Closed pipe resonance frequencies are odd harmonics of the fundamental

Steps:

• Calculate fundamental frequency: f₁ = v/(4L) = 340/(4×0.85) = 100 Hz.
• Resonance occurs at odd multiples: f = (2n+1)f₁, where n=0,1,2,...
• For n=0: 100 Hz; n=1: 300 Hz; n=2: 500 Hz.
• 200 Hz is even multiple (2f₁), not possible.

Why B is correct:

• In a closed pipe, stationary waves require a node at the closed end and antinode at the open end, allowing only odd harmonics per the boundary condition λ/4 = L for fundamental.

Why the others are wrong:

• A (100 Hz): Matches fundamental frequency f₁.
• C (300 Hz): Matches third harmonic 3f₁.
• D (500 Hz): Matches fifth harmonic 5f₁.

Final answer: B