---
title: "Derive or solve wave equation"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:derive-or-solve-wave-equation-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/derive-or-solve-wave-equation-two
site: BCT Engineering
---

# Derive or solve wave equation

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2081 Chaitra (new course), 4 marks
- Video: [Derivation of one dimensional wave equation | VTU Mathematics | Module 3](https://www.youtube.com/watch?v=c2IeHJOhSAg) — Dr Dhananjaya M
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/derive-or-solve-wave-equation-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Derive the one dimensional wave equation representing the vibration of the particles of a string.

## Short answer

Assume small transverse vibrations of a uniform string under tension T with linear density ρ. Net vertical force on an element is T(∂u/∂x|ₓ₊δₓ − ∂u/∂x|ₓ). Newton’s second law gives ρ δx ∂²u/∂t² = T(∂u/∂x|ₓ₊δₓ − ∂u/∂x|ₓ), leading to ∂²u/∂t² = c² ∂²u/∂x² with c² = T/ρ.

## Answer

**Result:** the one dimensional wave equation is ∂²u/∂t² = c² ∂²u/∂x², where c² = T/ρ.

**Assumptions.** A uniform, perfectly flexible string of mass ρ per unit length is stretched to tension T between two fixed points. It makes small transverse vibrations in a plane; gravity is neglected; the slope is small, so the horizontal motion of particles can be ignored and the tension T is constant.

**Derivation.** Consider an element PQ of the string between x and x + δx. Let u(x, t) be the displacement. The tension T acts along the tangent at each end, at small angles α (at P) and β (at Q) to the x-axis.

```text
 u ↑                  T ↗ β
   │             Q ●
   │          ╱
   │    P ●─╯   (string element PQ)
   │  T ↙ α
   └────┼─────────┼──────→ x
        x      x + δx
```

- Horizontal forces: T cos β − T cos α ≈ 0 (since the angles are small, cos ≈ 1). So T is constant.
- Vertical force on the element: T sin β − T sin α ≈ T(tan β − tan α), because for small angles sin ≈ tan.

Now tan α = (∂u/∂x)ₓ and tan β = (∂u/∂x)ₓ₊δₓ, so

net force = T[ (∂u/∂x)ₓ₊δₓ − (∂u/∂x)ₓ ]

By Newton's second law, mass × acceleration = force. The mass of the element is ρ δx:

ρ δx ∂²u/∂t² = T[ (∂u/∂x)ₓ₊δₓ − (∂u/∂x)ₓ ]

Divide by ρ δx and let δx → 0:

∂²u/∂t² = (T/ρ) ∂²u/∂x²

**∂²u/∂t² = c² ∂²u/∂x²,  c² = T/ρ**

Here c is the speed with which waves travel along the string.

## Check yourself

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

1. Which assumption lets the horizontal motion be ignored and T be treated as constant?
   A. String with variable density
   B. A string with no tension
   C. Large amplitude vibrations
   D. Small vibrations with small slopes
   Answer: D. Small angles make cos ~ 1, so horizontal forces cancel and T is constant. (Key point: Assumptions)

2. What is the net vertical force on element PQ?
   A. rho dx g
   B. T[(du/dx) at x+dx minus (du/dx) at x]
   C. T[(du/dt) at x+dx minus (du/dt) at x]
   D. T sin(alpha) only
   Answer: B. For small angles sin ~ tan = slope, so net force is T times the difference of slopes. (Key point: Net vertical force)

3. What is the mass of the element used in Newton's second law?
   A. rho dx
   B. rho
   C. T dx
   D. rho/dx
   Answer: A. Mass per unit length times length dx. (Key point: Newton's law)

4. What is c^2 in the wave equation?
   A. T rho
   B. sqrt(T/rho)
   C. rho/T
   D. T/rho
   Answer: D. Dividing by rho dx and letting dx tend to 0 gives u_tt = (T/rho) u_xx. (Key point: Result)

## Related in Mathematics III

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