Question

You have two tuning forks. One is a known frequency of 200 Hz. The second for his plate, there is a big frequency of five beats per second. What a piece of gum is placed on the second fork, the frequency increases three beats per second. What is the frequency of the second tuning fork?

Answer 1

sin w t + sin x t = 2 sin.5(w+x)t cos.5(w-x)t

beat frequency is
omega = 2 pi f = .5(w-x)
assume gum decreases x
thereby increasing w-x
so
w is our 200 Hz fork
w = 200*2 pi = 400 pi radians/s
w - x = 5*2 pi = 10 pi
so
x = 390 pi = 195 Hz

I suppose that is more than you wanted to know but that is where it comes from :)