A tube of length ℓ and diameter d is initially open at both ends, with fundamental frequency f₁. One end is then closed and the lowest resonance frequency is f₂. Including end correction, find f₂/f₁.

Question

G1 Physics · Accepted Answer

Step 1: For the open pipe, effective length is ℓ+0.6d, so f₁ = v/[2(ℓ+0.6d)]. Step 2: After one end is closed, effective length is ℓ+0.3d, so f₂ = v/[4(ℓ+0.3d)]. Step 3: f₂/f₁ = ½(ℓ+0.6d)/(ℓ+0.3d). Final: C.