Question

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

Answer 1

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.