Question

A fork A has a frequency 2\[\%\] more than the standard fork and B has a frequency 3\[\%\] less than the frequency of the same standard fork. The forks A and B when sounded together produced 6\[\text{beats/s}\]. The frequency of fork A is –
A) 116.4Hz
B) 120Hz
C) 122.4Hz
D) 238.8Hz

Answer 1

Hint: We can easily find the frequency of the forks which has deviation from the actual frequency by substituting in proper relations. The rate of beats gives the difference between the frequencies of the fork A and fork B, which makes the solution easier.

Complete step by step answer:
We are given two forks A and B which are supposed to have an equal frequency but are found to have deviations from the actual value. The percentage change of the frequencies of the two forks are given as 2\[\%\] more for fork A and 3\[\%\] less for fork B.
Now, let us relate our information to the standard value of frequency.
Let \[\eta \] be the frequency of the standard fork,
\[{{\eta }_{A}}\]be the frequency of fork A and
\[{{\eta }_{B}}\] be the frequency of fork B.
Now, we can relate these as –
\[\begin{align}
  & {{\eta }_{A}}=\eta +\dfrac{2}{100}\eta \text{ --(1)} \\
 & {{\eta }_{B}}=\eta -\dfrac{3}{100}\eta \text{ ---(2)} \\
\end{align}\]
Now, we are given that the beats per second when the two forks are used simultaneously is 6. This can be related as –
\[\begin{align}
  & {{\eta }_{A}}-{{\eta }_{B}}=6 \\
 & \Rightarrow \eta +\dfrac{2}{100}\eta -(\eta -\dfrac{3}{100}\eta )=6 \\
 & \Rightarrow \dfrac{5}{100}\eta =6 \\
 & \Rightarrow \eta =120Hz \\
\end{align}\]
So, we get the standard fork’s frequency as 120Hz.
The frequency of fork A can be found from (1) as –
\[\begin{align}
  & {{\eta }_{A}}=\eta +\dfrac{2}{100}\eta \text{ } \\
 & \Rightarrow {{\eta }_{A}}=120+\dfrac{2}{100}120 \\
 & \Rightarrow {{\eta }_{A}}=120+2.4 \\
 & \therefore {{\eta }_{A}}=122.4Hz \\
\end{align}\]
The frequency of fork A is 122.4Hz.

So, the correct answer is “Option C”.

Note: The frequency of tuning forks is required to be standardised for getting accurate outcomes for experiments. The forks needed to be calibrated using a standard fork for this purpose. The calibration is done by comparing the frequency of the given fork and the standard fork.