Resonance in a closed-end tube

Steps:

• In a closed tube, the fundamental mode has a node at the closed end and antinode at the open end, so length L = λ/4 and f = v/(4L).
• The problem states a node at the closed end and no nodes at the open end for frequency f, confirming the fundamental mode.
• The next resonance occurs at the third harmonic, where L = 3λ/4, so frequency = 3v/(4L) = 3f.
• Even harmonics are absent in closed tubes due to the boundary conditions requiring odd quarter-wavelengths.

Why D is correct:

• The third harmonic frequency is exactly 3f, as derived from the resonance condition L = (2n+1)λ/4 for odd integers n.

Why the others are wrong:

• A: 1.25f does not match any harmonic multiple for a closed tube.
• B: 1.50f is not a resonance frequency; only odd multiples occur.
• C: 2.00f would be the first even harmonic, impossible in a closed tube.

Final answer: D