Physics - Sound Wave Question with Solution | TestHub

Let the length of the pipe be $L$ and the velocity of sound in air be $v$.

For a pipe open at both ends, the fundamental frequency is given by:
$n_1=\frac{v}{2L}$

When the pipe is kept with ${\frac{3}{4}}^{\text{th}}$ of its length in water, the length of the air column above the water surface becomes:
$L^{\prime }=L-\frac{3L}{4}=\frac{L}{4}$

The pipe now acts as a pipe closed at one end (the water surface acts as the closed end). The fundamental frequency of this closed pipe is:
$n_2=\frac{v}{4L^{\prime }}$

Substituting $L^{\prime }=\frac{L}{4}$:
$n_2=\frac{v}{4\left(\frac{L}{4}\right)}=\frac{v}{L}$

The ratio of the frequencies is:
$\frac{n_1}{n_2}=\frac{\frac{v}{2L}}{\frac{v}{L}}=\frac{1}{2}$

Answer: $\frac{1}{2}$