---
title: "Derive one-dimensional wave equation"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:wave-eq-derive"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/wave-eq-derive
site: BCT Engineering
---

# Derive one-dimensional wave equation

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2070 Magh, 10 marks; 2071 Bhadra (asked 2 times)
- 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_4/applied_mathematics/wave-eq-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Derive the one dimensional wave equation (vibrating string).

## Short answer

Consider a small element of the string. The net vertical force is T[∂u/∂x|_{x+δx} − ∂u/∂x|_x]. Newton's second law yields ρδx ∂²u/∂t² = T ∂²u/∂x² δx, giving ∂²u/∂t² = c² ∂²u/∂x² with c = √(T/ρ).

## Answer

**Assumptions.** A uniform, perfectly flexible string of mass ρ per unit length is stretched to a large constant tension T between x = 0 and x = l. It vibrates in a plane (the xy-plane) with small displacements u(x, t), so the slope is small and gravity is negligible. Each particle moves only perpendicular to the x-axis.

**Forces on a small element.** Consider the element PQ between x and x + δx. Tension T acts along the tangent at each end, making angles θ₁ at P and θ₂ at Q with the x-axis.

```text
 u
 |            Q ○──→ T (angle θ₂)
 |          ╱
 |   P ○───╱
 |  ↙ T (angle θ₁)
 |_________________________ x
      x        x+δx
```

**Horizontal direction.** There is no horizontal motion, so T cos θ₂ − T cos θ₁ = 0. For small angles, cos θ ≈ 1, so this holds.

**Vertical direction.** The net upward force is

F = T sin θ₂ − T sin θ₁

For small angles sin θ ≈ tan θ = slope = ∂u/∂x. Hence

F = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ]

**Newton's second law.** Mass of the element = ρ δx. Its acceleration = ∂²u/∂t². So

```text
ρ δx · ∂²u/∂t² = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ]

∂²u/∂t² = (T/ρ) · [ (∂u/∂x)_{x+δx} − (∂u/∂x)ₓ ] / δx
```

Let δx → 0. The bracket divided by δx becomes ∂²u/∂x². Put c² = T/ρ:

**∂²u/∂t² = c² ∂²u/∂x²**

This is the one-dimensional wave equation. Here c = √(T/ρ) is the speed of wave propagation along the string.

## Check yourself

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

1. In horizontal direction
   A. forces balance, since cos theta ~ 1
   B. tension is zero
   C. motion occurs
   D. tension differs
   Answer: A. No horizontal motion. (Key point: Horizontal)

2. Net vertical force =
   A. T cos theta
   B. T[(u_x)(x+dx) - (u_x)(x)] using sin ~ tan ~ slope
   C. rho g dx
   D. T u_t
   Answer: B. Small slope. (Key point: Vertical force)

3. Element mass is
   A. T dx
   B. rho
   C. rho dx
   D. rho/dx
   Answer: C. Mass per unit length times dx. (Key point: Mass)

4. Wave speed c =
   A. sqrt(rho/T)
   B. T/rho
   C. rho T
   D. sqrt(T/rho)
   Answer: D. c^2 = T/rho. (Key point: Speed)

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