Physics - Sound Wave Question with Solution | TestHub

Let $n_o$ be the fundamental frequency of the open pipe of length $L$.

$n_o=\frac{v}{2L}$

The frequency of the second overtone of the open pipe is $n_2=3n_o$.

When the pipe is closed at one end, its fundamental frequency becomes $n_c=\frac{v}{4L}=\frac{n_o}{2}$.

The frequency of the third overtone of the closed pipe is $n_3^{\prime }=7n_c=7\left(\frac{n_o}{2}\right)=3.5n_o$.

According to the given condition, the third overtone of the closed pipe is higher in frequency by $200$ Hz than the second overtone of the open pipe.

$n_3^{\prime }-n_2=200$

$3.5n_o-3n_o=200$

$0.5n_o=200$

$n_o=400$ Hz

Answer: 400 Hz