Question

A tuning fork C produces 6 beats/secondwith another tuning fork D of frequency 320 Hz. When a little wax is put on the prongs of D, the number of beats reduces to 4 per second. Find the frequency of C.

Answer 1

1. Initially 6 beats per second are heard. Hence, the difference between the frequencies of the tuning forks is 6 Hz. As the frequency of fork D is 320 Hz, the frequency of fork C = 320 + 6 Hz = 326 Hz or 314 Hz

2. When the prongs of fork D are loaded with a little wax, the frequency of fork D decreases and becomes less than 320 Hz.

3. If the frequency of fork C is 326 Hz, the number of beats heard per second must increase. 

4. However, as the number of beats heard per second has decreased from 6 to 4, the frequency of fork C must be 314 Hz.