The Wave Equation — padho.wiki

In short

A travelling wave keeps its shape and moves at a constant speed v — mathematically, its displacement is y(x, t) = f(x - vt) for a wave moving in the +x direction (or f(x + vt) for the -x direction). The most important special case is the sinusoidal wave

y(x, t) = A\sin(kx - \omega t + \varphi),

where k = 2\pi/\lambda is the wave number, \omega = 2\pi f is the angular frequency, and the phase velocity is v = \omega/k = f\lambda. Every such wave — and every superposition of them — satisfies the wave equation

\frac{\partial^2 y}{\partial t^2} = v^2\,\frac{\partial^2 y}{\partial x^2},

a single partial differential equation whose solutions are exactly "anything that moves with speed v without changing shape."

Pluck a sitar string at its middle. A sharp pulse — a narrow hump of displacement — springs from the plucking point and races outward in both directions along the string. You watch it and see something remarkable: the hump does not spread, does not flatten, does not change its shape. It just moves. A thousandth of a second later the hump is one metre to the right, identical in every respect to what it was a thousandth of a second ago, only shifted.

That statement — "identical, only shifted" — is almost the entire wave equation. Write it in algebra and the equation writes itself. Any quantity that moves through space at a constant speed without distorting must depend on position and time only through the specific combination x - vt (if it is moving in the +x direction). From that single observation, one differential equation follows with iron necessity: the wave equation. Every wave — on a guitar string, through air, through the rail under an Indian Railways train — obeys it, with only the speed v changing from one situation to the next.

This article builds the equation. You will see exactly why the combination x - vt is the shape-preserving argument, derive the sinusoidal wave y = A\sin(kx - \omega t) by starting from a particle at the origin doing SHM, and differentiate twice to prove the wave equation. Then you will understand what "phase velocity" means and why it is not usually the speed of anything material.

The shape-preserving argument x - vt

Imagine the string at t = 0. The shape is some function f(x) — perhaps a Gaussian hump, perhaps a sharp triangle, perhaps a sinusoidal ripple. Label the displacement at that moment:

y(x, 0) = f(x)

Now let time pass. The pulse moves to the right at speed v. After time t, the entire shape has shifted to the right by a distance vt. The point that was at x = 0 is now at x = vt; the point that was at x = x_0 is now at x = x_0 + vt.

In other words, to find the height of the pulse at position x and time t, you ask: where was this height at t = 0? It was at x - vt (you have to "rewind" the pulse by a distance vt).

\boxed{y(x, t) = f(x - vt)}

That is the mathematical expression of "shape-preserving translation at speed v". Every undistorted travelling wave on a line looks like this. The specific function f can be anything — a Gaussian, a triangle, a sine wave, a delta function, whatever the initial shape was. The combination x - vt is the only constraint.

Three snapshots of the same travelling pulse. At $t = 0$ the hump sits at some position; at $t = T$ it has moved a distance $vT$ to the right; at $t = 2T$ it has moved $2vT$. The shape function $f$ never changes — only the argument $x - vt$ shifts.

Watch a pulse travel

A rough visualisation: the red dot is the peak of a pulse $f(x - 2t)$. It moves at $2$ m/s in $x$, leaving a trail that records its positions over time. Every point of the pulse moves rightward at the same $2$ m/s — the shape is preserved.

The sinusoidal wave — the periodic workhorse

The most important special case of a travelling wave is a sinusoidal wave. Why sinusoidal? Two reasons. First, any smooth periodic wave can be decomposed into sines and cosines (Fourier's theorem). Second, a sinusoidal wave is what you get when you drive one end of a medium with SHM — a tuning fork, a loudspeaker, a vibrating knot on a string.

Pick a particle of the medium at x = 0 and make it oscillate in SHM:

y(0, t) = A\sin(\omega t + \delta)

The disturbance at x = 0 at time t will reach position x after a delay x/v, because the wave travels at speed v. So the displacement at position x at time t is the same as the displacement at x = 0 a time x/v earlier:

y(x, t) = y(0, t - x/v) = A\sin\!\left[\omega(t - x/v) + \delta\right]

Expand the argument.

