Physics - Wave on String Question with Solution | TestHub

For a closed pipe, the frequency of the $p$-th overtone is given by $n_c=(2p+1)\frac{v}{4L_c}$.

For the fourth overtone ($p=4$), the frequency is $n_c=9\frac{v}{4L_c}$.

For an open pipe, the frequency of the $p$-th overtone is given by $n_o=(p+1)\frac{v}{2L_o}$.

For the fifth overtone ($p=5$), the frequency is $n_o=6\frac{v}{2L_o}$.

Given that the two frequencies are in unison:

$9\frac{v}{4L_c}=6\frac{v}{2L_o}$

$\frac{9}{4L_c}=\frac{3}{L_o}$

$\frac{L_c}{L_o}=\frac{9}{12}=\frac{3}{4}$

The ratio of the length of the closed pipe to that of the open pipe is $3:4$.

Answer: $3:4$