Question

Question #30ca9

Answer 1

Ultimately due to definitions

Explanation:

Consider perhaps the simplest example - a pure sine wave with frequency `f` and wavelength `lambda` that changes both in time `t` and a spatial direction `x`:
`f(t,x) = sin( (2 pi x)/lambda - 2pi f t)`.

The frequency `f` tells us how rapidly the wave oscillates if we keep position `x` fixed (fix position at e.g. `x = 0`).

The wavelength `lambda` tells us how far apart two peaks of the wave are, if we look at it at a fixed moment of time (fix time at e.g. `t = 0`).

If we allow the wave to vary in both space and time, then we can track for which times and positions the so-called phase of the wave is constant. This way we can find a relation to the speed of the wave `v`.

But first, what is phase? The phase tells us where in the cycle of the wave we are. For example, the phase is `90` degrees at a maximum (a peak) and `270` degrees at a minimum (trough), since `sin(90 deg) = 1` and `sin(270 deg) = -1`.

Now, back to the question. If we want to find a relation between `x` and `t` such that the phase is constant, then we require that `(2 pi x)/lambda - 2pi f t=0`.

Simplifying, we get that
`x/(tf) = lambda`.
But recall now that speed (if constant, as in this case) is given by `v = x/t`. Plugging this in, we get that
`v/f = lambda`,
which is the relation that you sought.

It can be tricky to imagine the two dimensions time and space at the same time. Perhaps someone else can provide some nice graphics? Oddly enough, I couldn't find any online.

From an intuitive standpoint, the wave speed is how far a wave-crest moves in space for a given interval of time. This is related to wavelength and frequency as follows:

For a fixed frequency holds that: the longer the wavelength, the faster the wave must travel in order to track a peak. For a fixed wavelength holds that: The higher the frequency, the faster must the wave must travel in order to be able to track a peak.