A closed organ pipe of length ' $L$ ' and an open organ pipe contain gases of densities $\rho_1$ and…
A closed organ pipe of length ' $L$ ' and an open organ pipe contain gases of densities $\rho_1$ and $\rho_2$ respectively. The compressibility of gases are equal in both the pipes. If the frequencies of their first overtones are same, then the length of the open organ pipe is
$\frac{4 L}{3} \sqrt{\frac{\rho_2}{\rho_1}}$
$\frac{4 L}{3} \sqrt{\frac{\rho_1}{\rho_2}}$
$\frac{4 L}{3}$
$\frac{L}{3}$
Solution
For closed organ pipe, Ist overtone $=3 \mathrm{rd}$ harmonic
$
f=f_2=3 f_1=\frac{3 \times v}{4 L}=\frac{3}{4 L} \times \sqrt{\frac{\gamma p}{\rho_1}}
$
For open organ pipe, Ist overtone $=2 \mathrm{nd}$ harmonics
$
f=f_2=2 f_1=\frac{2 v}{2 L^{\prime}}=\frac{v}{L^{\prime}}=\frac{1}{L^{\prime}} \times \sqrt{\frac{\gamma p}{\rho_2}}
$
As $B$ is same. So pressure $(\alpha \beta)$ is also same for both gas For same sound frequency
$
\begin{aligned}
& \frac{3}{4 L} \times \sqrt{\frac{\gamma p}{\rho_1}}=\frac{1}{L^{\prime}} \times \sqrt{\frac{\gamma p}{\rho_2}} \\
& \therefore \quad L^{\prime}=\frac{4 L}{3} \times \sqrt{\frac{\rho_1}{\rho_2}}
\end{aligned}
$