Question

If two tuning forks A and B are sounded together, they produce 4 beats per sec. A is then slightly loaded with wax and same no. of beats/sec. are produced again. If frequency of A is 256, the frequency of B would be?
A. 260
B. 262
C. 252
D. 260

Answer 1

Hint: The difference in the frequency of the two tuning forks is known as Beat frequency and this phenomenon is known as Beats. It is a phenomenon which occurs due to interference of sound waves. When two sound waves of different frequencies approach each other, then either constructive interference takes place or either destructive.

Complete step by step answer:
Since four beats per second are heard when tuning fork A and tuning fork B are sounded together, hence the difference of frequencies should be 4, let $\nu $ represent frequency of tuning forks B then, $256-v=4$.

Now, loading with wax decreases the frequency of the tuning fork.Waxing the tuning fork, A would mean the frequency of A has reduced such a way that it again produces 4 beats per second with B. Thus, it is now lower than B by 4Hz and initially it was higher than B by 4Hz.
$256-v=4$
$\therefore v=256-4=252Hz$

Hence the correct option C.

Note: Loading of the tuning fork with wax decreases its frequency as it increases the headiness of the tuning fork and thus slowing its movement. The vibrating object is the source of the disturbance that moves through the medium and when such a body vibrates which is to and fro motion it happens at a particular frequency.Always time period is to be taken in seconds and frequency in inverse of second commonly called hertz.