Two tuning forks P and Q when set vibrating, give 4 beats/s. If a prong of the fork P is filled, the beats…
Two tuning forks P and Q when set vibrating, give 4 beats/s. If a prong of the fork P is filled, the beats are reduced to 2 s$^{-1}$. Determine the frequency of P, if that of Q is 250 Hz.
Solution
Sol. There are four beats between P and Q, therefore the possible frequencies of P are 246 Hz or 254 Hz (i.e. 250 $\pm$ 4) Hz.
When the prong of P is filled, its frequency becomes greater than the original frequency.
If we assume that the original frequency of P is 254 Hz, then on filing, its frequency will be greater than 254 Hz. The beats between P and Q will be more than 4. But it is given that the beats are reduced to 2, therefore 254 Hz is not possible.
Therefore, the required frequency must be 246 Hz.
(This is true because on filing, the frequency may increase to 248 Hz, giving 2 beats with Q of frequency 250 Hz).
Answer: 246 Hz