Step 1. Distribute \omega through the parentheses.

y(x, t) = A\sin\!\left(\omega t - \frac{\omega}{v}x + \delta\right)

Step 2. Define the wave number k by

k = \frac{\omega}{v}

Why: the combination \omega/v appears as the coefficient of x in the phase. Giving it a name and units makes the equation cleaner and lets you state the relation to wavelength in a single line.

Step 3. Rewrite.

y(x, t) = A\sin(\omega t - kx + \delta)

Step 4. Conventionally, flip the sign of the phase so that "moving in the +x direction" has the kx - \omega t sign pattern.

\boxed{y(x, t) = A\sin(kx - \omega t + \varphi)}

Why: \sin(\omega t - kx + \delta) = \sin(-(kx - \omega t - \delta)) = -\sin(kx - \omega t - \delta). Absorbing the sign flip and the constant -\delta into a new phase constant \varphi gives the standard form. Both conventions — \sin(kx - \omega t) and \sin(\omega t - kx) — appear in textbooks; they describe the same physics.

You now have five quantities with precise meanings:

A — the amplitude. The maximum displacement of any particle from equilibrium.
k — the wave number (SI unit: rad/m). Measures spatial periodicity: k = 2\pi/\lambda where \lambda is the wavelength.
\omega — the angular frequency (SI unit: rad/s). Measures temporal periodicity: \omega = 2\pi/T where T is the period, and \omega = 2\pi f where f is the frequency in Hz.
v — the phase velocity: v = \omega/k = f\lambda.
\varphi — a constant phase, fixing where the wave's zero crossing is at x = t = 0.

Deriving k = 2\pi/\lambda

The wavelength \lambda is the spatial period — the distance over which the wave's shape repeats. At a fixed time t, freeze the wave. Then y(x, t) = A\sin(kx - \omega t + \varphi) is a pure sinusoid in x. Its spatial period is the \Delta x for which k\Delta x = 2\pi:

k\lambda = 2\pi \quad\Longrightarrow\quad \boxed{k = \frac{2\pi}{\lambda}}

Why: a sine function repeats every time its argument advances by 2\pi. If kx advances by 2\pi, then x advances by 2\pi/k, and that x-advance is one wavelength.

Deriving v = f\lambda

Combine k = 2\pi/\lambda and \omega = 2\pi f with v = \omega/k:

v = \frac{\omega}{k} = \frac{2\pi f}{2\pi / \lambda} = f\lambda

Why: this is the most famous wave formula. In one period T = 1/f, the wave advances by one wavelength \lambda. Speed is distance divided by time: v = \lambda/T = f\lambda.

A sinusoidal wave frozen in space at one instant. The horizontal distance between successive peaks is the wavelength $\lambda$. The vertical distance from axis to peak is the amplitude $A$.

Animating the travelling sinusoid

The static picture above is only half the story. Let time flow, and the sinusoid marches rightward at speed v. Every point of the string does SHM in place; the pattern — the collection of crests and troughs — moves as a whole.

The red dot is a crest of a wave $y = \sin(\pi x - \pi t)$ with $\lambda = 2$ m, $\omega = \pi$ rad/s, so $v = \omega/k = \pi/\pi = 1$ m/s. The dark dot is a trough, one half-wavelength behind. Both move rightward at the same phase velocity. Click replay.

The phase velocity v = \omega/k is the speed at which the pattern moves. It is not the speed at which any particle of the medium moves — each particle on a string just oscillates up and down in place, at a speed that varies with time. The phase velocity is the speed of the phase surface, the abstract locus of points where kx - \omega t has a given value.

The wave equation itself

The shape-preserving form y = f(x - vt) tells you which functions are travelling waves. The wave equation tells you which differential equation they solve. Every f(x - vt) satisfies one simple PDE.

Setting up partial derivatives

Let u = x - vt. Then y = f(u), so

\frac{\partial y}{\partial x} = f'(u)\cdot\frac{\partial u}{\partial x} = f'(u)\cdot 1 = f'(u)

\frac{\partial y}{\partial t} = f'(u)\cdot\frac{\partial u}{\partial t} = f'(u)\cdot(-v) = -v f'(u)

Why: the chain rule. f is a function of u, and u depends on both x and t. The partial derivative of y with respect to x holds t fixed, so \partial u/\partial x = 1; the partial derivative with respect to t holds x fixed, so \partial u/\partial t = -v.

