Question 432: Waves

Question

A closed organ pipe of length '$$\mathrm{L}_c$$' and an open organ pipe of length '$$\mathrm{L}_{\mathrm{o}}$$' 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?

Accepted Answer

For open organ pipe :
Fundamental frequency $$n=\frac{V}{2 L_o}$$
First overtone $$\mathrm{n}_1=2 \mathrm{n}=\frac{\mathrm{V}}{\mathrm{L}_{\mathrm{o}}}$$
For closed organ pipe
Fundamental frequency $$\mathrm{n}^{\prime}=\frac{\mathrm{V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}}$$
First overtone $$n_1^{\prime}=3 n^{\prime}=\frac{3 V^{\prime}}{4 L_c}$$
$$\begin{aligned}
\mathrm{n}_1 & =\mathrm{n}_1^{\prime} \quad 	herefore \frac{\mathrm{V}}{\mathrm{L}_{\mathrm{o}}}=\frac{3 \mathrm{~V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}} \
	herefore \frac{\mathrm{L}_0}{\mathrm{~L}_{\mathrm{c}}} & =\frac{4}{3} \frac{\mathrm{V}}{\mathrm{V}^{\prime}} \quad 	ext{.......(1)}\
\mathrm{V} & =\sqrt{\frac{\mathrm{k}}{ho_2}}, \mathrm{~V}^{\prime}=\sqrt{\frac{\mathrm{k}}{ho_{\mathrm{l}}}}
\end{aligned}$$
where $$\mathrm{k}$$ is the adiabatic bulk modulus, which is reciprocal of compressibility.
$$	herefore \frac{\mathrm{V}}{\mathrm{V}^{\prime}}=\sqrt{\frac{ho_1}{ho_2}}$$
Putting this value in Eq.(1)
$$\frac{\mathrm{L}_0}{\mathrm{~L}_{\mathrm{c}}}=\frac{4}{3} \sqrt{\frac{\mathrm{ho}_1}{ho_2}}$$

Answer

$$\frac{4 L_c}{3} \sqrt{\frac{\rho}{\rho_2}}$$

Answer

$$\frac{3 \mathrm{~L}_{\mathrm{c}}}{4} \sqrt{\frac{\rho_2}{\rho_1}}$$

Answer

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

Answer

$$\frac{2 L_c}{3} \sqrt{\frac{\rho_2}{\rho}}$$

Waves

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