---
title: "Average power of wave on a string"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:string-wave-average-power-linear-density"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/string-wave-average-power-linear-density
site: BCT Engineering
---

# Average power of wave on a string

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2071 Shrawan, 5 marks
- Video: [Power Transmitted by a Wave on a String, Schwartz, Cal Poly Physics updated](https://www.youtube.com/watch?v=FV-dEfM-Sys) — Peter Schwartz
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/string-wave-average-power-linear-density
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

A string has a linear density of 625 gm/m and is stretched with a tension 50N. A wave, whose frequency and amplitude are 160Hz and 10mm respectively, is travelling along the string. At what average rate is the wave transporting energy along the string?

## Short answer

Average power = ½ μ v ω² A². Wave speed v = √(50/0.625) = 8.94 m/s, ω = 2π×160 = 1005 rad/s, giving P_avg ≈ 282 W.

## Answer

Given: μ = 625 g/m = 0.625 kg/m, T = 50 N, f = 160 Hz, amplitude A = 10 mm = 0.010 m.

**Formulas**
```text
v = √(T/μ)        (wave speed)
ω = 2πf
P_avg = ½ μ v ω² A²
```

#### Steps
1. Wave speed: v = √(50 / 0.625) = √80 = 8.944 m/s.
2. Angular frequency: ω = 2π × 160 = 1005.3 rad/s, so ω² = 1.0106 × 10⁶ rad²/s².
3. A² = (0.010)² = 1.0 × 10⁻⁴ m².
4. Power:
```text
P_avg = ½ × 0.625 × 8.944 × 1.0106×10⁶ × 1.0×10⁻⁴
      = 0.3125 × 8.944 × 101.06
      = 282.5 W
```
**Answer: the wave transports energy at an average rate of about 2.8 × 10² W (≈ 282 W).**

## Check yourself

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

1. Which formula gives the average power transported by a harmonic wave on a string?
   A. P = μ v ω A
   B. P = ½ μ v ω² A²
   C. P = ½ μ v² ω A
   D. P = μ ω² A²
   Answer: B. P_avg = ½ μ v ω² A². (Key point: Average power formula)

2. For μ = 0.625 kg/m and T = 50 N, the wave speed is
   A. 8.94 m/s
   B. 80 m/s
   C. 0.0125 m/s
   D. 31.25 m/s
   Answer: A. v = √(50/0.625) = √80 = 8.944 m/s. (Key point: Wave speed)

3. What angular frequency and amplitude values should be used (f = 160 Hz, A = 10 mm)?
   A. ω = 160 rad/s, A = 10 m
   B. ω = 1005 rad/s, A = 0.010 m
   C. ω = 1005 rad/s, A = 10 m
   D. ω = 80 rad/s, A = 0.010 m
   Answer: B. ω = 2πf = 1005.3 rad/s and the amplitude must be in metres: 0.010 m. (Key point: Unit conversion)

4. The average rate of energy transport is about
   A. 28 W
   B. 2.8 W
   C. 282 W
   D. 2820 W
   Answer: C. 0.3125 × 8.944 × 1.0106×10^6 × 10^-4 ≈ 282 W. (Key point: Final power)

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