The key observation

Look at the two first derivatives: \partial y/\partial t = -v\cdot(\partial y/\partial x). The time derivative is -v times the spatial derivative. That makes sense — moving rightward by one unit looks the same as moving backward in time by 1/v units.

Second derivatives

Differentiate again.

\frac{\partial^2 y}{\partial x^2} = f''(u)\cdot 1 = f''(u)

\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t}[-v f'(u)] = -v\cdot f''(u)\cdot(-v) = v^2 f''(u)

Why: differentiate -v f'(u) with respect to t, applying the chain rule once more. The outer -v is a constant. The derivative of f'(u) with respect to t is f''(u)\cdot\partial u/\partial t = -v f''(u). Multiply by -v: get v^2 f''(u).

Put the pieces together

Both second derivatives are proportional to f''(u). Eliminating f''(u):

\frac{\partial^2 y}{\partial t^2} = v^2\,\frac{\partial^2 y}{\partial x^2}

\boxed{\frac{\partial^2 y}{\partial t^2} = v^2\,\frac{\partial^2 y}{\partial x^2}}

Why: this is the wave equation — a second-order linear partial differential equation. Every f(x - vt) satisfies it, which means every shape-preserving wave is a solution. So is every g(x + vt) (check: \partial y/\partial t = v g', \partial^2 y/\partial t^2 = v^2 g'', \partial^2 y/\partial x^2 = g'' — identity holds). And so is the sum of any such solutions — because the equation is linear.

Verifying for the sinusoidal wave

Check that y = A\sin(kx - \omega t + \varphi) solves the wave equation.

\frac{\partial y}{\partial x} = Ak\cos(kx - \omega t + \varphi), \qquad \frac{\partial^2 y}{\partial x^2} = -Ak^2\sin(kx - \omega t + \varphi) = -k^2 y

\frac{\partial y}{\partial t} = -A\omega\cos(kx - \omega t + \varphi), \qquad \frac{\partial^2 y}{\partial t^2} = -A\omega^2\sin(kx - \omega t + \varphi) = -\omega^2 y

Substitute into the wave equation:

-\omega^2 y = v^2\cdot(-k^2 y) \quad\Longrightarrow\quad \omega^2 = v^2 k^2 \quad\Longrightarrow\quad v = \frac{\omega}{k}

Why: the wave equation is satisfied exactly when \omega and k satisfy \omega = vk — the dispersion relation for the simplest wave equation. This is the same statement as v = f\lambda in disguise.

d'Alembert's general solution

Because the wave equation is linear and second-order in both x and t, its most general solution is

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

for arbitrary twice-differentiable functions f and g. This is called d'Alembert's formula. The physical content: the most general disturbance is a rightward-travelling wave of any shape plus a leftward-travelling wave of any shape. Everything else — standing waves, wave packets, reflection patterns — is built from these two streams by superposition.

Phase velocity — what it is and what it is not

The phase velocity v = \omega/k is the speed at which the crests and troughs travel. Pick a crest. At time t, the crest is at the position where kx - \omega t = \pi/2 (the argument of sine at which sine peaks). As t advances by dt, the crest advances by dx satisfying k\,dx - \omega\,dt = 0, i.e. dx/dt = \omega/k = v.

What phase velocity is NOT

It is not the speed of any piece of the medium. A dot on the string does simple harmonic motion up and down. Its maximum speed is A\omega, which has nothing to do with v = \omega/k. A wave on a railway track travels many metres per second, but no atom of the rail moves more than a fraction of a millimetre during the passage.
It is not the speed of information. For a pure sinusoid extending infinitely in both directions, nothing "new" ever arrives; the wave is the same at every moment. Information travels only when you modulate the wave — create a pulse or a wave packet — and that speed (the group velocity) is different in general.
It is not always less than the speed of light. In some media, optical waves of certain frequencies have phase velocities exceeding c. No contradiction with relativity: information still travels at the group velocity, which remains below c.

