Question 37: Waves

Question

Two tuning forks
A
and
B
are sounded together giving rise to 8 beats in 2 s. When fork
A
is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork
B
is 380 Hz then original frequency of tuning fork
A
is _________ Hz.

Accepted Answer

Before waxing the fork, A
The beat frequency is the number of beats per second.
$$ \mathrm{f}_{\mathrm{b}}=\frac{	ext { Number of beats }}{	ext { Time }}=\frac{8 	ext { beats }}{2 \mathrm{~s}}=4 \mathrm{~Hz} $$
This means the absolute difference between the frequencies of fork $A\left(f_Aight)$ and fork $B\left(f_Bight)$ is 4 Hz :
$$ \left|\mathrm{f}_{\mathrm{A}}-\mathrm{f}_{\mathrm{B}}ight|=4 \mathrm{~Hz} $$
Since $\mathrm{f}_{\mathrm{B}}=380 \mathrm{~Hz}$, the original frequency of fork A could be either :
Case 1 :
$\mathrm{f}_{\mathrm{A}}=380+4=384 \mathrm{~Hz}$
Case 2 :
$\mathrm{f}_{\mathrm{A}}=380-4=376 \mathrm{~Hz}$
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to decrease. Let the new frequency of fork $A$ be $f_A^{\prime}$, where $f_A^{\prime} < f_A$.
The new beat frequency ( $\mathrm{f}_{\mathrm{b}}^{\prime}$ ) is :
$$ \mathrm{f}_{\mathrm{b}}^{\prime}=\frac{4 	ext { beats }}{2 \mathrm{~s}}=2 \mathrm{~Hz} $$
Initially, $\mathrm{f}_{\mathrm{A}}$ is higher than $\mathrm{f}_{\mathrm{B}}$ by $4 \mathrm{~Hz}(384-380=4)$.
If $f_A$ decreases (to say 382 Hz ), the gap between $f_A$ and $f_B$ decreases $(382-380=2)$.
The original frequency of tuning fork A must have been higher than fork B to allow the beat frequency to decrease as $f_A$ dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is $\mathbf{3 8 4}$.

Waves

Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________ Hz.

Integer / Numeric Answer Type