Question

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384Hz$. When one of the prongs of the fork $A$ is filed and sounded with$B$, then beat frequency increases, then the frequency of the fork $A$ is
A.$380Hz$
B.$388Hz$
C.$379Hz$
D.$389Hz$

Answer 1

Hint:Filing is a process in which the frequency of a tuning fork is increased. Waxing is a process in which the frequency of a tuning fork is decreased. The number of beats produced when two frequencies are produced at the same time is the magnitude of the difference between the two given frequencies.

Formula used:
$N = {f_1} - {f_2}$
where $N$ is the number of beats, and
${f_1}$ and ${f_2}$ are the two frequencies involved such that ${f_1} > {f_2}$.

Complete answer:
It is given that the number of beats is $4$ beats per second and the frequency of $B$ is $384Hz$.
We know that
$N = {f_1} - {f_2}$
where $N$ is the number of beats, and
${f_1}$ and ${f_2}$ are the two frequencies involved such that ${f_1} > {f_2}$.
On filing tuning forks $A$ its frequency increases.
If $A$ is the tuning fork with maximum frequency then after filing beat frequency i.e. the number of beats produced will increase.
It is given that the same is happening therefore from basic reasoning we can conclude that $A$ is the tuning fork with greater frequency and $B$ is the tuning fork with lower frequency.
Hence on substituting the values in the above formula we get,
${f_1} - 384 = 4$
where ${f_1}$ is the frequency of the tuning fork $A$.
${f_1} = 388Hz$

Therefore the correct answer to the above question is (B) $388Hz$ .

Note:
Prong is the part of the tuning fork which produces sound when struck. In filing the prongs of the tuning fork are rubbed along a rough surface and made sharper and lighter which increases the tuning fork’s frequency. In waxing a substance is added to the prongs of the tuning fork which increases their mass and decreases the tuning fork’s frequency.