Resonance in closed pipe requires odd harmonics

Steps:

• Calculate fundamental frequency: f₁ = v/(4L) = 340/(4×0.85) = 100 Hz.
• Identify resonant frequencies: odd multiples of f₁, so 1f₁=100 Hz, 3f₁=300 Hz, 5f₁=500 Hz.
• Check 200 Hz: equals 2f₁, an even multiple.
• Confirm: even multiples do not satisfy boundary conditions for closed pipe (node at closed end, antinode at open end).

Why B is correct:

• 200 Hz is the second harmonic (even multiple), which cannot form a standing wave in a closed pipe per the quarter-wavelength resonance condition.

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