Question

When two tuning forks A and B are sounded together, $4$beats per second are heard. The frequency of the fork B is384$$$$Hz. When one of the prongs of the fork A is filed and sounded with B, then beat frequency increases, the frequency of the fork A is
(A) $380$ $Hz$
(B) 388$$$$Hz
(C) 379$$$$Hz
(D) 389$$$$Hz

Answer 1

Hint Beat is produced by interference of waves and its frequency is equal to the difference of two close frequencies (the two sound waves have less difference in their frequency but they are not equal).

Complete step by step answer
We have two tuning forks A and B. The frequency of fork B is {v_b} = 384$$$$Hz
Let us suppose that the frequency of fork A is$x$, that is {v_a} = x$$$$Hz
We know that the beat frequency is equal to the difference of the given two close frequencies so, the beat frequency is given by formula,
${v_{beat}} = {v_a} \sim {v_b}$
Now, applying the given values in the above formula we get,
$4 = x - 384$
Therefore, $x = 384 - 4$
Here $x$ can be either $380$ or$388$, since these values will give the difference of $4$
It is given in the question that when one of the prongs of fork A is filled and sounded with B, then the beat frequency increases. After filing, the mass of that prong decreases and so the frequency of fork A increases since a lower mass increases natural frequency.
Let us suppose that the frequency of fork A is increases by 1$$$$Hz
Now the frequency of fork A becomes either $381$ or $389$
Given that after filing the beat frequency increases so the difference of the frequencies of fork A and B should now be greater than $4$
If we take $381$ as the new frequency of A and we know the frequency B is $384$ so, the beat frequency is,
${v_{beat}} = 381 \sim 384$which is equal to $3$, thus beat frequency is decreases
But if we take $389$ as the new frequency of A then,
${v_{beat}} = 389 \sim 384$ which is equal to 5, thus beat frequency increases.

Hence, the original frequency of fork A before filing is 388$$$$Hz. Thus, the correct option is B.

Note While calculating the beat frequency, note that we are taking the ‘difference’ of two frequencies so it does not matter that we are subtracting a smaller value from a greater value or vice-versa because either way the difference remains the same.