Derive one-dimensional wave equation

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.

 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

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.