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

For a closed pipe, the fundamental resonance occurs when the pipe length equals one-quarter of the wavelength: L = λ/4. First, λ = v/f = 340/512 ≈ 0.664 m. Therefore L = 0.664/4 ≈ 0.166 m ≈ 0.17 m. Option B (0.33 m) is λ/2, applicable to an open pipe; C and D are higher harmonics.