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Energy in Simple Harmonic Motion

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Energy sloshes between two forms

In an ideal oscillator, energy trades continuously between elastic potential energy (½kx²) and kinetic energy (½mv²), while the total stays constant. At the extremes all the energy is potential (the mass is momentarily still); at equilibrium all of it is kinetic (the spring is relaxed). Everywhere in between it is a mix — but the sum never changes in the absence of friction.

Total energy is set by the amplitude

Because the mass stops instantaneously at maximum displacement x = A, all the energy there is elastic potential: E_total = ½kA². This means the total energy depends only on the spring constant and the amplitude. Since that same energy becomes entirely kinetic at equilibrium, ½kA² = ½mv_max², which lets you solve directly for the maximum speed.

Energy in SHM
E_total = ½kA² = ½mv_max²
All potential at the extremes, all kinetic at equilibrium. Setting the two equal gives v_max = A√(k/m).
Worked example

A 0.5 kg block on a spring (k = 200 N/m) oscillates with amplitude 0.1 m on a frictionless surface. Find the total energy and the maximum speed.

  1. 1.Total energy is all potential at the amplitude: E = ½kA² = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = 1 J.
  2. 2.At equilibrium all of this is kinetic: ½mv_max² = 1 J.
  3. 3.So ½ × 0.5 × v_max² = 1 → 0.25 × v_max² = 1 → v_max² = 4.
  4. 4.Take the square root: v_max = 2 m/s.
Answer: E_total = 1 J and v_max = 2 m/s
Tip

The fastest route to maximum speed is energy conservation: set ½kA² (all PE at the extreme) equal to ½mv_max² (all KE at equilibrium) and solve. No need to track the motion in between.

Checkpoint

A mass oscillates on a spring with amplitude A. At what point is all of the energy stored as elastic potential energy?

Checkpoint

A spring (k = 200 N/m) is compressed 0.1 m and released, launching a 0.5 kg block along a frictionless surface. What is the block’s maximum speed?

On the exam

The total energy of an oscillator scales with the square of the amplitude (E = ½kA²). Double the amplitude and you quadruple the energy — and the maximum speed only doubles, since v_max scales linearly with A.

Answer the 2 checkpoints as you read.

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