Question

When two tuning forks, fork $A$ and fork $B$ are sounded together, $4$ beats per second are heard. When the tuning fork $B$ is loaded with wax then $6$ beats per second are heard. If the frequency of fork $A$ is $200\text{Hz}$ then what was the original frequency of fork $B$?

• A. $209\text{Hz}$
• B. $196\text{Hz}$
• C. $202\text{Hz}$
• D. $200\text{Hz}$

Answer 1

The correct option is B $196\text{Hz}$

Given,
$f_b=4\text{Hz}$
We know, beat frequency
$f_b=|f_1-f_2|$
$\Rightarrow f_2=f_1\pm f_b$
Frequency of tuning fork $A$,
$f_1=200\text{Hz}$
$\Rightarrow f_2=200\pm 4\text{Hz}$
So, frequency of tuning fork $B$ can be,
$f_2=204\text{Hz}$ or $196\text{Hz}$
When tuning fork $B$ is loaded with wax then,
$f_b=6\text{Hz}$
$\Rightarrow f_2^{{}^{\prime }}=200\pm 6\text{Hz}$
So, frequency of tuning fork $B$ can be,
$f_2^{{}^{\prime }}=206\text{Hz}$ or $194\text{Hz}$

As the tuning fork $B$ is loaded with wax, therefore its time period should increase and frequency should decrease.
So, by intuition, the original frequency should be $196\text{Hz}$ which will decrease to $194\text{Hz}$ when loaded with wax.

OR

Given,
$f_b=4\text{Hz}$
We know, beat frequency
$f_b=|f_A-f_B|=4\text{Hz}$
On waxing B, its frequency reduces and beat requency increases.
Hence,
$f_A-f_B=4\text{Hz}$
$200-f_B=4\text{Hz}$
$f_B=196\text{Hz}$