Resonance tube: which value of the speed of sound should be quoted?
A student measures the speed of sound with a resonance tube: a plastic tube of internal diameter 4.0 cm, closed by a movable piston and driven by a loudspeaker at 512 Hz. The loudest resonances occur with the piston at 16.0 cm and 49.5 cm from the open end. Using the first value alone in v = 4fL gives 328 m/s, which is 4.6 % below the value of 344 m/s expected at the room temperature of 21 °C. The student repeats at 1024 Hz and finds resonances at 7.6 cm and 24.4 cm. The open end of the tube stands free in the room; the piston position can be read to plus or minus 0.1 cm and the thermometer reads 21.0 °C with an uncertainty of 0.5 °C. Before writing up, the group must decide whether to quote 328 m/s, 343 m/s or 344 m/s, and how to justify the choice.
1. Explain why v = 4fL with L = 16.0 cm underestimates the speed of sound. Calculate the end correction from the data and hence obtain the best value of the speed of sound.
Read the worked answer
The displacement antinode does not sit at the plane of the open end; it lies a small distance e outside it, so the effective length of the first resonance is L + e, not L. Ignoring e shortens the quarter-wavelength and therefore underestimates v by the ratio 16.0/16.75 ≈ 0.955, i.e. by about 4.5 % - exactly the observed shortfall.
The key move is to use the spacing of successive resonances, because the same e is added to both positions and cancels: L₂ − L₁ = λ/2 = 33.5 cm, so λ = 67.0 cm and v = fλ = 512 × 0.670 = 343 m/s. Then λ/4 = 16.75 cm gives e = 16.75 − 16.0 = 0.75 cm, which is of the expected order for a tube of radius 2.0 cm (rule of thumb e ≈ 0.6r).
The 1024 Hz pair checks this: spacing 16.8 cm gives λ = 33.6 cm and v = 344 m/s, and e = 8.40 − 7.6 = 0.80 cm, agreeing with 0.75 cm to within the 1 mm reading precision. Quoting 328 m/s is simply wrong; the correct value is about 343–344 m/s.
2. The 1024 Hz pair gives 344 m/s and the 512 Hz pair 343 m/s. Assess whether the two results are consistent, using the reading precision and the temperature uncertainty.
Read the worked answer
Position readings are ±0.1 cm, so a spacing of ΔL has uncertainty √2 × 0.1 = 0.14 cm. At 512 Hz, δλ/λ = 0.14/33.5 = 0.42 %, giving δv ≈ 1.4 m/s. At 1024 Hz, δλ/λ = 0.14/16.8 = 0.83 %, giving δv ≈ 2.9 m/s. So the two results are 343 ± 1 and 344 ± 3 m/s - consistent well within one standard uncertainty, with a combined best value near 343.5 m/s.
Temperature: v ∝ √T, and near 21 °C the sensitivity is about 0.61 m s⁻¹ per °C, so ±0.5 °C contributes only ±0.3 m/s. It is a systematic uncertainty (the whole room is at one temperature) and is negligible against the 1.4 m/s from length readings.
The expected 344.0 m/s is therefore consistent with the measurement, and the 16 m/s discrepancy of the naive method is an instrumental/model error, not a random one.
3. Explain why the separation of successive resonances gives λ/2 regardless of the end correction, and why the higher-frequency pair is not automatically the more accurate one.
Read the worked answer
Each resonance satisfies L_n + e = (2n − 1)λ/4, with the same e at both, so subtracting gives L₂ − L₁ = λ/2 with e eliminated. This is a differential measurement: it removes a systematic offset that would otherwise enter every single-position calculation.
Halving the wavelength is tempting because the resonances are sharper, but the spacing also halves, from 33.5 cm to 16.8 cm. The absolute reading uncertainty of ±0.1 cm does not shrink, so its fractional effect doubles: 0.42 % at 512 Hz against 0.83 % at 1024 Hz. The higher-frequency pair is therefore less precise for a single v determination, but valuable as a consistency check, because agreement of e at two frequencies justifies the assumption that e is a fixed property of the tube mouth and not a fudge factor.