A closed organ pipe of length ' $L_c$ ' and an open organ pipe of length ' $L_0$ ' contain different gases…
A closed organ pipe of length ' $L_c$ ' and an open organ pipe of length ' $L_0$ ' contain different gases of densities ' $\rho_1$ ' and ' $\rho_2$ ' respectively. The compressibility of the gases is the same in both the pipes. The gases are vibrating in their first overtone with the same frequency. What is the length of open organ pipe?
$\frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$
$\frac{3 L_c}{4} \sqrt{\frac{\rho_2}{\rho_1}}$
$\frac{4 L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}$
$\frac{2 L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}$
Solution
For open organ pipe:
Fundamental frequency $\mathrm{n}=\frac{\mathrm{V}}{2 \mathrm{~L}_{\text {o }}}$
First overtone $\mathrm{n}_1=2 \mathrm{n}=\frac{\mathrm{V}}{\mathrm{L}_{\text {。 }}}$
For closed organ pipe
Fundamental frequency $n^{\prime}=\frac{V^{\prime}}{4 L_c}$
First overtone $n_1{ }^{\prime}=3 n^{\prime}=\frac{3 \mathrm{~V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}}$
$\begin{aligned}
& \because \mathrm{n}_1=\mathrm{n}_1^{\prime} \\
& \therefore \frac{\mathrm{V}}{\mathrm{L}_{\mathrm{o}}}=\frac{3 \mathrm{~V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}} \\
& \therefore \frac{\mathrm{L}_{\mathrm{o}}}{\mathrm{L}_{\mathrm{c}}}=\frac{4}{3} \frac{\mathrm{V}}{\mathrm{V}^{\prime}} \\
& \mathrm{V}=\sqrt{\frac{\mathrm{k}}{\rho_2}}, \mathrm{~V}^{\prime}=\sqrt{\frac{\mathrm{k}}{\rho_1}}
\end{aligned}$
where $\mathrm{k}$ is the adiabatic bulk modulus, which is reciprocal of compressibility.
$\frac{L_0}{L_c}=\frac{4}{3} \sqrt{\frac{\rho_1}{\rho_2}}$
: