Wave Speed

Stand at the seaside and watch a wave roll in. It moves — a bump of water sets off far out and marches steadily towards the beach. Shout across a valley and your voice races off through the air to the far side. Flick on a torch and the light leaps to the wall in a flash. Every wave travels, and every wave travels at some speed.

Wave speed is simply how fast the wave carries its energy along — how far a crest travels each second. A ripple on a pond dawdles across at walking pace; sound sprints through the air at about 340\ \text{m/s}; light and every other electromagnetic wave scream along at a mind-bending 3\times10^{8}\ \text{m/s}. This page is about the one neat rule that ties a wave's speed to the two numbers you can measure on it — its frequency and its wavelength.

Speed the everyday way: distance over time

A wave crest is a moving thing, so its speed is worked out exactly like the speed of a car or a runner — the distance it covers divided by the time it takes:

v = \frac{\text{distance}}{\text{time}}

Here v is the wave speed in metres per second (m/s). If you spot a wave crest travel 20\ \text{m} along a rope in 4\ \text{s}, its speed is v = 20 \div 4 = 5\ \text{m/s}. This is always true and is how wave speed is often measured. But for waves there is a second, more powerful way to get the speed — one that uses the shape of the wave itself.

Two numbers that describe a wave

Every repeating wave carries two key measurements, and we need both to build the wave equation:

Wavelength \lambda (the Greek letter "lambda") — the length of one complete wave, measured for example from one crest to the very next crest. It is a distance, in metres (m).
Frequency f — how many complete waves pass a fixed point each second. It is a count-per-second, measured in hertz (Hz). One hertz means one whole wave every second; 50\ \text{Hz} means fifty waves whizz past every second.

Frequency has a close partner, the period T — the time for one whole wave to pass, in seconds. If five waves pass each second, each one takes a fifth of a second, so frequency and period are simply upside-down versions of each other:

T = \frac{1}{f}, \qquad f = \frac{1}{T}.

A wave of frequency f = 4\ \text{Hz} has period T = \tfrac{1}{4} = 0.25\ \text{s}: one wave every quarter-second.

The wave equation: v = f\lambda

Now put the two measurements together. In one second, f whole waves stream past you, and each one is \lambda metres long. So the total length of wave that sweeps past in that one second is f lots of \lambda — and length-per-second is exactly speed. That single line of reasoning is the wave equation.

For any wave, the speed equals the frequency multiplied by the wavelength:

v = f\lambda

v — the wave speed, in metres per second (m/s)
f — the frequency, in hertz (Hz)
\lambda — the wavelength, in metres (m)

Check the units and it makes sense at once: hertz is "per second" (\text{s}^{-1}) and wavelength is in metres, so f\lambda comes out in \text{m} \times \text{s}^{-1} = \text{m/s} — a speed. The equation is a multiplication: frequency times wavelength, never one divided by the other.

See it move: squeeze the waves in

Below is a snapshot of a wave travelling to the right along a stretch of 10\ \text{m}. Drag the wavelength slider: shorten \lambda and more waves squeeze into the same distance; stretch it and the waves spread out. Change the frequency too, and watch the live readout of v = f\lambda at the top climb or fall. The one rule holds at every setting: multiply the two numbers and you have the speed.

Notice the trade-off this reveals: if you keep the speed fixed, then packing in more waves per second (higher f) forces each wave to be shorter (smaller \lambda), because their product must stay the same. Frequency and wavelength pull in opposite directions for a fixed speed.

Rearranging for f or \lambda

Because v = f\lambda links three quantities, knowing any two hands you the third. Rearrange it just like any equation — divide both sides by whichever quantity you already know:

v = f\lambda, \qquad f = \frac{v}{\lambda}, \qquad \lambda = \frac{v}{f}.

A handy way to remember all three at once is the "magic triangle": write v on top with f and \lambda underneath. Cover the one you want, and the triangle shows you the sum — cover v and you see f \times \lambda; cover f and you see v \div \lambda.

