Question

A tuning fork $A$ produces $4$ beats per second with another tuning fork $B$ of frequency $320\,Hz$. On filling one of the prongs of $A$, $4$ beats per second are again heard when sounded with the same fork $B$. Then, what is the frequency of the fork $A$ before filling?
(A) $328\,Hz$
(B) $316\,Hz$
(C) $324\,Hz$
(D) $320\,Hz$

Answer 1

In this problem it is given that there are two forks and the frequency of the second fork is given. The first fork gives $4$ beats per second when it is tuned with the second fork. Again, the first fork gives $4$ beats per second when sounded with the same fork. It is a logical problem, so no formula is required to determine the solution.

Complete step by step answer:
A tuning fork $A$ produces $4$ beats per second with another tuning fork $B$, so there are $4$ beats between $A$ and $B$, therefore by possible frequency for $A$ is greater than or less than by $4$ with the frequency of $B$. So, the possible frequencies are $316\,Hz$or $324\,Hz$, in other terms $\left( {320 \pm 4} \right)\,Hz$. When one of the prongs of $A$ is filled its frequency will become greater than the original frequency. If the original frequency of $A$ is $324\,Hz$, filling its frequency will become greater than $324\,Hz$. The beats between $A$ and $B$ will be more than $4$. But it is given that the beats are again increased by $4$, therefore $324\,Hz$ is not possible. So, the required frequency must be $316\,Hz$. So, by the first statement given in the question, filling the frequency may increase so as to give $4$ beats with $B$ of frequency $320\,Hz$.
Thus, the frequency of the fork $A$ is $316\,Hz$ before filling.
Hence, option (B) is the correct answer.

Note: First we have to understand the question completely, a tuning fork $A$ produces $4$ beats per second with another tuning fork $B$ of frequency $320\,Hz$, which means the frequency of $A$ is greater than or less than the frequency of $B$ by $4$ beats.