For simple non-dispersive waves (on a string, sound in air at ordinary pressures), phase velocity and group velocity are equal, and v = f\lambda is the speed at which everything practical travels. The distinction matters for optics, quantum mechanics, and plasma physics — topics you meet later.

Worked examples

Example 1: Wave on a sitar string

A sitar player plucks the string. A pulse travels along the string with y(x, t) = 0.3\sin(8x - 40t)\ \text{cm}, where x is in metres and t in seconds. Find: (a) the amplitude, (b) the wavelength, (c) the period, (d) the frequency, (e) the wave speed, (f) the direction of travel.

The sitar-string wave $y = 0.3\sin(8x - 40t)$: amplitude $0.3$ cm, wavelength $\pi/4$ m, and moving rightward at $5$ m/s.

Step 1. Read off A, k, and \omega from the equation.

Comparing y = 0.3\sin(8x - 40t) with y = A\sin(kx - \omega t):

A = 0.3 cm, k = 8 rad/m, \omega = 40 rad/s.

Why: the coefficients of x and t inside the sine are the wave number and angular frequency. Matching signs confirms that the kx - \omega t form applies, so the wave moves in the +x direction.

Step 2. Amplitude.

A = 0.3 cm (given directly).

Step 3. Wavelength.

\lambda = \frac{2\pi}{k} = \frac{2\pi}{8} = \frac{\pi}{4}\ \text{m} \approx 0.785\ \text{m}

Why: k = 2\pi/\lambda, so \lambda = 2\pi/k.

Step 4. Period.

T = \frac{2\pi}{\omega} = \frac{2\pi}{40} = \frac{\pi}{20}\ \text{s} \approx 0.157\ \text{s}

Step 5. Frequency.

f = \frac{1}{T} = \frac{20}{\pi}\ \text{Hz} \approx 6.37\ \text{Hz}

Why: frequency is the reciprocal of the period. Equivalently, f = \omega/(2\pi) = 40/(2\pi) \approx 6.37 Hz.

Step 6. Wave speed.

v = \frac{\omega}{k} = \frac{40}{8} = 5\ \text{m/s}

Check: v = f\lambda = (20/\pi)(\pi/4) = 5 m/s. Matches.

Why: two independent formulas for v give the same answer — confirming consistency. This is a slow wave because the pulse is large and the frequency is audio-inaudible; a real sitar string has wave speeds of hundreds of metres per second, which is why its audible frequencies land in the hundreds of hertz.

Step 7. Direction.

The kx - \omega t form (negative relative sign) means motion in the +x direction.

Why: a crest at kx - \omega t = \pi/2 must satisfy kx = \omega t + \pi/2, i.e. x = (\omega/k)t + \text{const}. As t grows, x grows — the crest moves rightward.

Result: A = 0.3 cm, \lambda = \pi/4 \approx 0.785 m, T = \pi/20 \approx 0.157 s, f \approx 6.37 Hz, v = 5 m/s, moving in the +x direction.

What this shows: Reading a wave equation is almost entirely about identifying k and \omega. Once you know those, every other wave parameter falls out of v = \omega/k and \lambda = 2\pi/k.

Example 2: Sound in an auditorium

A tanpura drones at 220 Hz inside a music hall in Chennai. The speed of sound in the still air of the hall is 340 m/s. (a) Find the wavelength. (b) Find the wave number. (c) Write the equation of the sound wave as a pressure variation, assuming the peak pressure amplitude is 0.5 Pa and the wave travels in the +x direction from the drone toward a listener.

A 220 Hz sound wave in air: compressions (dark) and rarefactions (light) travel at 340 m/s from the tanpura to the listener, separated by the wavelength.

Step 1. Compute the wavelength.

\lambda = \frac{v}{f} = \frac{340}{220}\ \text{m} \approx 1.545\ \text{m}

Why: v = f\lambda rearranges to \lambda = v/f. At 220 Hz, each full pressure oscillation spans about 1.5 m — roughly half the length of an autorickshaw.

