Question

When a tuning fork A of frequency $100Hz$ is sounded with a tuning fork B, the number of beats per second is $2$. On putting the wax on the prongs of B, the number of beats per second becomes $1$. The frequency of fork B is
A)$98Hz$
B)$99Hz$
C)$101Hz$
D)$102Hz$

Answer 1

Hint: whenever two sources of sound are turned on together and if there is a slight difference in the frequencies of the sources, then beats are produced and the number of beats is equal to the difference in the frequencies of the individual sources.

Complete step by step answer:
We know that beats are produced when two sound waves of slightly different frequencies are turned on together. The number of beats per second is given by the difference in the frequencies of the sound waves.
Here, frequency of source A$ = 100Hz$
Let the frequency of source B$ = f$
In the first case the number of beats per second$ = 2$
Which means that the difference between frequencies of a and b is $2$.
From this, we can say that the frequency of source B could be $102Hz$ or $98Hz$. This is because both $102$ and $98$ have a difference of $2$ with $100$.

Now,
The fork B is covered in wax and the number of beats per second reduces to $1$.
We know that, on applying wax on the fork, its frequency decreases because of the increase in Solution:
weight in the branches of the fork. So, from this information we can conclude that the frequency of the fork B should be $102Hz$. We can say this because if we assume the frequency to be $98Hz$, then its frequency will become even less than $98Hz$and it produces even more than $2$beats per second with the fork A.
D) is correct.

Note: The purpose of wax is to increase the weight of the branches of the wax to decrease its frequency. If we decrease the weight of the branches by some method, we can increase the frequency of the fork.