Question

A tuning fork of unknown frequency makes eight beats per second with a standard fork of frequency 468.0 Hz. The beat frequency decreases when a small piece of wax is put on a prong of the fork of unknown frequency.
What is the frequency of this fork?

Answer 1

The beat frequency is just the difference in frequencies between the two forks.

That means that there are two possible frequencies for the unknown fork:

(468+ 8) = 476Hz

or

(468 - 8) = 460 Hz

We need to find out which it is. Putting a blob of wax on the unknown fork will lower its frequency slightly. If the unknown fork's frequency was 460 Hz, artificially making it a bit lower will make the frequency difference between the two forks bigger. That's not the case.

So the unknown fork must have a frequency of 476 Hz. Artificially reducing it slightly reduces the frequency difference between the two forks, and hence reduces the beat frequency, which is what is observed.