An organ pipe resonates at 400 Hz in a fixed mode when the gas inside it is at 300 K. The same gas is heated to 363 K without changing the pipe dimensions or the mode. If the new resonance frequency is N Hz, find N.
Physics · Wave Motion · organ pipe and resonance tube
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Question

An organ pipe resonates at 400 Hz in a fixed mode when the gas inside it is at 300 K. The same gas is heated to 363 K without changing the pipe dimensions or the mode. If the new resonance frequency is N Hz, find N.

Physics · Wave Motion · organ pipe and resonance tube

G1 Physics · Accepted Answer

For the same ideal gas, v ∝ √T. For a fixed pipe and the same resonance mode, f ∝ v. Therefore, f₂/f₁ = √(T₂/T₁) = √(363/300) = 1.1. Hence, f₂ = 1.1×400 = 440 Hz. Therefore, N = 440.