Physics - Wave on String Question with Solution | TestHub

Let $L$ be the original length of both pipes and $v$ be the speed of sound.

Frequency of the first overtone of the closed pipe, $n_{c1}=\frac{3v}{4L}$

Frequency of the first overtone of the open pipe, $n_{o1}=\frac{2v}{2L}=\frac{v}{L}$

Given that they produce 4 beats per second when vibrating simultaneously:

$n_{o1}-n_{c1}=4$

$\frac{v}{L}-\frac{3v}{4L}=4$

$\frac{v}{4L}=4\Rightarrow \frac{v}{L}=16$

Now, the length of the open pipe is made half, $L_o^{\prime }=\frac{L}{2}$.

New fundamental frequency of the open pipe, $n_o^{\prime }=\frac{v}{2L_o^{\prime }}=\frac{v}{2(L/2)}=\frac{v}{L}=16$ Hz

The length of the closed pipe is made two times, $L_c^{\prime }=2L$.

New fundamental frequency of the closed pipe, $n_c^{\prime }=\frac{v}{4L_c^{\prime }}=\frac{v}{4(2L)}=\frac{v}{8L}=\frac{1}{8}\left(\frac{v}{L}\right)=\frac{16}{8}=2$ Hz

Number of beats produced when vibrating in their fundamental modes simultaneously = $n_o^{\prime }-n_c^{\prime }=16-2=14$

Answer: 14