---
title: "Average KE is half total energy"
subject: "Engineering Physics"
chapter: 1
key: "semester_1/engineering_physics:shm-average-ke-half-total"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/shm-average-ke-half-total
site: BCT Engineering
---

# Average KE is half total energy

Engineering Physics · Chapter 1: Oscillation

- Asked in: 2070 Chaitra, 5 marks
- Video: [kinetic energy , potential energy ,total energy and their time average values](https://www.youtube.com/watch?v=iukYOiol-Qc) — bindas physics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/shm-average-ke-half-total
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-1

## Question

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

## 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.

## Answer

**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:
```text
KE = ½ m v² = ½ m ω² a² cos²(ωt + φ)
```
3. Average over one period T:
```text
<KE> = (1/T) ∫₀ᵀ ½ m ω² a² cos²(ωt + φ) dt
```
4. Use cos²θ = (1 + cos 2θ)/2:
```text
<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:
```text
<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.

## Check yourself

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

1. KE in SHM is?
   A. (1/2)m w^2 a^2 sin^2
   B. m w a
   C. (1/2)m w a^2
   D. (1/2)m w^2 a^2 cos^2(wt + phi)
   Answer: D. v = aw cos(...). (Key point: Instantaneous KE)

2. Average of cos^2 over a period?
   A. 1/2
   B. 1
   C. 1/4
   D. 0
   Answer: A. (1 + cos 2theta)/2 averages to 1/2. (Key point: Average of cos^2)

3. <KE> =?
   A. (1/2)m w^2 a^2
   B. (1/4)m w^2 a^2 = E/2
   C. (1/8)m w^2 a^2
   D. E
   Answer: B. Half of total. (Key point: Result)

4. <PE> =?
   A. E/4
   B. E
   C. E/2
   D. 0
   Answer: C. sin^2 also averages 1/2. (Key point: PE)

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