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
The total stays the same: 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.
Energy against time
A mass on a vertical spring: Hanging on a spring, the mass has a third store: gravitational PE. On the way up, kinetic energy and the spring's elastic PE turn into gravitational PE; on the way down it reverses. The total of all three stays the same.
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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.
Total energy = PE at the amplitude: 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)
Where it comes from: It is the data booklet's elastic potential energy EH = ½kΔx² (topic A.3) worked out at the biggest stretch, Δx = 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. At HL the booklet also gives the total energy as ET = ½mω²x₀² (micro 3.1.5). 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)
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.
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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 answer broken into clear points, with guidance on how each point earns marks.