The wave equation

Pluck a stretched string and a bump runs down it without changing shape. Blow a sharp pulse of air and a pressure kink races off at the speed of sound. Wiggle a charge and an electromagnetic kink races off at the speed of light. All three are the same mathematics: a single differential equation, the wave equation, that Newton's second law forces onto any medium where a restoring force is proportional to curvature. This node derives it from scratch for a stretched string — the cleanest possible setting — by isolating a small element, writing down the net tension force on it, and applying $F=ma$ . Out falls not just an equation but its entire solution family: any shape that translates rigidly at a fixed speed $v$ , sinusoidal or not. Once you have this equation and its d'Alembert solution in hand, you can recognise the same structure wherever it reappears in the syllabus.

The core insight

Newton's second law applied to a string element under tension gives $\partial^2y/\partial t^2=v^2\,\partial^2y/\partial x^2$ with $v=\sqrt{T/\mu}$ , whose general solution $y=f(x-vt)+g(x+vt)$ describes any fixed shape rigidly translating at speed $v$ .

The core idea

Setting up the element. Take a string under uniform tension $T$ , mass per unit length $\mu$ , lying close to the $x$ -axis, and let $y(x,t)$ be its small transverse displacement. Isolate a short element between $x$ and $x+dx$ , of mass $dm=\mu\,dx$ . Because the slope $\partial y/\partial x$ is small, the vertical component of the tension pulling on either end of the element is approximately $T\,\partial y/\partial x$ evaluated at that end, so the two ends do not quite cancel when the string is curved.

Net force from curvature. The net upward force on the element is the difference of these end-slopes,

dF_y = T\left[\frac{\partial y}{\partial x}\bigg|_{x+dx}-\frac{\partial y}{\partial x}\bigg|_{x}\right]\approx T\,\frac{\partial^2 y}{\partial x^2}\,dx,

where: $T$ is the tension (N), $\partial^2y/\partial x^2$ is the curvature of the string at $x$ , and $dx$ is the element length. A straight segment ( $\partial^2y/\partial x^2=0$ ) feels no net force — it is curvature, not slope, that accelerates the element.

Newton's second law closes the loop. Setting $dF_y=dm\,\partial^2y/\partial t^2=\mu\,dx\,\partial^2y/\partial t^2$ and cancelling $dx$ gives $T\,\partial^2y/\partial x^2=\mu\,\partial^2y/\partial t^2$ , i.e. the wave equation with $v^2=T/\mu$ : a stiffer or lighter string produces a faster wave. The figure below shows how the two unequal end-tensions on a curved element produce the net restoring force that curvature drives.

Figure

A curved string element between x and x plus dx showing unequal tension slopes at its two ends producing a net upward restoring force

The general solution. Any twice-differentiable function of the single variable $x-vt$ or $x+vt$ solves the wave equation:

y(x,t)=f(x-vt)+g(x+vt).

$f(x-vt)$ is a fixed shape $f$ translating rigidly in the $+x$ direction at speed $v$ ; $g(x+vt)$ is a shape moving in $-x$ . Substituting $u=x-vt$ gives $\partial^2f/\partial t^2=v^2f''(u)$ and $\partial^2f/\partial x^2=f''(u)$ , so the equation holds for any functional form — the equation fixes only the speed and rigidity of propagation, never the shape.

Harmonic waves. The special case $y(x,t)=A\cos(kx-\omega t+\phi)$ is of this d'Alembert form with $u=x-(\omega/k)t$ , so it solves the wave equation provided $v=\omega/k$ . Substituting directly gives $\partial^2y/\partial t^2=-\omega^2y$ and $\partial^2y/\partial x^2=-k^2y$ , so $-\omega^2y=v^2(-k^2y)\Rightarrow v=\omega/k$ , consistent with $\omega=2\pi f$ , $k=2\pi/\lambda$ , $v=f\lambda$ .

Derivation ADeriving the 1D wave equation for a stretched string

The goal is to obtain $\partial^2y/\partial t^2=v^2\,\partial^2y/\partial x^2$ directly from Newton's second law applied to an infinitesimal element of a stretched string, using only the small-slope approximation, and to read off $v=\sqrt{T/\mu}$ as the natural speed scale that falls out of the tension and the mass density — this is the derivation the course wants reproduced, not a formula to memorise.

Step 1. Isolate an element of the string between $x$ and $x+dx$ , mass $dm=\mu\,dx$ , with $\mu$ the mass per unit length. The tension $T$ is uniform along the string; only its direction changes where the string is curved, since the string is nearly horizontal and displacement $y$ is small.

Step 2. Write the vertical (transverse) component of tension at each end. For small slope $\theta\approx\partial y/\partial x$ , $\sin\theta\approx\tan\theta\approx\partial y/\partial x$ , so the vertical pull at the right end is $T\,\partial y/\partial x|_{x+dx}$ and at the left end $T\,\partial y/\partial x|_{x}$ (pulling the other way).

Step 3. Subtract to get the net force, and Taylor-expand the difference in slope over the small interval $dx$ : $dF_y=T\left[\partial y/\partial x|_{x+dx}-\partial y/\partial x|_x\right]\approx T\,(\partial^2y/\partial x^2)\,dx$ . This is where curvature, the second spatial derivative, enters — a straight or uniformly sloped segment gives zero net force.

Step 4. Apply Newton's second law to the element, $dF_y=dm\,\partial^2y/\partial t^2$ : $T(\partial^2y/\partial x^2)\,dx=\mu\,dx\,(\partial^2y/\partial t^2)$ . The $dx$ cancels on both sides, leaving $\partial^2y/\partial t^2=(T/\mu)\,\partial^2y/\partial x^2$ .