Worked examples

Example 1 — find the speed of a sound wave. A musical note has frequency f = 170\ \text{Hz} and wavelength \lambda = 2\ \text{m}. How fast does the sound travel?

v = f\lambda = 170 \times 2 = 340\ \text{m/s}.

That is the everyday speed of sound in air — just what we expect.

Example 2 — find the frequency of a water ripple. Ripples cross a pond at v = 0.6\ \text{m/s} with a wavelength of \lambda = 0.2\ \text{m}. What is their frequency? Rearrange to make f the subject:

f = \frac{v}{\lambda} = \frac{0.6}{0.2} = 3\ \text{Hz}.

Three complete ripples pass a floating leaf every second.

Example 3 — find the wavelength of a radio wave. A radio station broadcasts at f = 100\ \text{MHz} = 1\times10^{8}\ \text{Hz}. Radio is an electromagnetic wave, so it travels at v = 3\times10^{8}\ \text{m/s}. What is its wavelength?

\lambda = \frac{v}{f} = \frac{3\times10^{8}}{1\times10^{8}} = 3\ \text{m}.

Every colour of light, every radio and TV signal, every microwave and X-ray is an electromagnetic wave, and they all travel through space at the very same 3\times10^{8}\ \text{m/s} — so for them, a higher frequency always means a shorter wavelength.

Measuring wave speed in the lab

You do not need fancy kit to find a wave's speed — you just need to measure two of the three quantities and use the equation.

Water ripples in a tank: use a strobe or a video to count how many ripples pass a mark each second (that is f), and a ruler to measure the crest-to-crest distance (that is \lambda). Then v = f\lambda.
Sound with an echo: stand a measured distance from a wall, clap, and time the echo. The sound travels there and back, so v = \dfrac{2 \times \text{distance}}{\text{time}}.
Waves on a rope: shake one end, time a single crest travelling the length of the rope, and use v = \dfrac{\text{distance}}{\text{time}}.

This is the single most-tested tripwire on the whole topic. When a wave passes from one material into another — sound from air into water, light from air into glass — its speed changes, and so its wavelength changes too. But its frequency stays exactly the same.

Why? The frequency is set by whatever made the wave — the source. The waves are still arriving at the boundary at the same rate they were made, so the same number per second must carry on into the new material. Nothing can pile up or vanish at the boundary.
Since v = f\lambda and f is fixed, if v changes then \lambda must change in step with it. Light slowing down in glass gets a shorter wavelength, but the same frequency — which is exactly why it bends (refracts).

So the reflex "the wave slows down, so its frequency drops" is wrong. Speed down, wavelength down, frequency unchanged.

Two smaller traps that quietly cost marks:

The equation is v = f\lambda, a multiplication. Not v = f/\lambda and not v = \lambda/f. Sanity-check with units: hertz (per second) times metres gives metres per second, a speed. Dividing them would not.
Hertz means "per second". A frequency of 5\ \text{Hz} is 5 waves every second, not 5 waves in total. And keep your units SI: f in Hz and \lambda in metres, or v comes out wrong. A wavelength given in centimetres must be turned into metres first (20\ \text{cm} = 0.2\ \text{m}), and a frequency in kilohertz or megahertz into hertz (3\ \text{kHz} = 3000\ \text{Hz}).

Both are made at the same instant by the same flash — yet the light reaches you almost immediately while the thunder rumbles in seconds later. It is all down to wave speed. Light travels at 3\times10^{8}\ \text{m/s}, so from a storm a couple of kilometres away it arrives in a millionth of a second. Sound crawls along at a mere 340\ \text{m/s} — nearly a million times slower — so it takes about three seconds to cover each kilometre.

That gives you a free rule for how far away a storm is: count the seconds between the flash and the bang, and divide by three to get the distance in kilometres. Ten seconds between them? The storm is roughly 10 \div 3 \approx 3\ \text{km} away — and using v = \text{distance}/\text{time} in reverse is exactly how you worked it out.