The Wave Equation

One relationship ties together the three quantities from the previous lesson and the speed the wave travels at, and it is used more often in the knowledge tests than any other wave fact. By the end of this lesson you should be able to calculate a wave's speed from its frequency and wavelength, rearrange the relationship to find either of those from the speed, and explain why frequency and wavelength must trade off against each other when the speed cannot change.

Where the relationship comes from

Stand at a fixed point and watch a wave go past. In one period — the time for one complete cycle, which the previous lesson called T — exactly one full repeat of the wave passes you, which means the pattern has advanced by exactly one wavelength. That is the whole derivation in one sentence, and everything else is arithmetic. Speed, from the mechanics module, is distance divided by time; here the distance is one wavelength and the time is one period, so the speed of the wave is the wavelength divided by the period. And since the previous lesson established that 1 divided by the period is the frequency, dividing by the period is the same as multiplying by the frequency. So the speed is the frequency multiplied by the wavelength.

Check that with the units, because it is a good habit and because it makes the formula memorable. A hertz is one per second, so multiplying hertz by metres gives metres per second — which is a speed. Multiplying the two the wrong way round is not possible here, since they are multiplied, but dividing them gives nonsense units, and that is one way of catching the commonest error in the topic.

A plain-English reading helps as well: the number of crests emitted each second, multiplied by the length of each one, is how much wave gets laid down per second — which is how far the front of it has advanced. Fifty crests a second, six metres apiece, is three hundred metres of wave per second.

The two rearrangements

Both of them come up. Divide both sides of speed = frequency × wavelength by the wavelength: on the right the wavelengths cancel, leaving $\text{frequency} = \frac{\text{speed}}{\text{wavelength}}$. Divide the original by the frequency instead and the frequencies cancel, giving $\text{wavelength} = \frac{\text{speed}}{\text{frequency}}$. As with every formula in this module, that is one fact with three faces, and deriving the face you need on the spot is safer than memorising three separate lines — particularly here, where the two divisions look alike and are easy to mix up.

Why the trade-off happens

The reason frequency and wavelength move in opposite directions is worth understanding rather than accepting, because it explains a great deal in the two lessons that follow. What sets the speed of a wave is the medium it is travelling through — how closely its particles are packed and how strongly they pull on one another — and not the source that made the wave. So for a given material the speed is a fixed number, and the product of frequency and wavelength has to come out as that number no matter what the source does.

Picture what happens if you shake the end of a rope twice as fast. Each crest still runs along the rope at the same speed, because the rope has not changed, but you are now sending crests off twice as often — so the next crest is launched when the previous one has covered only half the distance it used to, and the crests end up half as far apart. Twice the frequency, half the wavelength, same speed. Nothing was done to the crests in flight; the spacing was determined the moment they were made.

There is one honest qualification. When a wave crosses from one material into another, the speed does change, because the medium has changed — and it is the frequency that stays fixed, because the crests arriving at the boundary have to leave it at the same rate they arrive, or they would pile up. So at a boundary the wavelength changes and the frequency does not. Inside one uniform material, the speed is fixed and it is the source that decides how the fixed product is split between frequency and wavelength.

Why it matters in aviation

Aviation radio is organised around exactly this trade-off. The waves involved all travel at one speed, so choosing a frequency chooses a wavelength, and the wavelength settles both how large an aerial has to be and how the wave behaves on the way. The band used for ordinary air-to-ground speech runs from roughly 118 to 137 megahertz, which puts its wavelengths at a couple of metres — short enough for a small aerial, and short enough that the waves travel essentially in straight lines. Their range is therefore limited by the horizon rather than by transmitter power, and it increases with altitude, which is why a station unusable on the ground becomes readable once the aircraft has climbed.

Much lower frequencies have wavelengths measured in tens of metres, and waves that long are not confined to line of sight in the same way: they can follow the curve of the Earth and be reflected back down to it, so a transmission can be received thousands of miles from the transmitter. That is why long-range communication away from ground stations, over oceans and remote areas, uses those lower frequencies despite the poorer clarity. One relationship between frequency and wavelength, two quite different operational consequences.

What the test asks

The trap is arithmetic rather than conceptual: multiplying when the item wanted a division, or dividing the wrong pair. The defence is to look at what you have been given rather than reaching for a remembered line. If you have been given a frequency and a wavelength, the only thing you can do with them is multiply, and the answer is a speed. If you have been given a speed and one of the other two, you divide the speed by what you have, and the answer is the one you were missing. A second and subtler trap offers an answer with the right digits and the wrong unit — 2 hertz where the answer is 2 metres — so always name the quantity you have calculated. And expect at least one item that changes the frequency and asks what happens to the wavelength within the same medium, where the answer is always the opposite change by the same factor, never the same change.