Resonance Tube: Which Length Difference Is Trustworthy?
A student uses a vertical resonance tube open at the top and closed by a water surface. A 512 Hz tuning fork is held above the open end. Resonance is loudest when the air-column lengths are 16.2 cm and 49.0 cm. Room temperature is 22.0°C, and the accepted speed of sound is about 344 m/s. The tube diameter is 3.0 cm. Student A computes v = 2f(L2 − L1) = 336 m/s. Student B uses only L1 = λ/4 and gets 332 m/s. Student C says the tube is open at both ends, so L1 = λ/2, and gets 166 m/s. The students argue about whether the 8 m/s discrepancy from 344 m/s is due to temperature, the tuning fork, or the open end. The water level is adjusted slowly, and lengths are read to the nearest 0.1 cm.
1. Identify the correct boundary conditions at the water surface and at the open end. Which student's resonance interpretation is correct, and what is the wavelength from the two resonances? Calculate the speed.
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The water surface is a displacement node because the water is essentially rigid; the open top is a displacement antinode. A closed-open pipe resonates when L + e = (2n − 1)λ/4, so the first two resonances are separated by λ/2. Therefore Student A's interpretation is correct: L2 − L1 = λ/2, so λ = 2(0.490 − 0.162) = 0.656 m, and v = fλ = 512(0.656) = 335.9 m/s. Student B's first-resonance-only formula ignores the end correction. Student C is wrong because an open-open pipe would have antinodes at both ends and a different resonance spacing; moreover, the water surface cannot be a displacement antinode. The observed ratio 49.0/16.2 ≈ 3.0 is consistent with the closed-open pattern.
2. Estimate the end correction for this tube and calculate the speed that would be inferred from the first resonance alone with and without that correction. Why is the two-resonance method more reliable?
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A common estimate is e ≈ 0.6r = 0.6(0.015 m) = 0.009 m, about 0.9 cm. With the correction, L1 + e = 0.171 m = λ/4, so λ = 0.684 m and v = 350.2 m/s. Without the correction, λ = 4(0.162) = 0.648 m and v = 331.8 m/s. The two-resonance difference cancels the end correction because both resonances are shifted by the same effective length; the first-only method does not. The 4% spread between the corrected and uncorrected first-resonance values shows that the end correction is not negligible here. The difference method is therefore more reliable, provided both resonance positions are identified consistently.
3. The accepted speed at 22.0°C is about 344 m/s. Decide whether the remaining discrepancy is more likely due to temperature, tuning-fork frequency, or measurement uncertainty. What additional measurement would settle it?
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The discrepancy is 344 − 336 = 8 m/s, or 2.4%. Temperature cannot explain it: using v ≈ 331.3 + 0.606T, a speed of 336 m/s corresponds to about 7.8°C, not 22°C. A tuning-fork frequency of f = v/(2ΔL) = 344/(0.656) ≈ 524 Hz would explain it, a 2.4% error; check the fork with a frequency counter or another fork. Measurement uncertainty in ΔL is also relevant: ±0.5 cm gives δv ≈ 2fδ(ΔL) ≈ 5.1 m/s, so 336 ± 5 m/s is still about 1.6 standard deviations low. The settling measurement is to verify the fork frequency and repeat with several forks or a longer tube to see whether the discrepancy persists.