Wave equation — derivation for string

WHAT we are deriving

HOW: derivation from first principles

We take a tiny element of string between $x$ and $x+dx$ and apply $F=ma$ in the transverse (vertical, $y$) direction.

Step 1 — Set up the element

The element has length $\approx dx$ (for small slopes) and mass $dm = \mu\, dx.$ Why this step? Mass = (mass per length)×(length). We assume small displacements so the stretched length is still $\approx dx$.

Step 2 — Forces at the two ends

The tension $T$ acts along the string at each end. At the right end the string makes angle $\theta(x+dx)$ with the horizontal; at the left end angle $\theta(x)$.

Vertical components of tension:

• Right end: $+T\sin\theta(x+dx)$ (pulls up-and-right)
• Left end: $-T\sin\theta(x)$ (pulls down-and-left)

Why this step? Only the vertical component of $T$ causes transverse motion; the horizontal components nearly cancel (more below).

Step 3 — Small-angle approximation

For small slopes, $\theta$ is tiny, so $\sin\theta \approx \tan\theta = \frac{\partial y}{\partial x}.$ Why this step? $\tan\theta$ is literally the slope of the string, which is $\partial y/\partial x$. This converts geometry into derivatives. (Also $\cos\theta\approx 1$, so the horizontal tension components are equal and cancel — tension stays $\approx T$ everywhere.)

So net vertical force: $F_y = T\left[\left.\frac{\partial y}{\partial x}\right|_{x+dx} - \left.\frac{\partial y}{\partial x}\right|_{x}\right].$

Step 4 — Turn the difference into a derivative

A difference of a function at $x+dx$ and $x$, divided implicitly by $dx$, is a derivative: $\left.\frac{\partial y}{\partial x}\right|_{x+dx} - \left.\frac{\partial y}{\partial x}\right|_{x} \approx \frac{\partial^2 y}{\partial x^2}\,dx.$ Why this step? This is the definition of the second derivative: how the slope changes across the element. A changing slope means the string is curved — and curvature is what produces a net force. So $F_y = T\,\frac{\partial^2 y}{\partial x^2}\,dx.$

Step 5 — Newton's second law

The transverse acceleration is $\partial^2 y/\partial t^2$. Then $F_y = dm \cdot a_y$: $T\,\frac{\partial^2 y}{\partial x^2}\,dx = (\mu\, dx)\,\frac{\partial^2 y}{\partial t^2}.$ Why this step? $F=ma$ for the element. The same $dx$ on both sides cancels — the result is independent of how small a piece we chose, which is why it holds for the whole string.

Step 6 — Cancel and identify $v$

$\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu}\,\frac{\partial^2 y}{\partial x^2}.$ Comparing with the standard form $y_{tt}=v^2 y_{xx}$ gives $\boxed{v=\sqrt{\dfrac{T}{\mu}}}$ Why this step? Dimensional sense: $[T]=\text{N}=\text{kg·m/s}^2$, $[\mu]=\text{kg/m}$, so $T/\mu$ has units $\text{m}^2/\text{s}^2$ — a speed squared. ✓

WHY any $f(x-vt)$ solves it (Forecast-then-Verify)

Forecast: A pulse that keeps its shape while sliding right at speed $v$ is $y=f(x-vt)$. Let's verify it satisfies the equation.

Let $u=x-vt$.

• $\dfrac{\partial y}{\partial t}=f'(u)\cdot(-v)$, so $\dfrac{\partial^2 y}{\partial t^2}=f''(u)\cdot v^2$.
• $\dfrac{\partial y}{\partial x}=f'(u)$, so $\dfrac{\partial^2 y}{\partial x^2}=f''(u)$.

Then $y_{tt}=v^2 f''(u)=v^2 y_{xx}$. ✓ Verified. Any rigidly-moving shape works.

Worked examples

Common mistakes (Steel-man + fix)

Active recall

Connections

• Transverse vs Longitudinal Waves
• Wave speed on a string — v = sqrt(T over mu)
• Standing waves on a string
• Travelling wave function y = A sin(kx - omega t)
• Dispersion relation omega = vk
• Newton's Second Law
• Partial derivatives and curvature
• Energy carried by a wave

Concept Map

Hinglish (regional understanding)