A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

Sol. Group – I Group – II Given V = 350 v = 350 λ base 1 = 32 cm λ base 2 = 32.2 cm = 32 × 10^–2 m = 32.2 × 10^–2 m So η base 1 = frequency = 1093 Hz η base 2 = 350 / 32.2 × 10^–2 = 1086 Hz So beat frequency = 1093 – 1086 = 7 Hz.

Beats per second = 4 Frequency of the fork given= 256HzLet the frequency of the unknown fork be x.Possible frequencies 252Hz or 260Hz...So acc. To ques. On loading with wax the beat per second increases to 6....Since the beat per second increases with the loading of the fork with wax hence the answer= 252Hz (as we know the frequency decreases on loading the fork with wax)