Question

A tuning fork of unknown frequency gives 4 beats/s. With another fork of frequency 310 Hz, it gives the same number of beats/s when loaded with wax. Find the unknown frequency.

Answer 1

Let us name the tuning fork of unknown frequency as A and the tuning fork of known frequency i.e., 310 Hz as B. When A and B are sounded together, 4 beats are heard per second. 

Thus, frequency of A, i.e.,v is either (310 + 4) Hz = 314 Hz

or v = (310 – 4) Hz = 306 Hz

In case, the frequency of A is 306 Hz, on loading it with wax, its frequency decreases and may become 305 Hz, 304 Hz, 303 Hz etc. In that case, the number of beats given by it per second when sounded with B will be (310 – 305) = 5, (310 – 304) = 6, (310 – 303) = 7 etc. i.e., the number of beats/ s is more than 4. Thus, v = 306 Hz.If v = 314 Hz, on loading A with wax, its frequency may decrease to 306 Hz and then it will give (310 – 306) = 4 beats/s with B. Thus, the unknown frequency is 314 Hz.