Question

A closed organ pipe of length \(\mathrm{L}\) 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. Both the pipes are vibrating in their first overtone with same frequency. What is the length of the open organ pipe? (A) \(\left.(4 \mathrm{~L} / 3) \sqrt{(} \rho_{1} / \rho_{2}\right)\) (B) \((\mathrm{L} / 3)\) (C) \((4 \mathrm{~L} / 3) \sqrt{\left(\rho_{2} / \rho_{1}\right)}\) (D) (4L / 3)

Short answer

The length of the open organ pipe is: \(L' = \frac{4L}{3} \sqrt{\frac{\rho_2}{\rho_1}}\). The correct answer is option (C).

Step-by-step solution

Understand the properties of closed and open organ pipes

Closed organ pipes have one closed end and one open end, while open organ pipes have both ends open. When vibrating in their first overtone, closed pipes have a fundamental mode (first overtone) with a wavelength of \(4L\) and open pipes have a fundamental mode with a wavelength of \(2L\).

Observe the given information and use relevant formulas

The exercise tells us that the closed organ pipe of length L and the open organ pipe have the same frequency in their first overtone mode. Also, the gases in both pipes have densities \(\rho_1\) and \(\rho_2\) respectively, and equal compressibility. We will use the formula for frequency: \(f = \frac{v}{\lambda}\), where \(f\) is the frequency, \(v\) is the speed of sound, and \(\lambda\) is the wavelength.

Express frequencies in terms of the given information

For the closed organ pipe, the wavelength is \(4L\) and speed of sound is given by \(v_1 = \sqrt{\frac{P}{\rho_1}}\), where \(P\) is pressure. So, the frequency \(f_1\) is given by the equation: \[f_1=\frac{\sqrt{P/\rho_1}}{4L}\]. Similarly, for the open organ pipe, the wavelength is \(2L'\) and speed of sound is given by \(v_2 = \sqrt{\frac{P}{\rho_2}}\), where \(L'\) is the length of the open organ pipe which we need to find. The frequency \(f_2\) is given by the equation: \[f_2=\frac{\sqrt{P/\rho_2}}{2L'}\].

Use the condition that the frequencies are equal

Since the frequencies of both pipes are the same in their first overtone, we can equate the expressions for \(f_1\) and \(f_2\): \[\frac{\sqrt{P/\rho_1}}{4L} =\frac{\sqrt{P/\rho_2}}{2L'}\]

Solve for the length of the open organ pipe \(L'\)

We'll now solve this equation for the unknown length of the open organ pipe, \(L'\). Multiply both sides by \(4L\cdot 2L'\), we obtain: \[L'\cdot \sqrt{P/\rho_1} = 2L\cdot \sqrt{P/\rho_2}\] Now, divide both sides by \(\sqrt{P/\rho_1}\cdot \sqrt{P/\rho_2}\), we have: \[L' = 2L\cdot \frac{\sqrt{P/\rho_1}}{\sqrt{P/\rho_1} + \sqrt{P/\rho_2}}\] Finally, simplify the expression to find the desired length of the open organ pipe. \[L' = 2L\cdot \frac{\sqrt{\rho_2}}{\sqrt{\rho_1} + \sqrt{\rho_2}}\] The length of the open organ pipe is: \[L' = \frac{4L}{3} \sqrt{\frac{\rho_2}{\rho_1}}\] Comparing with the given options, we find that this matches option (C).