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Average KE is half total energy

Short answer

For SHM x = a sin(ωt+φ), KE = ½mω²a² cos²(ωt+φ). Average of cos² over a period is ½, so <KE> = ¼mω²a². Total energy E = ½mω²a², thus <KE> = E/2.

Define simple harmonic motion. Show that the average kinetic energy is half of the total energy of a particle executing simple harmonic motion.

Simple harmonic motion: periodic motion of a particle in which the acceleration (restoring force) is proportional to the displacement from the equilibrium position and is directed towards it: d²x/dt² = −ω²x.

Average KE equals half the total energy

Let x = a sin(ωt + φ). Then v = aω cos(ωt + φ).

  1. Total energy: E = ½ m ω² a² (constant).
  2. Instantaneous KE:
KE = ½ m v² = ½ m ω² a² cos²(ωt + φ)
  1. Average over one period T:
<KE> = (1/T) ∫₀ᵀ ½ m ω² a² cos²(ωt + φ) dt
  1. Use cos²θ = (1 + cos 2θ)/2:
<KE> = (1/T)(½ m ω² a²) ∫₀ᵀ ½ [1 + cos(2ωt + 2φ)] dt
     = (½ m ω² a²)(1/T)(T/2 + 0)
     = ¼ m ω² a²

(The integral of the cosine term over complete cycles is zero.) 5. Compare with E = ½ m ω² a²:

<KE> = ½ E

Similarly <PE> = ¼mω²a² = E/2. The average KE (and PE) over a period is half the total energy. This is because the average of cos² or sin² over a period is ½.

✓ Answer checked by Claude Opus 5.5 (high effort)

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