---
title: "Maximum wave speed in steel wire"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:steel-wire-maximum-wave-speed"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/steel-wire-maximum-wave-speed
site: BCT Engineering
---

# Maximum wave speed in steel wire

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2070 Chaitra, 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/steel-wire-maximum-wave-speed
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

The elastic limit of steel forming a piece of wire is equal to 2.70×10⁸ Pa. What is the maximum speed at which transverse wave pulses can propagate along this wire without exceeding this stress? (density of steel = 7.89×10³ kg/m³)

## Short answer

v_max = √(σ/ρ) = √(2.70×10⁸ / 7.89×10³) ≈ 1.85×10² m/s (185 m/s).

## Answer

Given: elastic limit (maximum stress) σ = 2.70 × 10⁸ Pa, density of steel ρ = 7.89 × 10³ kg/m³.

**Formula.** For a wire of cross-sectional area S, tension T = σS and linear density μ = ρS. The wave speed is
```text
v = √(T/μ) = √(σ S / ρ S) = √(σ/ρ)
```
The area cancels, so the maximum speed depends only on the material.

#### Steps
1. σ/ρ = 2.70 × 10⁸ / 7.89 × 10³ = 3.422 × 10⁴ m²/s².
2. v_max = √(3.422 × 10⁴) = 185 m/s.

**Answer: v_max ≈ 1.85 × 10² m/s (about 185 m/s).**

## Check yourself

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

1. For a wire with tension T = σS and linear density μ = ρS, the wave speed is
   A. √(σρ)
   B. σ/ρ
   C. √(σ/ρ)
   D. √(σS/ρ)
   Answer: C. v = √(T/μ) = √(σS/ρS) = √(σ/ρ). (Key point: Speed in terms of stress)

2. Why does the cross-sectional area not appear in the maximum speed?
   A. Area is zero for wires
   B. It cancels because both T and μ are proportional to S
   C. It is replaced by length
   D. It is always 1 m²
   Answer: B. T = σS and μ = ρS both contain S, so S cancels, leaving a material-only result. (Key point: Area cancels)

3. With σ = 2.70×10⁸ Pa and ρ = 7.89×10³ kg/m³, v_max is about
   A. 3.42×10⁴ m/s
   B. 585 m/s
   C. 18.5 m/s
   D. 185 m/s
   Answer: D. σ/ρ = 3.42×10⁴ m²/s², whose square root is about 185 m/s. (Key point: Maximum speed value)

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