Question

A closed organ pipe of length $L$ and an open organ pipe contain gases of densities ${\rho }_1$ and ${\rho }_2$ respecticely. The compressibility of gases are eqaul in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is $\frac{x}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}$. Where $x$ is (Round off to the Nearest Integer)

Answer 1

Speed of sound:$$v=\sqrt{\frac{B}{\rho }}=\sqrt{\frac{1}{K\rho }}$$ (compressibility $K=\frac{1}{B}$)

For closed pipe $f_1=\frac{3v}{4L}=\frac{3}{4L}\sqrt{\frac{1}{K_1{\rho }_1}}$

Let the length of the open pipe be $L_2$.
For open pipe $f_2=\frac{2v}{2L_2}=\frac{1}{L_2}\sqrt{\frac{1}{K_2{\rho }_2}}$

from the question, $f_1=f_2$
$\frac{3}{4L}\sqrt{\frac{1}{K_1{\rho }_1}}=\frac{1}{L_2}\sqrt{\frac{1}{K_2{\rho }_2}}$

$\Rightarrow \frac{3}{4L}\sqrt{\frac{{\rho }_2}{{\rho }_1}}=\frac{1}{L_2}$

$\Rightarrow L_2=\frac{4L}{3}\sqrt{\frac{{\rho }_1}{{\rho }_2}}$

Now, comparing with given equation: $\frac{x}{3}L\sqrt{\frac{{\rho }_1}{{\rho }_2}}=\frac{4L}{3}\sqrt{\frac{{\rho }_1}{{\rho }_2}}$
We get, $x=4$