Physics - Sound Wave Question with Solution | TestHub

For an open pipe, the frequency of the $n$-th harmonic is given by $f=n\frac{v}{2L_0}$.
The fundamental frequency corresponds to $n=1$. The fifth overtone corresponds to the 6th harmonic, so $n=6$.
Frequency of the fifth overtone of the open pipe is $f_o=6\frac{v}{2L_0}=\frac{3v}{L_0}$.

For a pipe closed at one end, the frequency of the $m$-th harmonic is given by $f=m\frac{v}{4L_c}$, where $m$ takes only odd integer values ($m=1,3,5,\dots$).
The fundamental frequency corresponds to $m=1$. The first overtone is $m=3$, the second is $m=5$, and so on. The fifth overtone corresponds to $m=2(5)+1=11$.
Frequency of the fifth overtone of the closed pipe is $f_c=11\frac{v}{4L_c}$.

Since the two overtones are in unison, their frequencies are equal:
$\frac{3v}{L_0}=\frac{11v}{4L_c}$

Rearranging to find the ratio of $L_0$ to $L_c$:
$\frac{L_0}{L_c}=\frac{3\times 4}{11}=\frac{12}{11}$

Answer: $12:11$