Physics - Sound Wave Question with Solution | TestHub

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

For the third overtone ($p=3$), the frequency is:
$n_c=(2(3)+1)\frac{v}{4L_c}=\frac{7v}{4L_c}$

For a pipe open at both ends, the frequency of the $p$-th overtone is given by $n_o=(p+1)\frac{v}{2L_0}$.

For the sixth overtone ($p=6$), the frequency is:
$n_o=(6+1)\frac{v}{2L_0}=\frac{7v}{2L_0}$

Given that the frequency of the third overtone of the closed pipe is equal to the frequency of the sixth overtone of the open pipe:
$\frac{7v}{4L_c}=\frac{7v}{2L_0}$

$\frac{1}{4L_c}=\frac{1}{2L_0}$

$\frac{L_c}{L_0}=\frac{2}{4}=\frac{1}{2}$

The ratio $L_c:L_0$ is $1:2$.

Answer: $1:2$