Pluck a guitar string and it springs to life: a shape ripples up and down the wire, bounces off the far end, races back, and the whole thing sings a clean, definite note. Flick a long rope you're holding and a hump of rope shoots away from your hand and travels down the line, keeping its shape as it goes. In both cases nothing material travels the length of the string — each little bit of wire or rope only bobs up and down about its resting place — yet a disturbance sweeps along at a well-defined speed. What law governs that travelling shape?

The astonishing answer is that a single equation captures all of it — the guitar note, the rope flick, and (with the right symbols swapped in) sound in air, ripples on water, light in a vacuum, and the vibration of a bridge. It is the wave equation, and this page builds the simplest version — a stretched string moving in one dimension — from nothing more than Newton's second law applied to a tiny slice of string. Once we have it, we'll read what it says, and then meet d'Alembert's beautiful discovery that any shape at all is a legal wave.

The setup: a stretched string that wiggles a little

Picture a string pulled taut with tension T (measured in newtons) and made of some material with a linear mass density \mu (mass per unit length, in \text{kg/m}). At rest it lies flat along the x-axis. Now nudge it, so that at position x and time t the string is displaced sideways by a small amount y(x,t). The whole game is to find the equation that y(x,t) must obey.

We make one honest simplifying assumption: the displacements are small, so the string is only ever gently sloped. "Small" has a precise pay-off — the slope \partial y/\partial x is tiny, so the angle \theta the string makes with the horizontal is small too, and we may use \sin\theta \approx \tan\theta = \partial y/\partial x and \cos\theta \approx 1. This keeps the tension essentially constant along the string and lets each little element move purely up and down, not along its length.

Newton's second law on a single slice

Zoom in on one short element of the string, stretching from x to x + dx. Two forces act on it, one from the string pulling at each end. At the right end the neighbouring string pulls with tension T directed along the tangent there; at the left end it pulls with tension T along its tangent, in the opposite sense. Because the string is curved, those two tangents point in slightly different directions — and that mismatch is the whole story. Reveal the figure step by step.

Horizontal balance. Both tensions are nearly horizontal (\cos\theta \approx 1), and they point opposite ways, so their horizontal components cancel. The element does not accelerate along x — good, that matches our picture of pure up-and-down motion.

Vertical (transverse) force. The vertical component of each tension is T\sin\theta \approx T\,\dfrac{\partial y}{\partial x}, evaluated at that end. The net upward force on the element is the right-end pull minus the left-end pull:

That difference of a function across a gap dx is exactly what a derivative measures. Divide and multiply by dx:

So the net transverse force is tension times curvature times the element's width. A string that is curved concave-up (a valley) feels a net upward force; concave-down (a hill) feels a net downward force. The string is always pushed back toward straightness — and the more sharply it is bent, the harder the push.

Putting F = ma together

Our element has mass (density times length) dm = \mu\,dx, and its upward acceleration is the second time-derivative of its displacement, \partial^2 y/\partial t^2. Newton's second law, F_y = dm\cdot a_y, reads

The width dx cancels from both sides — a sign we've found something local and universal, not a fact about our particular slice. Dividing by \mu gives the star of the show:

Notice that v^2 = T/\mu has units \text{N} / (\text{kg/m}) = (\text{kg·m/s}^2)/(\text{kg/m}) = \text{m}^2/\text{s}^2, so v really is a speed. Everything checks out.

d'Alembert's solution: any shape is a wave

Here is the magic. Take any smooth shape you like — call it f(u) — and let it slide bodily to the right at speed v by feeding it the combination x - vt:

As t grows, the value the string had at x=0 is now found at x = vt: the whole profile marches to the right, rigidly, keeping its shape. Drag the time slider below and watch the pulse glide. It need not be a sine wave — a single bump, a triangle, a jagged spike, it all travels.

Does it really solve the equation? Let u = x - vt and use the chain rule. Differentiating in space, \partial u/\partial x = 1, so

Differentiating in time, \partial u/\partial t = -v, so each time-derivative brings down a factor of -v:

Compare the two second derivatives: \partial^2 y/\partial t^2 = v^2 f''(u) and v^2\,\partial^2 y/\partial x^2 = v^2 f''(u). They are equal — for every choice of f. The same works for a left-mover g(x + vt) (now the factor is (+v)^2, still v^2). Because the equation is linear, you can add the two:

Worked examples

Example 1 — verifying a given shape. Does y(x,t) = A\,e^{-(x - vt)^2} solve the wave equation? It has the form f(x - vt) with f(u) = A e^{-u^2}, so by d'Alembert's result it is a right-moving solution automatically — no grinding through derivatives needed. It is a bell-shaped bump that slides right at speed v without changing shape. Any function of the single combination x - vt passes the test; y = \sin(x)\cos(t) at first glance does not look like it — but a trig identity splits it into \tfrac12\sin(x-t) + \tfrac12\sin(x+t), a right-mover plus a left-mover, so it solves the equation with v = 1.

Example 2 — computing the wave speed. A guitar string has tension T = 80\ \text{N} and linear density \mu = 5.0\times 10^{-3}\ \text{kg/m}. Its wave speed is

Waves travel that fast along the wire. (Tighten the string — raise T — and v rises, so the pitch climbs; that's exactly what a tuning peg does.)

Example 3 — working backwards. A rope carries transverse waves at v = 20\ \text{m/s} under a tension of T = 50\ \text{N}. What is its linear density? Rearranging v = \sqrt{T/\mu} gives \mu = T/v^2: