Question

A fork of unknown frequency gives $4beats/\sec$ when sounded with another fork of frequency $256Hz$. The fork is now loaded with a piece of wax and again $4beats/\sec$ are heard. Then the frequency of the unknown fork is:
(A) $256Hz$
(B) $252Hz$
(C) $264Hz$
(D) $260Hz$

Answer 1

Hint
To answer this question, we need to see the effect of waxing on the frequency of the fork. Applying the formula for the number of beats will give two values of the frequency, one of which will be rejected.
$\Rightarrow f = \dfrac{{{{1.875}^2}}}{{2\pi {l^2}}}\sqrt {\dfrac{{EI}}{{\rho A}}}$
Where $f$ is the frequency of a tuning fork, $l$ is the length of the prongs, $E$ is the Young’s modulus, $\rho$ is the density of fork material.

Complete step by step answer
Let the frequency of the unknown fork be $f$
The frequency of the other fork is given to be equal to $256Hz.$
We know that the number of beats is equal to the difference in frequency of the two sources.
So, two cases are possible.
Case I: When the frequency of the unknown fork is greater.
In this case, the number of beats is given as
$\Rightarrow n = f - 256$
Substituting$n = 4$
$\Rightarrow f - 256 = 4$
$\Rightarrow f = 260Hz$
Case II: When the frequency of the unknown fork is smaller.
In this case, the number of beats is given as
$\Rightarrow n = 256 - f$
Substituting$n = 4$
$\Rightarrow 256 - f = 4$
$\Rightarrow f = 252Hz$
So, we have two possible values of the frequency of the unknown fork, $252Hz$ and $252Hz$.
Now, the unknown fork is loaded with the wax. So, the mass of the fork will increase which will increase its density $\rho$. We know that the frequency is given by the relation
$\Rightarrow f = \dfrac{{{{1.875}^2}}}{{2\pi {l^2}}}\sqrt {\dfrac{{EI}}{{\rho A}}}$
So, with the increase in the density $\rho$ of the fork, its frequency decreases.
Let us assume that the second case is correct, that is, the frequency of the unknown fork is smaller. So the unknown fork frequency is $4Hz$ less than the frequency of the known fork. After the frequency is decreased, the unknown fork frequency will become further smaller; making the magnitude of the difference between the two frequencies, or equivalently the number of beats increase. But according to the question, the number of beats is the same.
So, Case II is rejected.
Thus, Case I is correct, that is, the frequency of the unknown fork is $260Hz.$
Hence the correct answer is option (D), $260Hz.$

Note
We need not remember the formula for the frequency of the fork for studying the effect of waxing on the frequency. As we know that the frequency of a wave in a string is given by
$\Rightarrow f = \sqrt {\dfrac{T}{\mu }}$
Where $\mu$ is the linear mass density. So, the frequency in this case also is proportional to the inverse root of density. So, density may be linear, superficial, or volumetric. The proportionality doesn’t change.