91Ó°ÊÓ

\(f_{closed}\) = the frequency of the closed organ pipe \(f_{open}\) = the frequency of the open organ pipe \(L_{closed}\) = the length of the closed organ pipe \(L_{open}\) = the length of the open organ pipe The formula for the frequency of a closed organ pipe is given by: \(f_{closed} = \frac{(2n - 1) v}{4L_{closed}}\) The formula for the frequency of an open organ pipe is given by: \(f_{open} = \frac{2n v}{2L_{open}}\) Where: - \(n\) is the mode number - \(v\) is the speed of sound in the gas Since the question mentioned the first overtone frequency, which is the second harmonic frequency, the value of n is 2 in both cases.

Given that the first overtone frequency of the both organ pipes is the same, we can write: \(\frac{(2 \cdot 2 - 1) v}{4L} = \frac{2 \cdot 2 v}{2L_{open}}\)

To solve for the length of the open organ pipe, we will simplify the equation: \(\frac{3v}{4L} = \frac{4v}{2L_{open}}\) Divide both sides by \(v\): \(\frac{3}{4L} = \frac{2}{L_{open}}\) Now, we can rearrange for \(L_{open}\): \(L_{open} = \frac{8}{3} L\) However, the question also states that both gases have the same compressibility but different densities (\(\rho_1\) and \(\rho_2\)). In this case, we should also take the densities into account: The speed of sound in a gas is given by the formula \(v = \sqrt{\frac{\kappa P}{\rho}}\), where \(\kappa\) is the adiabatic index, \(P\) is the pressure, and \(\rho\) is the density of the gas. As both gases have the same compressibility, their speeds of sound are related by \(\sqrt{\frac{\rho_1}{\rho_2}}\). We can use this relationship to obtain the length of the open organ pipe in terms of the closed organ pipe:

\(L_{open} = \frac{8}{3} L \sqrt{\frac{\rho_1}{\rho_2}}\) Comparing our answer to the provided multiple-choice options, we find that the correct answer is: (B) \(L_{open} = \frac{4L}{3} \sqrt{\frac{\rho_{1}}{\rho_{2}}}\)