Question 402: Waves

Question

A closed organ pipe of length '$$L_1$$' and an open organ pipe contain diatomic gases of densities '$$\rho_1$$' and '$$\rho_2$$' respectively. The compressibilities of the gases are same in both pipes, which are vibrating in their first overtone with same frequency. The length of the open organ pipe is (Neglect end correction)

Accepted Answer

Given both gases are vibrating in the first overtone with same frequency, we get
$$\begin{aligned}
& \mathrm{f}_{	ext {closed }}=\mathrm{f}_{	ext {open }} \
& \Rightarrow \frac{3 \mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2}
\end{aligned}$$
According to Laplace's correction
$$\begin{aligned}
& v=\sqrt{\frac{\gamma \mathrm{P}}{ho}} \
& \frac{3}{4 \mathrm{~L}_1} 	imes \sqrt{\frac{\gamma \mathrm{P}}{ho_1}}=\frac{1}{\mathrm{~L}_2} 	imes \sqrt{\frac{\gamma \mathrm{P}}{ho_2}} \
	herefore \quad \mathrm{L}_2 & =\frac{4 \mathrm{~L}_1}{3} \sqrt{\frac{ho_1}{ho_2}}
\end{aligned}$$

Answer

$$\frac{4 \mathrm{~L}_1}{3}$$

Answer

$$\frac{4 L_1}{3} \sqrt{\frac{\rho_1}{\rho_2}}$$

Answer

$$\frac{4 L_1}{3} \sqrt{\frac{\rho_2}{\rho_1}}$$

Answer

$$\frac{3}{4 L_1} \sqrt{\frac{\rho_1}{\rho_2}}$$

Waves

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