Interference beats and Tartini tones

Two waves of similar amplitude produce interference beats if the frequency difference is small, and Tartini tones if the difference is larger. Both are useful important in practice for measuring frequencies and for tuning musical instruments. It is also worth looking at for the insight it gives to Heisenberg's Uncertainty Principle, as we shall see below.

graph of dB scale

Let's consider two waves with the same amplitude A, and frequencies f1 and f2 that are not very different. Before we do any maths, we can see what will happen by looking at the diagram. In this plot, the wave depicted by the red graph plus that depicted by the purple one gives that represented by the blue curve. The horizontal axis is usually time. Suppose that the waves start out in phase, so that they add up, as shown at the left of the diagram. The wave depicted by the red graph has a slightly higher frequency than that depicted by the purple one, so it gradually gains on it, and eventually gets one half cycle out of phase. At this point, the two component waves cancel out, and the amplitude of the blue curve is near zero. After an equal interval of time, they get back in step again, as shown.

If the waves are sound waves, what will this sound like? Well, provided that the difference in frequency is small enough, the resultant wave will sound loud when the two components are in phase and soft or absent when they are out of phase. The frequency of the blue wave is, if you look carefully, about halfway between that of the red and the purple. So we should hear a wave of intermediate frequency, getting regularly louder and softer. This is the acoustical example of the phenomenon of interference beats.

(If the difference in frequencies is greater than a few tens of cycles per second, we won't recognise the very short period of cancellation as being soft, and in fact we'll get some interesting effects, to which we return later.)

Now let's get quantitative. For the purple and red waves respectively, let's write
    y1 = A cos (2π f1)t     and y2 = A cos (2π f2t),     so

    ytotal = y1 + y2 = A {cos (2π f1t) + cos (2π f2t)}    (1).

To get any further, we need the trigonometric identity that
    cos (a+b) = cos a * cos b - sin a * sin b,     from which it follows that
    cos (a-b) = cos a * cos b + sin a * sin b.
Adding these two equations gives us
    cos (a+b) + cos (a-b) = 2 cos a * cos b    (2).
We now use this identity by making the substitions
    a = 2π (f1t + f2t)/2   and   b = 2π (f1t - f2t)/2,     so
    a + b = 2π f1t       and       a - b = 2π f2t.
We now substitute this into equation (2) to get
    cos (2π f1t) + cos (2π f2t) = 2 cos (2π (f1t + f2t)/2) * cos (2π (f1t - f2t)/2)     (3)
Now we recognise (f1 + f2)/2) as the average frequency fav and (f1 - f2) as the frequency difference Δf.
Finally, we multiply (3) by A to get:
    ytotal = y1 + y2 = 2A {cos (2π Δf/2)t} * {cos (2π fav)t}     (4).
On the diagram below, we see that the maximum amplitude of the compound wave is twice that of the two interfering waves. The carrier wave indeed has the average frequency, as you can verify by counting cycles on the diagram.