---
title: "Transverse wave equation on a string"
subject: "Engineering Physics"
chapter: 2
key: "semester_1/engineering_physics:transverse-wave-string-velocity"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/transverse-wave-string-velocity
site: BCT Engineering
---

# Transverse wave equation on a string

Engineering Physics · Chapter 2: Wave motion

- Asked in: 2074 Chaitra, 5 marks
- Video: [Wave Equation Derivation - Transverse Waves on a String](https://www.youtube.com/watch?v=ORXRbkAelbU) — Physics Ninja
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/transverse-wave-string-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-2

## Question

Define transverse wave. Develop a differential equation of the wave in a stretched string and then find the velocity of the transverse wave.

## Short answer

A transverse wave has particle displacement perpendicular to propagation. The wave equation is ∂²y/∂x² = (μ/T) ∂²y/∂t², yielding velocity v = √(T/μ).

## Answer

A **transverse wave** is a wave in which the particles of the medium vibrate perpendicular to the direction of propagation of the wave, e.g. waves on a stretched string.

#### Wave equation for a stretched string
Take a string of linear mass density μ (mass per unit length) under tension T, lying along the x-axis. Displacements y are small, so T is nearly constant.
```text
          T          y
   θ1  /----\ θ2     |
 ------A--dx--B----> x
```
1. Consider a small element AB of length dx. Tension T pulls at each end along the tangent, making small angles θ1 at A and θ2 at B with the x-axis.
2. Horizontal components cancel (cosθ ≈ 1). The net vertical force:
```text
F = T sinθ2 − T sinθ1 ≈ T (tanθ2 − tanθ1)
```
3. tanθ = ∂y/∂x, so tanθ2 − tanθ1 = (∂²y/∂x²) dx. Hence F = T (∂²y/∂x²) dx.
4. Mass of element = μ dx; its acceleration is ∂²y/∂t². Newton's second law:
```text
μ dx ∂²y/∂t² = T (∂²y/∂x²) dx
```
```text
∂²y/∂x² = (μ/T) ∂²y/∂t²
```
This is the differential equation of the wave on a string.

#### Velocity
5. Compare with the standard wave equation ∂²y/∂x² = (1/v²) ∂²y/∂t²:
```text
1/v² = μ/T   →   v = √(T/μ)
```
The speed of the transverse wave depends only on the tension and the mass per unit length of the string: it increases with T and decreases with μ.

## Check yourself

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

1. In a transverse wave the particles of the medium vibrate
   A. parallel to the direction of propagation
   B. perpendicular to the direction of propagation
   C. in circles only
   D. not at all
   Answer: B. Transverse means particle motion is perpendicular to the wave direction, as on a stretched string. (Key point: Transverse wave definition)

2. For a small string element dx, the net vertical force is
   A. T(∂²y/∂x²) dx
   B. T(∂y/∂x)
   C. μ dx
   D. T dx
   Answer: A. F = T(tanθ2 - tanθ1) and the difference in slope over dx is (∂²y/∂x²)dx. (Key point: Net force on element)

3. Applying Newton's law with mass μ dx gives
   A. ∂²y/∂x² = (T/μ) ∂²y/∂t²
   B. ∂²y/∂t² = μ T ∂²y/∂x²
   C. ∂²y/∂x² = (μ/T) ∂²y/∂t²
   D. ∂y/∂x = (μ/T) ∂y/∂t
   Answer: C. μ dx ∂²y/∂t² = T(∂²y/∂x²)dx rearranges to ∂²y/∂x² = (μ/T)∂²y/∂t². (Key point: Wave equation)

4. Comparing with ∂²y/∂x² = (1/v²)∂²y/∂t², the wave speed on the string is
   A. √(μ/T)
   B. T/μ
   C. √(Tμ)
   D. √(T/μ)
   Answer: D. 1/v² = μ/T gives v = √(T/μ): faster with higher tension, slower with greater mass per length. (Key point: Wave velocity)

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