Topic 13 is about anything that swings, bounces or vibrates: simple harmonic motion, springs and pendulums, energy exchange, damping and resonance.
a = −ω²x · x = A cos(ωt) · vmax = ωA · amax = ω²AT = 2π√(m/k) for a mass–spring · T = 2π√(l/g) for a simple pendulum
The definition, in one line: SHM occurs when the acceleration is directly proportional to the displacement from equilibrium and is always directed towards equilibrium. That minus sign in a = −ω²x is the physics — it is what makes the motion oscillate rather than run away.
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Topic 13 · defining SHM
The defining equation of SHM
a = − ω² xthe minus sign: acceleration always points BACK towards equilibrium
Starting from maximum displacement (x = A at t = 0), the solutions are:
Displacement: x = A cos(ωt)
Velocity: v = −Aω sin(ωt), so vmax = ωA — greatest at the equilibrium position, zero at the extremes.
Acceleration: a = −Aω² cos(ωt), so amax = ω²A — greatest at maximum displacement, zero at equilibrium.
Also useful: v = ±ω√(A² − x²), which gives the speed at any displacement.
Because ω = 2π/T = 2πf, the period is independent of amplitude — the defining feature of isochronous motion, and the reason pendulum clocks work.
Phase relationships: velocity is π/2 (a quarter cycle) ahead of displacement; acceleration is π (half a cycle) out of phase with displacement — exactly antiphase.
Quick check
Is it SHM?
?Which condition defines simple harmonic motion?
Topic 13 · springs & pendulums
The two standard oscillators
mass–spring: T = 2π√(m/k) · simple pendulum: T = 2π√(l/g)notice what is ABSENT from each: neither contains the amplitude
Mass–spring: the restoring force is F = −kx (Hooke's law), so ma = −kx, giving ω² = k/m. Add mass → slower. Use a stiffer spring → faster. Gravity does not appear — this system has the same period on the Moon.
Simple pendulum: the restoring force is the component of weight along the arc, −mg sin θ. For small angles (under about 10°), sin θ ≈ θ and the motion is SHM with ω² = g/l. Note the mass of the bob cancels — but g does not, so a pendulum runs slower on the Moon.
Worked example — mass on a spring
m = 0.25 kg, k = 40 N m⁻¹.
T = 2π√(0.25 ÷ 40) = 2π√(0.00625) = 2π × 0.0791 = 0.497 s
2A simple pendulum has a length of 0.80 m. Taking g = 9.81 m s⁻², calculate its period for small oscillations. Give your answer in s to 3 significant figures.
3For that same 0.25 kg mass on the 40 N m⁻¹ spring, calculate the angular frequency ω. Give your answer in rad s⁻¹ to 3 significant figures.
rad s⁻¹
Hint: ω = √(k/m) = √(40 ÷ 0.25) = √160.
Calculate
Your turn — maximum speed
4The same mass–spring system (ω = 12.6 rad s⁻¹) oscillates with an amplitude of 0.060 m. Calculate its maximum speed. Give your answer in m s⁻¹ to 3 significant figures.
m s⁻¹
Hint: v_max = ωA = 12.65 × 0.060.
Calculate
Your turn — maximum acceleration
5For the same system (ω = 12.6 rad s⁻¹, A = 0.060 m), calculate the maximum acceleration. Give your answer in m s⁻² to 3 significant figures.
Etotal = ½ k A² = ½ m ω² A²constant — provided there is no damping
Energy sloshes continuously between kinetic and potential, but the total stays constant:
At maximum displacement (x = ±A): v = 0, so KE = 0 and PE is maximum.
At equilibrium (x = 0): speed is maximum, so KE is maximum and PE = 0.
Both KE and PE complete two full cycles for every one cycle of the displacement — they depend on x² and v², and squaring doubles the frequency.
Worked example — check it two ways
k = 40 N m⁻¹, A = 0.060 m → E = ½kA² = ½ × 40 × 0.0036 = 0.072 J
Or: E = ½mvmax² = ½ × 0.25 × 0.759² = ½ × 0.25 × 0.576 = 0.072 J ✓
Calculate
Your turn — total energy
6The mass–spring system (k = 40 N m⁻¹) oscillates with amplitude 0.060 m. Calculate the total energy of the oscillation. Give your answer in mJ.
mJ
Hint: E = ½kA² = 0.5 × 40 × (0.060)² = 0.5 × 40 × 0.0036 = 0.072 J. Convert to mJ.
Quick check
Where is the energy?
?A mass on a spring oscillates in SHM. At the point of maximum displacement, which statement is correct?
Topic 13 · damping & resonance
Damping, forced vibrations and resonance
Free vibration: an object oscillating at its own natural frequency f₀, with no external driver and no energy loss.
Damping: a resistive force (air resistance, friction, a dashpot) removes energy, so the amplitude decays. Crucially, the frequency changes very little.
Light damping: amplitude decays slowly, over many oscillations (exponential envelope).
Critical damping: returns to equilibrium in the shortest possible time without oscillating — car suspension, analogue meter needles, self-closing doors.
Heavy (over)damping: returns to equilibrium without oscillating, but more slowly than critical.
Forced vibration: an external periodic driver at frequency f. When f = f₀, energy is transferred most efficiently and the amplitude peaks — this is resonance.
What damping does to resonance: increasing the damping lowers the peak amplitude, broadens the peak, and shifts the peak to a slightly lower frequency. Examples of resonance: pushing a swing in time, a wine glass shattering, a radio tuned to a station, MRI scanners, and the Millennium Bridge wobble (later fixed with dampers).
Quick check
Critical damping
?A car suspension system is designed to be critically damped. What does that achieve?
Quick check
Resonance
?A system with natural frequency f₀ is driven by an external periodic force. Increasing the damping has what effect on the resonance curve?
Sort it
What changes the period?
Tap a change, then tap its effect. Think hard — some of these are traps.
📈 Increases the period of a mass–spring
⏸️ No effect on the period
🕰️ Increases the period of a pendulum
Match it
Match the quantity to its equation
Tap an item on the left, then its partner on the right.
Quantity
Equation
Recap
The big ideas to know
SHM defined: a = −ω²x — acceleration ∝ displacement, always towards equilibrium
Equations: x = A cos(ωt) · v = ±ω√(A² − x²) · v_max = ωA · a_max = ω²A · ω = 2π/T
Mass–spring: T = 2π√(m/k) — no g · Pendulum: T = 2π√(l/g) — no mass, small angles only
Period is independent of amplitude (isochronous)
Energy: E = ½kA² total; KE max at equilibrium, PE max at the extremes; total constant if undamped
Damping: light → slow decay · critical → fastest return with no oscillation · heavy → slow, no oscillation
Resonance: maximum amplitude when the driving frequency equals f₀; damping lowers and broadens the peak
That is Edexcel Topic 13 — and with it, all thirteen topics of the 9PH0 specification. Press Finish to see your score.
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