Physics - Sound Wave Question with Solution | TestHub

Let the length of the pipe be $L$ and the speed of sound be $v$.

The fundamental frequency of the open pipe is $f_o=\frac{v}{2L}$.

The first overtone of the open pipe is $f_{o1}=2f_o=\frac{v}{L}$.

When the pipe is closed at one end, its fundamental frequency becomes $f_c=\frac{v}{4L}$.

The second overtone of the closed pipe corresponds to the fifth harmonic, so $f_{c2}=5f_c=\frac{5v}{4L}$.

According to the given condition, $f_{c2}-f_{o1}=100$.

$\frac{5v}{4L}-\frac{v}{L}=100$

$\frac{v}{4L}=100$

The fundamental frequency of the open pipe is $f_o=\frac{v}{2L}=2\times \left(\frac{v}{4L}\right)$.

$f_o=2\times 100=200$ Hz.

Answer: $200$ Hz