An oscillation is simple harmonic when the acceleration is proportional to the displacement from a fixed point and always directed towards that point: a=−ω2x. The angular frequency is ω=2πf=T2π. The solution is sinusoidal: x=x0sinωt (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=−ω2x.
Use x=x0sinωt and x=x0cosωt.
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=−ω2x. x=x0sinωt; v=±ωx02−x2; period independent of amplitude.
1. Defining SHM
Displacement, velocity and acceleration in SHMVelocity 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
x=x0sinωt
Starting at the centre.
v=ωx0cosωt
Differentiate.
a=−ω2x0sinωt=−ω2x
Differentiate again.
v2=ω2x02(1−sin2ωt)=ω2(x02−x2)
cos2=1−sin2.
vmax=ωx0 at x=0; amax=ω2x0 at the extremes.
2. Mass–spring and pendulum
A mass on a spring oscillatingT=2πkm. A heavier mass or softer spring oscillates more slowly.
Mass on a spring
ma=−kx
Hooke's law is the restoring force.
a=−mkx⇒ω2=mk
Compare with a=−ω2x.
T=ω2π=2πkm
Period.
Pendulum (small angles): T=2πgl.
Worked examplem = 0.50 kg, k = 50 N m⁻¹, amplitude 4.0 cm
ω, T, max speed, max acceleration, speed at 2.0 cm.
Goal
Five values.
Strategy
Formulas.
1
ω=100=10,T=0.63 s
k/m.
2
vmax=10×0.04=0.40 m s−1
ωx0.
3
amax=100×0.04=4.0 m s−2
ω2x0.
4
v=100.0016−0.0004=0.35 m s−1
At x=0.02 m.
Answer10 rad s−1; 0.63 s; 0.40; 4.0; 0.35.
✓ CheckSpeed at half amplitude is 23vmax ✓.
⚠️ Common misconceptionDegrees in sin ωtωt is in radians.
Pause and think
Pendulum length quadrupled: period?
Summary
a=−ω2x.
v=±ωx02−x2.
Key words
Isochronous
Period independent of amplitude.
Displacement, velocity and acceleration in SHMVelocity leads displacement by 2π; acceleration is in antiphase with displacement, as a=−ω2x requires.
📜 HistoryGalileo's chandelier.The story goes that in 1583 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 1656.
🤯 Fun factYour quartz watch.A quartz watch contains a tiny tuning fork of quartz crystal vibrating at 32768 Hz, which is 215. A circuit halves this frequency 15 times to get exactly one tick per second.
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