Physics - Sound Wave Question with Solution | TestHub

The fundamental frequency of a pipe open at both ends of length $L$ is given by $f=\frac{v}{2L}$.

When the pipe is dipped vertically in water such that $\frac{1}{3}$ of its length is submerged, the length of the air column becomes $L^{\prime }=L-\frac{L}{3}=\frac{2L}{3}$.

The pipe now acts as a closed pipe at one end (the water surface acts as a rigid boundary). The fundamental frequency of a closed pipe of length $L^{\prime }$ is $f^{\prime }=\frac{v}{4L^{\prime }}$.

Substituting $L^{\prime }=\frac{2L}{3}$, we get:

$f^{\prime }=\frac{v}{4\left(\frac{2L}{3}\right)}=\frac{3v}{8L}$

Since $f=\frac{v}{2L}$, we can rewrite $f^{\prime }$ as:

$f^{\prime }=\frac{3}{4}\left(\frac{v}{2L}\right)=\frac{3}{4}f$

Comparing this with $f^{\prime }=xf$, we get $x=\frac{3}{4}$.

Answer: $\frac{3}{4}$