Question

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

• A. $\frac{4 L}{3} \sqrt{\frac{\rho_{2}}{\rho_{1}}}$
• B. $\frac{4 L}{3} \sqrt{\frac{\rho_{1}}{\rho_{2}}}$
• C. $\frac{4 L}{3}$
• D. $\frac{L}{3}$

Answer 1

Answer: (B)

For closed organ pipe,
$f =f_{2}=3 f_{1}$
$=\frac{3 \times V}{4 L}=\frac{3}{4 L} \times \sqrt{\frac{\gamma p}{\rho_{1}}}$
For open organ pipe,
$f =f_{2}=2 f_{1}=\frac{2 v}{2 L'}=\frac{V}{L'}$
$=\frac{1}{L'} \times \sqrt{\frac{\gamma p}{\rho_{2}}}$
For same sound frequency,
$\frac{3}{4 L} \times \sqrt{\frac{\gamma P}{\rho_{1}}} =\frac{1}{L'} \times \sqrt{\frac{\gamma p}{\rho_{2}}}$
$L' =\frac{4 L}{3} \times \sqrt{\frac{\rho_{1}}{\rho_{2}}}$