---
title: "Power supplied to a string by a rod"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:string-wave-average-power-rod"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/string-wave-average-power-rod
site: BCT Engineering
---

# Power supplied to a string by a rod

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2068 Shrawan, 5 marks
- 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-rod
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

A rod vibrating at 12 Hz generates harmonic waves with amplitude of 1.5mm in a string of linear mass density 2 gm/m. If the tension in the string is 15N, what is the average power supplied by the source?

## Short answer

Average power = ½ μ v ω² A². v = √(15/0.002) = 86.6 m/s, ω = 2π×12 = 75.4 rad/s, A = 1.5 mm → P_avg ≈ 1.11×10⁻³ W (1.11 mW).

## Answer

Given: f = 12 Hz, amplitude A = 1.5 mm = 1.5 × 10⁻³ m, μ = 2 g/m = 2 × 10⁻³ kg/m, T = 15 N.

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

#### Steps
1. Wave speed: v = √(15 / 2×10⁻³) = √7500 = 86.6 m/s.
2. ω = 2π × 12 = 75.40 rad/s; ω² = 5684.9 rad²/s².
3. A² = (1.5×10⁻³)² = 2.25 × 10⁻⁶ m².
4. Power:
```text
P_avg = ½ × 2×10⁻³ × 86.6 × 5684.9 × 2.25×10⁻⁶
      = 1.0×10⁻³ × 86.6 × 0.012791
      = 1.11 × 10⁻³ W
```
**Answer: average power supplied ≈ 1.1 × 10⁻³ W (1.11 mW).**

## Check yourself

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

1. For T = 15 N and μ = 2×10⁻³ kg/m, the wave speed is
   A. 7500 m/s
   B. 2.7 m/s
   C. 86.6 m/s
   D. 0.0115 m/s
   Answer: C. v = √(15/0.002) = √7500 = 86.6 m/s. (Key point: Wave speed)

2. Which values must be used in SI units in the power formula?
   A. μ = 2 kg/m, A = 1.5 m
   B. μ = 2×10⁻³ kg/m, A = 1.5×10⁻³ m
   C. μ = 2×10⁻³ kg/m, A = 1.5 mm
   D. μ = 2 g/m, A = 1.5×10⁻³ m
   Answer: B. g/m and mm must be converted to kg/m and m before substituting. (Key point: Unit handling)

3. The average power supplied by the source is given by
   A. ½ μ v ω² A²
   B. μ v ω A
   C. ½ μ ω A
   D. T ω A
   Answer: A. P_avg = ½ μ v ω² A² with ω = 2πf. (Key point: Power formula)

4. The average power is about
   A. 1.1 W
   B. 1.1×10⁻³ W
   C. 1.1×10⁻⁶ W
   D. 86.6 W
   Answer: B. ½ × 2×10⁻³ × 86.6 × 5685 × 2.25×10⁻⁶ ≈ 1.11×10⁻³ W. (Key point: Final power)

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