Physics - Sound Wave Question with Solution | TestHub

The fundamental frequency of an organ pipe open at both ends is given by $n=\frac{v}{2l}$, where $v$ is the velocity of sound and $l$ is the length of the pipe.

For the first pipe of length $L$, the frequency is:
$n_1=\frac{v}{2L}$

For the second pipe of length $L+L_1$, the frequency is:
$n_2=\frac{v}{2(L+L_1)}$

The beat frequency is the difference between the two frequencies:
$n_b=n_1-n_2$

$n_b=\frac{v}{2L}-\frac{v}{2(L+L_1)}$

$n_b=\frac{v}{2}\left(\frac{1}{L}-\frac{1}{L+L_1}\right)$

$n_b=\frac{v}{2}\left(\frac{L+L_1-L}{L(L+L_1)}\right)$

$n_b=\frac{v{L}_1}{2L(L+L_1)}$

Answer: $\frac{\mathrm{v}{\mathrm{L}}_1}{2\mathrm{L}\left(\mathrm{L}+{\mathrm{L}}_1\right)}$