An organ pipe resonates at 400 Hz in a fixed mode when it contains an ideal gas at 300 K. The same gas is heated to 363 K while the same mode remains excited. Neglect thermal expansion of the pipe. What is the new resonance frequency?
Options

Question

An organ pipe resonates at 400 Hz in a fixed mode when it contains an ideal gas at 300 K. The same gas is heated to 363 K while the same mode remains excited. Neglect thermal expansion of the pipe. What is the new resonance frequency?

Options

G1 Physics · Accepted Answer

For the same ideal gas, v ∝ √T. For a fixed pipe and the same mode, the wavelength is fixed by the effective length. Hence, f ∝ v ∝ √T. Therefore, f₂/f₁ = √(T₂/T₁) = √(363/300) = √1.21 = 1.1. Thus, f₂ = 1.1(400) = 440 Hz. Correct option: B.