Question:

A closed organ pipe of length \( L_c \) and an open organ pipe of length \( L_o \) 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?

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Remember: For a closed pipe, only odd harmonics exist. First overtone = third harmonic (3f₁). For an open pipe, first overtone = second harmonic (2f₁). Speed of sound in a gas is inversely proportional to the square root of density when bulk modulus is constant.
Updated On: Jun 8, 2026
  • \(\frac{4L_c}{3} \sqrt{\frac{\rho}{\rho_2}}\)
  • \(\frac{3L_c}{4} \sqrt{\frac{\rho_2}{\rho_1}}\)
  • \(\frac{4L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}\)
  • \(\frac{2L_c}{3} \sqrt{\frac{\rho_2}{\rho}}\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
A closed pipe and an open pipe contain different gases (densities \(\rho_1\) and \(\rho_2\)) but same compressibility (hence same bulk modulus \(B\)). Both vibrate in their first overtone with identical frequency. We need \(L_o\) in terms of \(L_c\) and densities.

Step 2: Key Formula or Approach:
For a closed pipe, first overtone = third harmonic: \(f_c = \frac{3v_1}{4L_c}\).
For an open pipe, first overtone = second harmonic: \(f_o = \frac{v_2}{L_o}\).
Speed of sound in a gas: \(v = \sqrt{\frac{B}{\rho}}\). Since \(B\) is same, \(v \propto \frac{1}{\sqrt{\rho}}\).

Step 3: Detailed Explanation:
Let \(v_1 = \sqrt{B/\rho_1}\) and \(v_2 = \sqrt{B/\rho_2}\). Equate frequencies: \[ \frac{3v_1}{4L_c} = \frac{v_2}{L_o} \quad \Rightarrow \quad L_o = \frac{4L_c}{3} \cdot \frac{v_2}{v_1}. \] Now \(\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}}\). Thus: \[ L_o = \frac{4L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}. \] In the given options, option (A) writes \(\sqrt{\frac{\rho}{\rho_2}}\) where \(\rho\) stands for \(\rho_1\). Hence (A) is correct.

Step 4: Final Answer:
The length of the open organ pipe is \(\frac{4L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}\), which corresponds to option (A).
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