Two tuning forks P and Q when set vibrating give 4 beats per second. If a prong of the fork P is filed, the beats are reduced to 2 per second. Determine the frequency of P, if that of Q is 250 Hz.

Explanation

There are four beats between P and Q, therefore the possible frequencies of P are 246 or 254 (that is `250 pm 4` ) Hz .
When the prong of P is filed, its frequency becomes greater then the original frequency. If we assume that the original frequency of P is 254, then on filing its frequency will be greater than 254. The beats between P and Q will be more than 4. But it is given that the beats are reduced to 2, therefore 254 is not possible
Therefore, the requrired frequency must be 246 Hz. (This is true, because on filing the frequency may increases to 248, given 2 beats with Q of frequency 250 H