g}$,

$$L=g\left(\frac{T}{2\pi}\right)^2=9.8\times\left(\frac{2.0}{2\pi}\right)^2=0.99\ \text{m}.$$
Notice the amplitude never entered – a wide or narrow swing keeps the same time, which is what makes pendulums good clocks.

A set of tuning forks of different sizes Tuning forks: a longer, heavier fork vibrates more slowly, giving a lower frequency

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
period 周期 zhōu qī
frequency 频率 pín lǜ
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 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.

Source: College Board AP Course and Exam Description

The displacement varies sinusoidally: $x(t)=A\cos(\omega t)$ (or sine), where $A$ is the amplitude 振幅 (maximum displacement) and $\omega=2\pi f$ is the angular frequency 角频率. Reading the motion:

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

  • At the extremes ($x=\pm A$): displacement and restoring force are maximum, so acceleration is maximum, but velocity is zero.
  • At equilibrium 平衡位置 ($x=0$): force and acceleration are zero, but speed is maximum.

Velocity and acceleration are also sinusoidal, shifted in phase 相位 from the displacement – velocity leads displacement by a quarter cycle, and acceleration is exactly opposite to displacement.

Displacement, velocity, and acceleration in SHM, each a quarter-cycle apart Displacement, velocity, and acceleration in SHM, each a quarter-cycle apart

Read the figure as a story: where $x$ is largest (the turning point), $v$ has just fallen to zero and $a$ is at its most negative, hauling the mass back; a quarter-cycle later the mass races through the middle at top speed with zero acceleration.

Vocabulary Train
English Chinese Pinyin
amplitude 振幅 zhèn fú
angular frequency 角频率 jiǎo pín lǜ
phase 相位 xiàng wèi
equilibrium 平衡位置 píng héng wèi zhì
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.
    • 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

Energy conservation: KE ⇄ PE

Energy sloshes between kinetic and potential while the total stays constant (no friction):

$$E=\tfrac{1}{2}kA^2 = \tfrac{1}{2}kx^2+\tfrac{1}{2}mv^2.$$
At the extremes it is all potential; at equilibrium it is all kinetic (maximum speed). 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. A $0.50\ \text{kg}$ mass on a spring of stiffness $k=200\ \text{N/m}$ oscillates with amplitude $A=0.10\ \text{m}$. Find its maximum speed. All the energy is kinetic at the equilibrium point, so $\tfrac12 kA^2=\tfrac12 mv_{\max}^2$:

$$v_{\max}=A\sqrt{\frac{k}{m}}=0.10\times\sqrt{\frac{200}{0.50}}=0.10\times20=2.0\ \text{m/s}.$$
The energy method is faster than tracking the sine functions when you only need the greatest speed.

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.

Exercise sheet
7.4

Exam tips

  • Test for SHM: the acceleration must be proportional to the displacement and directed back toward the middle ($a=-\tfrac{k}{m}x$).
  • The period does not depend on amplitude — use $T=2\pi\sqrt{m/k}$ (spring) or $T=2\pi\sqrt{L/g}$ (pendulum, small angles).
  • Speed is maximum at the middle (all kinetic) and zero at the extremes (all potential); acceleration is largest at the extremes.
  • Use energy ($\tfrac12 kA^2 = \tfrac12 kx^2 + \tfrac12 mv^2$) to find the maximum speed quickly: $v_{\max}=A\sqrt{k/m}$.
  • Total energy $\propto A^2$, so doubling the amplitude quadruples the energy.

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