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Learn9702Paper 4 · A Level Structured QuestionsOscillationsSimple harmonic motion
Oscillations · Subtopic 1 of 3

Simple Harmonic Motion

An oscillation is simple harmonic when the acceleration is proportional to the displacement from a fixed point and always directed towards that point: a=−ω2xa=-\omega^2x. The angular frequency is ω=2πf=2πT\omega=2\pi f=\dfrac{2\pi}T. The solution is sinusoidal: x=x0sin⁡ωtx=x_0\sin\omega t (starting from equilibrium). The period doesn't depend on the amplitude. A mass on a spring and a small-angle pendulum are both approximately simple harmonic.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use displacement, amplitude, period, frequency, angular frequency and phase difference.
  • State the defining condition for SHM and use a=−ω2xa=-\omega^2x.
  • Use x=x0sin⁡ωtx=x_0\sin\omega t and x=x0cos⁡ωtx=x_0\cos\omega t.
  • Find phase differences from graphs, in radians.
Pause and think

A child on a swing is pushed higher. Does it take longer for each swing?

Section 2 of 14

Core idea

Start hereClocks and springs

A pendulum clock keeps time because its swing period doesn't depend on how wide it swings. Masses on springs share that property. Both obey one rule: the pull back to the middle is proportional to how far out you are.

The big idea

SHM: a=−ω2xa=-\omega^2x. x=x0sin⁡ωtx=x_0\sin\omega t; v=±ωx02−x2v=\pm\omega\sqrt{x_0^2-x^2}; period independent of amplitude.

1. Defining SHM

Displacement, velocity and acceleration in SHM
24681012−2−1120timexva
Velocity is a quarter-cycle ahead of displacement; acceleration is exactly opposite to displacement. Fastest at the centre; momentarily stopped at the extremes.
Displacement, velocity, acceleration
  1. x=x0sin⁡ωtx=x_0\sin\omega t
    Starting at the centre.
  2. v=ωx0cos⁡ωtv=\omega x_0\cos\omega t
    Differentiate.
  3. a=−ω2x0sin⁡ωt=−ω2xa=-\omega^2x_0\sin\omega t=-\omega^2x
    Differentiate again.
  4. v2=ω2x02(1−sin⁡2ωt)=ω2(x02−x2)v^2=\omega^2x_0^2(1-\sin^2\omega t)=\omega^2(x_0^2-x^2)
    cos⁡2=1−sin⁡2\cos^2=1-\sin^2.
vmax⁡=ωx0v_{\max}=\omega x_0 at x=0x=0; amax⁡=ω2x0a_{\max}=\omega^2x_0 at the extremes.

2. Mass–spring and pendulum

A mass on a spring oscillating
equilibrium
T=2πmkT=2\pi\sqrt{\dfrac mk}. A heavier mass or softer spring oscillates more slowly.
Mass on a spring
  1. ma=−kxma=-kx
    Hooke's law is the restoring force.
  2. a=−kmx⇒ω2=kma=-\frac kmx\Rightarrow\omega^2=\frac km
    Compare with a=−ω2xa=-\omega^2x.
  3. T=2πω=2πmkT=\frac{2\pi}\omega=2\pi\sqrt{\frac mk}
    Period.
Pendulum (small angles): T=2πlgT=2\pi\sqrt{\frac lg}.
Worked examplem = 0.50 kg, k = 50 N m⁻¹, amplitude 4.0 cm

ω\omega, TT, max speed, max acceleration, speed at 2.0 cm.

Goal
Five values.
Strategy
Formulas.
  1. ω=100=10, T=0.63 s\omega=\sqrt{100}=10,\ T=0.63\text{ s}
    k/m\sqrt{k/m}.
  2. vmax⁡=10×0.04=0.40 m s−1v_{\max}=10\times0.04=0.40\text{ m s}^{-1}
    ωx0\omega x_0.
  3. amax⁡=100×0.04=4.0 m s−2a_{\max}=100\times0.04=4.0\text{ m s}^{-2}
    ω2x0\omega^2x_0.
  4. v=100.0016−0.0004=0.35 m s−1v=10\sqrt{0.0016-0.0004}=0.35\text{ m s}^{-1}
    At x=0.02x=0.02 m.
Answer1010 rad s−1^{-1}; 0.630.63 s; 0.400.40; 4.04.0; 0.350.35.
✓ CheckSpeed at half amplitude is 32vmax⁡\frac{\sqrt3}2v_{\max} ✓.
Common misconceptionDegrees in sin ωt ωt\omega t is in radians.
Pause and think

Pendulum length quadrupled: period?

Summary

  • a=−ω2xa=-\omega^2x.
  • v=±ωx02−x2v=\pm\omega\sqrt{x_0^2-x^2}.
Key words
Isochronous
Period independent of amplitude.
Displacement, velocity and acceleration in SHM
time tvaluedisplacement xvelocity vacceleration a
Velocity leads displacement by π2\tfrac{\pi}{2}; acceleration is in antiphase with displacement, as a=−ω2xa=-\omega^2x requires.
HistoryGalileo's chandelier. The story goes that in 15831583 a young Galileo, bored in Pisa cathedral, timed a swinging chandelier against his pulse and noticed that big swings and small swings took the same time. Christiaan Huygens used this to build the first accurate pendulum clock in 16561656.
Fun factYour quartz watch. A quartz watch contains a tiny tuning fork of quartz crystal vibrating at 32 76832\,768 Hz, which is 2152^{15}. A circuit halves this frequency 1515 times to get exactly one tick per second.
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