Step 2. Wave number.

k = \frac{2\pi}{\lambda} = \frac{2\pi}{1.545} \approx 4.07\ \text{rad/m}

Why: direct application of k = 2\pi/\lambda.

Step 3. Angular frequency.

\omega = 2\pi f = 2\pi \times 220 \approx 1382\ \text{rad/s}

Step 4. Pressure wave equation.

p(x, t) = (0.5\ \text{Pa})\,\sin(4.07\,x - 1382\,t)

Why: the sinusoidal wave A\sin(kx - \omega t) in the +x direction, with A being the pressure amplitude. Sound waves in air are longitudinal pressure variations, but the mathematical form is identical to a transverse wave on a string.

Step 5. Sanity check.

v = \frac{\omega}{k} = \frac{1382}{4.07} \approx 339.6 \approx 340\ \text{m/s}\ \checkmark

Why: the computed \omega/k returns the given sound speed. Any small discrepancy is from rounding \lambda to 4 significant figures.

Result: \lambda \approx 1.55 m, k \approx 4.07 rad/m, p(x, t) = 0.5\sin(4.07\,x - 1382\,t) Pa.

What this shows: The wave equation structure y = A\sin(kx - \omega t) applies to any travelling wave, whether transverse (string) or longitudinal (sound). The only thing that changes between media is the value of v — the structure of the equation is universal.

Example 3: Verifying a proposed solution

A student writes down the function y(x, t) = 5\cos(3x - 12t) (in SI units) and claims it is a travelling wave. Verify that it satisfies the wave equation and find the speed of travel.

A cosine travelling wave with $k = 3$ rad/m and $\omega = 12$ rad/s. Speed: $v = 12/3 = 4$ m/s.

Step 1. Identify k and \omega.

k = 3 rad/m, \omega = 12 rad/s. Amplitude 5 m. The function uses cosine, but \cos\theta = \sin(\theta + \pi/2), so this is a sinusoidal wave with a \pi/2 phase offset — the same physics.

Step 2. Compute \partial^2 y/\partial x^2.

\frac{\partial y}{\partial x} = -5\cdot 3\sin(3x - 12t) = -15\sin(3x - 12t)

\frac{\partial^2 y}{\partial x^2} = -15\cdot 3\cos(3x - 12t) = -45\cos(3x - 12t) = -9y

Why: differentiating cosine once gives -\sin times the chain-rule factor k = 3; differentiating again gives -\cos times another factor of 3. Combined, you pick up a factor of -k^2 = -9. In general, any y of the form A\cos(kx - \omega t) satisfies \partial^2 y/\partial x^2 = -k^2 y.

Step 3. Compute \partial^2 y/\partial t^2.

\frac{\partial y}{\partial t} = -5\cdot(-12)\sin(3x - 12t) = 60\sin(3x - 12t)

\frac{\partial^2 y}{\partial t^2} = 60\cdot(-12)\cos(3x - 12t) = -720\cos(3x - 12t) = -144 y

Why: differentiating with respect to t pulls down a factor of -\omega = -12 each time (two minus signs cancel in the first derivative because the chain rule introduces one, and cosine differentiates to minus sine). The final second-time-derivative is -\omega^2 y = -144 y.

Step 4. Check the wave equation.

\frac{\partial^2 y}{\partial t^2} = v^2\,\frac{\partial^2 y}{\partial x^2} \quad\Longrightarrow\quad -144 y = v^2\cdot(-9 y) \quad\Longrightarrow\quad v^2 = 16

v = 4\ \text{m/s}

Why: divide both sides by y (and by -1). The wave equation holds for any y, and uniquely determines v = \omega/k = 12/3 = 4 m/s — matching the direct formula.

Result: The function solves the wave equation with v = 4 m/s in the +x direction.

What this shows: Verifying a candidate wave solution is mechanical: differentiate twice, plug in, check. The dispersion relation \omega = vk pops out automatically. The same procedure applies to any function of the form f(x - vt) — not just sines and cosines.

