Question

Derive expression for Beat frequency.​

Answer 1

For two diff waves we can 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

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).

But how often do the beats occur? Let's write (4) this way:

ytotal = {2A cos (2π Δf/2)} * cos (2π fav) (4).

So the beat frequency is simply Δf: the number of beats per second equals the difference in frequency between the two interfering waves.