Step 5. Identify $v^2=T/\mu$ , giving the boxed result $\partial^2y/\partial t^2=v^2\,\partial^2y/\partial x^2$ , with $v=\sqrt{T/\mu}$ having units of $\sqrt{\text{N}/(\text{kg/m})}=\sqrt{\text{m}^2/\text{s}^2}=\text{m/s}$ , confirming it is indeed a speed.

Worked example

A string with $\mu=2.5\times10^{-3}\ \text{kg/m}$ is under tension $T=90\ \text{N}$ . A transverse pulse of fixed shape $y(x,0)=f(x)$ , peaked at $x=0$ , is launched at $t=0$ travelling in the $+x$ direction. A second string carries the harmonic wave $y(x,t)=0.02\cos(15.7\,x-3140\,t)$ (SI units).

Step 1. Wave speed for the pulse string: $v=\sqrt{T/\mu}=\sqrt{90/(2.5\times10^{-3})}=\sqrt{36000}\approx189.7\ \text{m/s}$ .

Step 2. Since the pulse keeps its shape while translating at $v$ , $y(x,t)=f(x-189.7\,t)$ ; the peak (at argument zero) sits at $x-189.7t=0$ , so at $t=0.01\ \text{s}$ it is at $x=1.897\ \text{m}$ — the whole shape has shifted 1.897 m right, undistorted, exactly as d'Alembert's solution predicts.

Step 3. For the harmonic wave, compare $y=A\cos(kx-\omega t)$ to read off $A=0.02\ \text{m}$ , $k=15.7\ \text{rad/m}$ , $\omega=3140\ \text{rad/s}$ , so $v=\omega/k=3140/15.7=200\ \text{m/s}$ , $f=\omega/2\pi\approx500\ \text{Hz}$ , $\lambda=2\pi/k\approx0.4\ \text{m}$ . Check: $f\lambda=500\times0.4=200\ \text{m/s}=v$ , consistent.

Step 4. If this second string has $\mu=4\times10^{-3}\ \text{kg/m}$ , its tension follows from $v^2=T/\mu$ : $T=\mu v^2=4\times10^{-3}\times(200)^2=160\ \text{N}$ . Answer: pulse speed $\approx189.7\ \text{m/s}$ (peak shifts to 1.897 m at $t=0.01$ s); harmonic wave has $v=200\ \text{m/s}$ , $f\approx500\ \text{Hz}$ , $\lambda\approx0.4\ \text{m}$ , and requires $T=160\ \text{N}$ .

Common traps

Forgetting linearity: only functions of $x-vt$ or $x+vt$ (or sums of these) solve the wave equation — an arbitrary function of $x$ and $t$ separately generally does not.
Sign confusion in $y=A\cos(kx-\omega t)$ : a $+\omega t$ term ( $\cos(kx+\omega t)$ ) describes a wave moving in the $-x$ direction, not $+x$ — check the sign every time.
Treating $v=\omega/k$ as an independent input rather than a consequence of substituting the harmonic form into the wave equation, where $v^2=T/\mu$ is already fixed by the string's physical properties.
Assuming the wave equation is specific to strings — sound pressure, electromagnetic field components and many other quantities obey the identical equation, each with its own expression for $v$ .
Confusing the wave speed $v$ with the transverse speed of a piece of the string, $\partial y/\partial t$ ; the string material itself only moves up and down, while $v=\sqrt{T/\mu}$ is the speed at which the disturbance pattern travels along $x$ — the two are numerically unrelated in general.

Ask the tutor: If the small-slope approximation or the step from the tension-difference to $T\,\partial^2y/\partial x^2\,dx$ feels shaky, ask the AI tutor to redo that Taylor-expansion step slowly or to check your own derivation line by line.

Why does only curvature, not slope, produce a restoring force on the string element?

A uniformly sloped but straight segment has the same tension direction pulling at both ends in a way that just balances — imagine tilting the whole picture; nothing about a constant slope makes one end pull harder upward than the other pulls downward. It is only when the slope itself changes across the element, i.e. when $\partial^2y/\partial x^2\neq0$ , that the two end-forces fail to cancel and a net transverse force appears. That is exactly why the second derivative, not the first, shows up in the wave equation.

Does the wave equation require the wave to be sinusoidal?

No. The wave equation itself, $\partial^2y/\partial t^2=v^2\,\partial^2y/\partial x^2$ , is satisfied by any twice-differentiable function of $x-vt$ or $x+vt$ — a single pulse, a square-ish kink, a superposition of many shapes, or a pure sinusoid all qualify. The harmonic wave $A\cos(kx-\omega t)$ is just the most convenient special case, useful because it is easy to Fourier-decompose other shapes into sums of such terms, not because the equation demands sinusoids.

Why is the small-slope approximation needed, and what happens if the string is displaced a lot?

The step $\sin\theta\approx\tan\theta\approx\partial y/\partial x$ requires the slope angle $\theta$ to be small, which keeps the tension magnitude effectively constant along the string (only its direction changes) and keeps the equation linear. If the displacement is large, higher-order terms in $\theta$ survive, the equation becomes nonlinear, tension itself may vary with position, and the clean d'Alembert solution $f(x-vt)+g(x+vt)$ no longer exactly applies — real large-amplitude strings show mild dispersion and distortion that this idealised derivation ignores.

The wave equation

The core idea

Worked example

Common traps

Before this

Where this leads