Physics - Sound Wave Question with Solution | TestHub

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

The frequency of the second overtone of the open pipe corresponds to the third harmonic, so $f_{o2}=3f_o$.

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

The frequency of the third overtone of the closed pipe corresponds to the seventh harmonic, so $f_{c3}=7f_c=7\left(\frac{f_o}{2}\right)=3.5f_o$.

Given that the third overtone of the closed pipe is higher in frequency by $150\text{ Hz}$ than the second overtone of the open pipe:

$f_{c3}-f_{o2}=150$

$3.5f_o-3f_o=150$

$0.5f_o=150$

$f_o=300\text{ Hz}$

Answer: $300\text{Hz}$