The big idea: Pull a mass on a spring aside and let go: at the ends it stops dead for an instant, through the middle it blurs past at full speed. Its energy is swapping between two forms.
Kinetic energy (KE) = the energy of moving; potential energy (PE) = stored energy — in a spring, the energy stored when it is stretched or squashed.
While they swap, the total energy stays the same (no friction).
New words — amplitude & equilibrium: The equilibrium position is the middle of the swing — the resting point it passes through fastest.
The amplitude (A) is the greatest displacement from that middle — the turning point, where it stops for an instant before coming back.
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Where each one peaks: KE is biggest at the centre (moving fastest) · PE is biggest at the ends (the amplitude, momentarily still). At every point in between, KE + PE add up to the same total.
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At the very ends of the swing the object stops for an instant, so its KE is zero — at that moment all the energy is potential. So the total energy equals the PE stored at the amplitude. For a mass on a spring of stiffness k stretched to amplitude A, that stored energy is ½kA².
- total energy of the oscillation (J) — stays constant
- spring constant / stiffness (N m⁻¹)
- amplitude — the greatest displacement from the centre (m)
Not in the data booklet — so remember it: Etotal = ½kA² is not given — you have to know it.
Memory aid: it is just the elastic PE of a spring (½kx²) worked out at the biggest stretch, x = A — the point where the object is still and all the energy is stored.
What the booklet does give you: The data booklet (C.1) gives the SHM relations the energy is built from — the defining rule a = −ω²x and the period link T = 1/f = 2π/ω. From these the total energy can also be written ET = ½mω²A². Tap the formula to see its booklet badge.
- acceleration (m s⁻²)
- angular frequency (rad s⁻¹) — how fast it oscillates
- displacement from the centre (m)
- period — time for one full oscillation (s)
- frequency — oscillations per second (Hz)
A 0.50 kg mass on a spring of stiffness k = 200 N m⁻¹ oscillates with an amplitude of 0.10 m. Find the total energy of the oscillation.
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How this is tested — SHM energy questions test the swap and the constant total:
Paper 1A
- A quick MCQ — where the KE or PE is greatest, or what the energy graph looks like.
Paper 2
- Link the energies — e.g. find the maximum speed by setting the maximum KE equal to the total energy.
The classic trap: The maximum speed is at the centre (where KE peaks), not at the ends — at the ends the object is momentarily at rest.
The key move: At the centre the PE is zero, so the maximum KE equals the total energy:
½mvmax² = Etotal = ½kA². Cancel the ½ and solve for vmax.
Animated graph
Watch the graph build step by step in study mode.
The same 0.50 kg mass on the k = 200 N m⁻¹ spring (amplitude 0.10 m, total energy 1.0 J) oscillates back and forth. Find its maximum speed.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.