Chapter 24 · Waves · Lesson 15.1
Simple Harmonic Motion
The oscillation behind sinusoidal waves
An ideal spring oscillator and a small-angle pendulum can be modelled with simple harmonic motion. Sinusoidal waves connect this repeating motion across space; arbitrary pulses need a more general description.
Read the relationship. Check its limits.
Lesson Overview
Simple harmonic motion is oscillation in which acceleration is proportional to displacement and always directed toward the equilibrium position.
Pull a mass on a spring down twice as far and the restoring force doubles — it oscillates with the same period, just a larger swing.
Learning Objectives
- Recognise a restoring acceleration
- Relate period, frequency and angular frequency
- Compare displacement, velocity and acceleration
- Explain energy transfer in ideal SHM
Key Concepts
Explore the oscillator
Change one input. Compare the graph and its readout before drawing a conclusion.
Watch the acceleration mirror the displacement with opposite sign — that is the SHM condition a = −ω²x.
Displacement x0 cm
Velocity-15.7 cm/s
Acceleration0 cm/s²
ω = 2π/T3.14 rad/s
Ideal, undamped, sinusoidal motion. The two-period time window rescales when T changes. Pendulum SHM is a small-angle approximation.
Key Formulae
- Defining condition
- a = −ω²xAcceleration mirrors displacement with a minus sign: the further from centre, the harder it is pulled back.ω = 2πf = 2π/T
- Displacement
- x = A sin(ωt)The position traces a sine curve in time — A is the amplitude, the largest displacement reached.Or A cos(ωt), depending on the start
- Mass–spring and pendulum
- T = 2π√(m/k), T = 2π√(L/g)Heavier masses and softer springs swing slower; longer pendulums swing slower, and amplitude appears in neither expression. For the pendulum that holds only while the swing stays small — at large angles the period creeps up.For ideal SHM the period is independent of amplitude. A simple pendulum behaves approximately as SHM for small angular displacements
Use SI units in calculations: metres, seconds, hertz and radians. The lab labels any centimetre, millimetre or nanometre inputs explicitly.
Practice Questions
Build the physical model.
Motion is simple harmonic when the restoring force — and so the acceleration — is proportional to the displacement and points back toward equilibrium. That single condition produces sinusoidal motion whose period, for ideal simple harmonic motion, does not depend on amplitude.
Restoring force
At equilibrium the net force is zero. Displace the object and a force appears pulling it back — overshooting equilibrium keeps the cycle going.
Energy exchange
Kinetic energy peaks at the centre; potential energy peaks at the extremes. The total stays constant in the ideal, undamped case.
Why it matters for waves
In an ideal sinusoidal mechanical wave, a component of each particle's displacement can follow SHM about equilibrium. Real water waves combine vertical and horizontal motion, and arbitrary pulses are not SHM. Electromagnetic waves need no material medium: their fields oscillate instead.
Know what the model can tell you.
Assumptions & boundaries
Ideal, undamped, sinusoidal motion. The two-period time window rescales when T changes. Pendulum SHM is a small-angle approximation.
Larger amplitude does not mean a longer period: in SHM the object travels further but proportionally faster, so the period depends on the system rather than on how hard you start it. That is exact for a mass on an ideal spring. A pendulum only approximates SHM, and once the swing is large its period does grow — about 0.7% longer at 20°.
A pendulum's length is quadrupled. What happens to its period?
Choose an answer to test the idea.
Worked Examples
State the assumptions, convert units, calculate and check the physical meaning. Check which ideal-wave assumptions apply.
EasyA pendulum completes 20 swings in 30 s. Find its period and frequency.
- T = 30 ÷ 20 = 1.5 s.
- f = 1/T ≈ 0.67 Hz.
T = 1.5 s, f ≈ 0.67 Hz
MediumA 0.50 kg mass on a spring of constant k = 20 N/m oscillates. Find the period and frequency.
- T = 2π√(m/k) = 2π√(0.50/20).
- T = 2π × 0.158 ≈ 0.99 s.
- f = 1/T ≈ 1.0 Hz.
T ≈ 0.99 s, f ≈ 1.0 Hz
HardA 0.20 kg mass on a spring oscillates with amplitude 5.0 cm and period 0.80 s. Find its maximum speed and maximum acceleration.
- ω = 2π/T = 7.85 rad/s.
- v(max) = Aω = 0.05 × 7.85 ≈ 0.39 m/s (at the centre).
- a(max) = Aω² = 0.05 × 61.7 ≈ 3.1 m/s² (at the extremes).
v(max) ≈ 0.39 m/s; a(max) ≈ 3.1 m/s²
ChallengingA 'seconds pendulum' (T = 2.000 s) is moved to a mountain where g is 0.20% smaller. How much time does the clock lose per day?
- T ∝ 1/√g: fractional period change = ½ × 0.20% = 0.10%.
- Slower tick by 1.001 factor → loses 0.10% of 86 400 s.
- ≈ 86 s — nearly a minute and a half per day.
≈ 86 s/day slow
Put the ideas to work.
Suspension systems
A spring provides a restoring force; real suspension also needs damping to settle after a disturbance.
Timekeeping
A small-amplitude pendulum provides a nearly constant period. Large swings require a more complete model.
Before you move on.
- Recognise a restoring acceleration.
- Relate period, frequency and angular frequency.
- Compare displacement, velocity and acceleration.
Quick Quiz
Five questions. Answers stay in this open page only.
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