Resonance Tube and the End Correction
A student uses a resonance tube to measure the speed of sound in air. A tuning fork of frequency 512 Hz is held above the open end of a tube whose water level can be adjusted. The first loud resonance is found with the air column length L1 = 0.155 m and the second with L2 = 0.490 m. Lengths are read from a metre rule with an uncertainty of ±0.002 m. The frequency is given as 512 Hz with an uncertainty of ±1 Hz. The student first calculates λ = 4L1, obtaining 0.620 m, and then v = fλ = 317 m/s. This is 8% below the expected value of about 343 m/s at room temperature, so the student concludes that the tuning fork is mislabelled. A classmate suggests using the difference between the two resonance lengths instead.
1. Explain why λ = 4L1 is not correct for a real tube, and derive the relation between L1, L2 and the end correction e.
Read the worked answer
In a real tube the displacement antinode is not exactly at the open end; it lies a small distance e outside the tube. The first resonance therefore has L1 + e = λ/4, and the second has L2 + e = 3λ/4.
Subtracting the two equations eliminates e: (L2 + e) − (L1 + e) = 3λ/4 − λ/4, so L2 − L1 = λ/2 and λ = 2(L2 − L1).
Using λ = 4L1 assumes e = 0. That is why the single-length calculation gives a wavelength that is too short by about 4e: 4L1 = λ − 4e. The error is not a property of the tuning fork.
2. Calculate e and the corrected speed of sound from the two lengths. Compare your value with the expected value.
Read the worked answer
λ = 2(0.490 − 0.155) = 2 × 0.335 = 0.670 m.
v = fλ = 512 × 0.670 = 343 m/s.
e = (L2 − 3L1)/2 = (0.490 − 3 × 0.155)/2 = (0.490 − 0.465)/2 = 0.0125 m.
The corrected value, 343 m/s, agrees with the expected speed of sound at room temperature (about 331 + 0.6 × 20 = 343 m/s). The naive value was 317 m/s because it ignored the end correction.
3. Determine the percentage uncertainty in the corrected speed and evaluate whether the data support the student's claim that the tuning fork is mislabelled.
Read the worked answer
Uncertainty in the difference: ΔL = 0.335 m with uncertainties ±0.002 m on each length. Combining in quadrature gives ±√(0.002² + 0.002²) = ±0.0028 m, which is 0.0028/0.335 = 0.85%.
Frequency uncertainty: 1/512 = 0.20%.
Combined percentage uncertainty in v: √(0.85² + 0.20²) ≈ 0.87%, i.e. about ±3 m/s.
So v = (343 ± 3) m/s. The expected value 343 is well within this range. The data do not support the claim that the tuning fork is mislabelled; the discrepancy in the student's first calculation is fully explained by the neglected end correction, not by a wrong frequency.