Common confusions

"y(x, t) = A\sin(kx)\sin(\omega t) is a travelling wave." No — this is a standing wave, not a travelling wave. The argument is not of the form x - vt or x + vt. A standing wave is the sum of two counter-travelling waves: \sin(kx)\sin(\omega t) = \frac{1}{2}[\cos(kx - \omega t) - \cos(kx + \omega t)]. Travelling waves have the spatial and temporal parts entangled in one sinusoid; standing waves factor them apart.
"The wave number k is the frequency." No — k is the spatial periodicity (2\pi/\lambda), while f is the temporal periodicity (1/T). They are linked by v = f\lambda = \omega/k, but they are different quantities with different units (rad/m versus Hz).
"Phase velocity is the speed of the medium." No — the medium does not flow with the wave. Each particle oscillates in place. Phase velocity is the speed of the pattern (the crests), not the speed of any material.
"The wave equation only describes sinusoidal waves." No — the wave equation is satisfied by any function of the form f(x - vt), regardless of whether f is a sine, a Gaussian, a triangle, or a delta function. Sines are simply the cleanest solutions to work with, and every other solution can be built from sines by superposition (Fourier's theorem).
"A wave equation with v depending on frequency is impossible." Actually, it is common. Media where different frequencies travel at different speeds are called dispersive — optical glass, deep water, plasmas. The simple wave equation here is non-dispersive. Adding frequency-dependent speed requires more elaborate equations (Klein-Gordon, Schrödinger, telegrapher's), which you meet in advanced courses.
"y = A\sin(kx - \omega t) and y = A\sin(\omega t - kx) describe different waves." They describe the same physical wave — just with opposite sign conventions for the phase. Both travel in the +x direction at speed \omega/k. The difference is only a minus sign inside the sine, which corresponds to a \pi phase shift — a symmetry of the medium. Use whichever convention your textbook prefers.

If you came here to read a wave equation, identify A, k, and \omega, compute the speed, and picture the motion, you are done. What follows is for readers who want the derivation of the wave equation from a physical system, the connection to Maxwell's equations and light, and the general three-dimensional form.

Deriving the wave equation on a string from Newton's second law

The wave equation \partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2 emerged above from the kinematic assumption "the shape translates without distorting." But where does the equation come from in physical reality? For a stretched string, you can derive it from Newton's second law applied to a tiny element.

Consider a string under tension T, with linear mass density \mu (kg/m). Pick a small element stretching from x to x + dx. Its mass is \mu\,dx. The tension pulls it at both ends — to the left at angle \theta(x) below horizontal, to the right at angle \theta(x+dx) above horizontal.

For small displacements, the tension is approximately the same at both ends, T, and \theta \approx \tan\theta = \partial y/\partial x. The net vertical force on the element is

F_y = T\sin\theta(x+dx) - T\sin\theta(x) \approx T\left.\frac{\partial y}{\partial x}\right|_{x+dx} - T\left.\frac{\partial y}{\partial x}\right|_{x}

\approx T\,\frac{\partial^2 y}{\partial x^2}\,dx

Why: the difference of a quantity evaluated at two points separated by dx is approximately (\partial/\partial x) of the quantity, times dx. Here the quantity is \partial y/\partial x, so the difference is (\partial^2 y/\partial x^2)\,dx.

By Newton's second law, F_y = (\mu\,dx)\,(\partial^2 y/\partial t^2). Equate:

T\,\frac{\partial^2 y}{\partial x^2}\,dx = \mu\,\frac{\partial^2 y}{\partial t^2}\,dx

\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu}\,\frac{\partial^2 y}{\partial x^2}

Comparing with \partial^2 y/\partial t^2 = v^2\,\partial^2 y/\partial x^2, you read off

\boxed{v = \sqrt{\frac{T}{\mu}}}

Why: the wave speed on a string is set by the ratio of the restoring force (tension) to the inertia (linear density). Tighter string = faster waves; heavier string = slower waves. This is why a tightened sitar string rings at a higher frequency than a loosened one: the same wavelength, but the speed has increased, so f = v/\lambda increases.

This is not just a kinematic coincidence — it is Newton's second law, applied to the string, giving you the wave equation with v = \sqrt{T/\mu} popping out of the algebra. See Speed of Waves on a String for the careful version.

