The Wave Equation

Sine waves are only one kind of wave. A flick of a rope sends a single bump; a clap sends a short pulse of sound. What do all of these have in common? Each one obeys the same law, the wave equation, a statement about partial derivatives of $y(x, t)$.

Key idea

A wave moving at speed $v$ in one dimension satisfies $$\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\,\frac{\partial^2 y}{\partial t^2}.$$ Any twice-differentiable shape moving at speed $v$, $y = f(x - vt)$ or $y = g(x + vt)$, solves it, and so does any sum of such shapes. For a string, $v = \sqrt{F_T/\mu}$.

Why it works

Let $u = x - vt$ and $y = f(u)$. By the chain rule, $\frac{\partial y}{\partial x} = f'(u)$ and $\frac{\partial^2 y}{\partial x^2} = f''(u)$. In time, $\frac{\partial y}{\partial t} = f'(u)\cdot(-v)$ and $\frac{\partial^2 y}{\partial t^2} = v^2 f''(u)$. So $\frac{\partial^2 y}{\partial t^2} = v^2\,\frac{\partial^2 y}{\partial x^2}$, whatever the shape $f$ is. The same steps with $u = x + vt$ give the same result, because $(+v)^2 = (-v)^2$.

Where the equation comes from

Take a short piece of string of length $\Delta x$ and mass $\mu\,\Delta x$, under tension $F_T$. The tension pulls along the string at each end. If the string's slope is $\frac{\partial y}{\partial x}$, the sideways part of the pull is about $F_T\,\frac{\partial y}{\partial x}$ (small slopes). The two ends pull with slightly different slopes, so the net sideways force is $$F_T\left[\frac{\partial y}{\partial x}\Big|_{x+\Delta x} - \frac{\partial y}{\partial x}\Big|_{x}\right] \approx F_T\,\frac{\partial^2 y}{\partial x^2}\,\Delta x.$$ Newton's second law sets this equal to $(\mu\,\Delta x)\,\frac{\partial^2 y}{\partial t^2}$. Divide by $F_T\,\Delta x$: $$\frac{\partial^2 y}{\partial x^2} = \frac{\mu}{F_T}\,\frac{\partial^2 y}{\partial t^2}.$$ Matching with the wave equation gives $\frac{1}{v^2} = \frac{\mu}{F_T}$, so $v = \sqrt{F_T/\mu}$. The curvature of the string is what accelerates it.

Two different shapes moving at $v = 1.5$ m/s in opposite directions. Each solves the wave equation, and because the equation is linear, so does their sum (dashed).

Worked examples

Worked example: Speed from a pulse

A pulse on a rope is $$y = (0.20\ \text{m})\,e^{-(2.0x - 8.0t)^2}$$ with $x$ in m and $t$ in s. Find its speed and direction.

Factor the 2.0 out of the inside: $2.0x - 8.0t = 2.0(x - 4.0t)$.
So $y$ is a function of $x - 4.0t$ alone: the shape moves toward $+x$ at $4.0$ m/s.
Check with the rule for any $f(ax - bt)$: $v = b/a = 8.0/2.0 = 4.0$ m/s.

Answer: $4.0$ m/s toward $+x$

Worked example: Check a candidate

Does $y = \sin(3x)\cos(6t)$ satisfy the wave equation? If so, for what $v$?

$\frac{\partial^2 y}{\partial x^2} = -9\sin(3x)\cos(6t)$.
$\frac{\partial^2 y}{\partial t^2} = -36\sin(3x)\cos(6t)$.
The equation needs $-9 = \frac{1}{v^2}(-36)$, so $v^2 = 4$.

Answer: Yes, with $v = 2$ m/s (it is a standing wave, a sum of two waves moving each way)

Worked example: A string's wave speed

A guitar string has $\mu = 1.2 \times 10^{-3}$ kg/m and tension $F_T = 75$ N. Find the speed of waves on it.

$v = \sqrt{F_T/\mu} = \sqrt{\frac{75\ \text{N}}{1.2 \times 10^{-3}\ \text{kg/m}}}$.
$= \sqrt{6.25 \times 10^{4}\ \text{m}^2/\text{s}^2}$. Units: N/(kg/m) = (kg·m/s²)(m/kg) = m²/s².

Answer: $v = 250$ m/s

Common mistake

Thinking only sine waves solve the wave equation. Any shape that keeps its form as it moves works: a bump, a step, a triangle. Sines matter because every shape can be built from sines (Fourier's idea), and because each sine keeps its shape even in materials where the speed depends on frequency.

Common mistake

Reading the speed straight off the coefficient of $t$. In $e^{-(2x - 8t)^2}$ the speed is $8/2 = 4$ m/s, not 8 m/s: write the inside as a multiple of $(x - vt)$ first.

Two pulses pass through each other. Where they overlap the displacements add, and afterward each continues unchanged: that is what a linear wave equation predicts.

The Wave Equation

Where the equation comes from

Worked examples

Practice

Builds on

Leads to

Sources and further reading