Physics - Wave on String Question with Solution | TestHub

Let $n_A$ and $n_B$ be the frequencies of tuning forks A and B respectively.

For the closed pipe, the fundamental frequency is $n_A=\frac{v}{4L_A}$.

$n_A=\frac{v}{4\times 20}=\frac{v}{80}$

For the open pipe, the fundamental frequency is $n_B=\frac{v}{2L_B}$.

$n_B=\frac{v}{2\times 40.5}=\frac{v}{81}$

Since $80<81$, $n_A>n_B$.

The beat frequency is given as $n_A-n_B=5$.

$\frac{v}{80}-\frac{v}{81}=5$

$v\left(\frac{81-80}{80\times 81}\right)=5$

$\frac{v}{6480}=5\Rightarrow v=32400$ cm/s

Substituting the value of $v$:

$n_A=\frac{32400}{80}=405$ Hz

$n_B=\frac{32400}{81}=400$ Hz

Answer: 405 Hz, 400 Hz