Deriving the wave equation for sound

The same argument works for sound, with the tension-to-density ratio replaced by the pressure-stiffness-to-density ratio. In an ideal gas,

v_{\text{sound}} = \sqrt{\frac{B}{\rho}} = \sqrt{\frac{\gamma p}{\rho}}

where B is the bulk modulus, p is the pressure, \rho is the density, and \gamma = C_p/C_v is the adiabatic index (\gamma = 7/5 for dry air at standard conditions). Plugging in atmospheric pressure and air density gives v \approx 340 m/s — the number you used in Example 2.

The wave equation and Maxwell's equations — why c is the speed of light

Maxwell's equations of electromagnetism (which you meet in class 12 and in JEE Advanced) contain, as a mathematical consequence, the statement that the electric and magnetic fields in vacuum satisfy

\frac{\partial^2 E}{\partial t^2} = \frac{1}{\mu_0\varepsilon_0}\,\frac{\partial^2 E}{\partial x^2}

This is the wave equation, with v^2 = 1/(\mu_0\varepsilon_0). Plug in \mu_0 = 4\pi\times 10^{-7} H/m and \varepsilon_0 = 8.854\times 10^{-12} F/m, and you get v \approx 3\times 10^{8} m/s — the speed of light. Maxwell's discovery that electromagnetic waves must travel at this speed was the recognition that light is an electromagnetic wave. The speed of light is not a separately postulated constant; it falls out of the wave equation derived from electromagnetism.

Three-dimensional wave equation

In three dimensions, the wave equation is

\frac{\partial^2 y}{\partial t^2} = v^2\left(\frac{\partial^2 y}{\partial x^2} + \frac{\partial^2 y}{\partial y^2} + \frac{\partial^2 y}{\partial z^2}\right) = v^2\nabla^2 y

where \nabla^2 is the Laplacian operator. Spherical waves (sound from a point source), cylindrical waves (ripples on a pond from a thrown stone), and plane waves (distant sound, light from the sun) are all solutions. The sphere ripples out at speed v from its source, its amplitude dropping as 1/r to conserve energy.

The dispersion relation as the real content of a wave equation

For the simple wave equation, the dispersion relation is \omega = vk — a straight line through the origin, with v the same for every frequency. This is what "non-dispersive" means.

Many physically important wave equations have curved dispersion relations. For the Klein-Gordon equation \partial^2\psi/\partial t^2 = c^2\nabla^2\psi - (mc^2/\hbar)^2\psi, the relation is \omega^2 = c^2 k^2 + (mc^2/\hbar)^2. For a water wave in deep water, \omega = \sqrt{gk}. For a quantum free particle, \omega = \hbar k^2/(2m). Each curve tells you how different frequencies travel — constant phase velocity only if the relation is linear. Everything else is called dispersion, and it is why a water wave entering shallow water changes shape, and why white light splits into colours in a prism.

Fourier's theorem — why sines are enough

Any travelling wave pulse can be written as a superposition of sines:

y(x, t) = \int_{-\infty}^{\infty} A(k)\,e^{i(kx - \omega(k)\,t)}\,dk

For the simple wave equation (non-dispersive), \omega(k) = vk and the integrand factors to e^{ik(x - vt)}. The integral becomes a function of x - vt alone — the shape-preserving form. For dispersive equations, \omega(k) is a more complex function, the components travel at different speeds, and the pulse spreads as it propagates. This is why lightning flashes from a thunderstorm arrive as a sharp crack (non-dispersive sound in air) while distant lightning flashes arrive as a long low rumble (the higher frequencies were absorbed or scattered away over the kilometres).

Where this leads next

Speed of Waves on a String — the physics behind v = \sqrt{T/\mu}, derived carefully from Newton's second law on a stretched string.
Principle of Superposition — two waves passing through each other add; this linearity is what makes the wave equation solvable and interference possible.
Introduction to Waves — the overview article: transverse versus longitudinal, mechanical versus electromagnetic, wave parameters.
Equation of SHM and Phase — the single-particle version of sinusoidal motion, which becomes the wave equation's sinusoidal solution when extended in space.
Superposition of SHMs — adding SHMs at a single point; the wave equation extends this to adding SHMs across space.