Physics - Sound Wave Question with Solution | TestHub

Let$f_0=\frac{v_0}{2L}$be the fundamental frequency of the open pipe
$\therefore$is second overtone is$3f_0=\frac{3v_0}{2L}$
Let$f=\frac{v_0}{4L}$be the fundamental frequency of close pipe.
And third overtone of the organ pipe closed at one end only is,${\left(f_3\right)}_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}}\mathrm{ }=\frac{7}{4} \frac{v_0}{L}$
As given$\frac{7v_0}{4L}-\frac{3v_0}{2L}=150$
$\left(\frac{7}{4}-\frac{6}{4}\right) \frac{v_0}{L}=150$
$\therefore \frac{v_0}{4L}=150$
$\frac{v_0}{2\mathrm{L }}=\; 300\; Hz$