A vibrating tuning fork of frequency 512 Hz is held over a resonance tube. If the speed of sound in air is 340 m/s, what is the shortest length of air column that will resonate with the tuning fork? (Neglect end correction)

For a closed pipe, resonance first occurs at a length of λ/4. The wavelength λ = v/f = 340/512 ≈ 0.664 m, so the shortest resonating length = λ/4 ≈ 0.664/4 ≈ 0.166 m. However, for an open pipe the fundamental is λ/2 = 0.332 m ≈ 0.33 m. The standard resonance tube experiment uses a closed tube, giving λ/4 ≈ 0.17 m for the first resonance. The answer 0.17 m corresponds to the closed-tube first resonance (λ/4), making A the expected answer; however, 0.33 m is λ/2 for an open tube. For a closed resonance tube, the shortest resonating length = λ/4 = 340/(4×512) ≈ 0.166 m ≈ 0.17 m.