- 2069 Asar · 5 marks
Progressive wave energy per unit volume
Short answer
For y = A sin(ωt−kx), kinetic and potential energy per unit volume are ½ρA²ω² cos²(ωt−kx). Their time averages are constant, each ¼ρA²ω².
In the progressive wave, show that the potential energy and kinetic energy of every particle will change with time but the average K.E. per unit volume and P.E. per unit volume remains constant.
Let a plane progressive wave in a medium of density ρ be y = A sin(ωt − kx), with speed v = ω/k.
Kinetic energy per unit volume
Particle velocity ∂y/∂t = Aω cos(ωt − kx).
KE per volume = ½ ρ (∂y/∂t)² = ½ ρ A² ω² cos²(ωt − kx)
Potential energy per unit volume
The energy of deformation per unit volume is ½ (elastic constant) × (strain)², where the strain is ∂y/∂x = −Ak cos(ωt − kx) and the elastic constant is ρv² (for a string, the tension is T = μv²).
PE per volume = ½ ρ v² (∂y/∂x)² = ½ ρ v² A² k² cos²(ωt − kx)
= ½ ρ A² ω² cos²(ωt − kx) (since vk = ω)
So at every point KE and PE are equal and in phase.
Variation and averages
- Both depend on (ωt − kx), so they change with time (and position) between zero and ½ρA²ω²; the energy of a given particle is not constant. Particles at the mean position have maximum KE and PE; at the extremes both are zero.
- Averaging over one period, <cos²(ωt − kx)> = ½, so
<KE per volume> = ¼ ρ A² ω² , <PE per volume> = ¼ ρ A² ω²
These are constants, independent of time and position.
- The total average energy density is ½ ρ A² ω², a constant, and the two averages are equal.
energy | . . . . KE and PE: same curve, 2ω frequency
density|.' '.' '.' '. (always >= 0)
|-------------- average = ¼ρA²ω² each
|______________ t
✓ Answer checked by Claude Opus 5.5 (high effort)
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