---
title: "Progressive wave energy per unit volume"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:progressive-wave-energy-per-volume"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/progressive-wave-energy-per-volume
site: BCT Engineering
---

# Progressive wave energy per unit volume

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2069 Asar, 5 marks
- Video: [56. Energy Density, Power and Intensity of a Harmonic Wave in a Stretched String](https://www.youtube.com/watch?v=1C0uZ9UKVEo) — Ramesh Sir's JEE PHYSICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/progressive-wave-energy-per-volume
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

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.

## 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.

## Answer

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).
```text
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).
```text
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
```text
<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.

```text
 energy |  .  .  .  .   KE and PE: same curve, 2ω frequency
 density|.' '.' '.' '.   (always >= 0)
        |-------------- average = ¼ρA²ω² each
        |______________ t
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For y = A sin(ωt - kx), the kinetic energy per unit volume is
   A. ½ρA²ω² cos²(ωt - kx)
   B. ½ρA²ω² sin²(ωt - kx)
   C. ρA²ω²
   D. ½ρAω
   Answer: A. Particle velocity is Aω cos(ωt - kx), so KE per volume = ½ρ(∂y/∂t)². (Key point: Kinetic energy density)

2. How does the PE per unit volume compare with the KE per unit volume at any point?
   A. PE is zero when KE is maximum
   B. PE is twice KE
   C. They are equal and in phase
   D. They are 90° out of phase
   Answer: C. PE = ½ρv²A²k²cos²(ωt - kx) = ½ρA²ω²cos²(ωt - kx), identical to KE. (Key point: Potential energy density)

3. The time-average of KE per unit volume over a period is
   A. ½ρA²ω²
   B. ¼ρA²ω²
   C. ρA²ω²
   D. zero
   Answer: B. The average of cos² is ½, giving ¼ρA²ω². (Key point: Average value)

4. The average total energy density of the wave is
   A. ¼ρA²ω²
   B. ρA²ω²
   C. ½ρA²ω²
   D. ρAω
   Answer: C. Average KE plus average PE = ¼ + ¼ = ½ρA²ω², which is constant. (Key point: Total energy density)

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