Resonance tube: trusting two readings and an end correction
A student measures the speed of sound with a resonance tube: a glass tube whose lower end is closed by a water surface that can be raised and lowered. Holding a 512 Hz tuning fork at the open top, she finds the air column resonates when the water surface is 0.155 m below the top and again when it is 0.495 m below the top. The room is at 21 °C. She calculates a wavelength from the two lengths and obtains a speed. A second student objects that the first resonance should be a quarter of a wavelength and that the tube diameter has been ignored, so 'the answer must be wrong'. The internal diameter of the tube is about 5 cm. The fork is stamped 512 Hz at 20 °C. The metre rule is read to the nearest millimetre.
1. Calculate the speed of sound from the two lengths. Show why the end correction does not affect this calculation, then find the end correction and test it against the tube diameter.
Read the worked answer
Lengths: consecutive resonances in a tube closed at one end are separated by half a wavelength. l2 - l1 = 0.495 - 0.155 = 0.340 m = λ/2, so λ = 0.680 m.
v = fλ = 512 x 0.680 = 348 m/s.
Why the end correction cancels: the antinode sits a distance e beyond the open top, so l1 + e = λ/4 and l2 + e = 3λ/4. Subtracting gives l2 - l1 = λ/2 with e eliminated. Measuring the difference of two resonances avoids having to know the end correction.
End correction: e = λ/4 - l1 = 0.170 - 0.155 = 0.015 m. The rule of thumb e ≈ 0.3d gives d ≈ 0.015 / 0.3 = 0.05 m, i.e. 5 cm, which matches the stated tube diameter. The numbers are self-consistent, which is strong evidence that the two readings are genuine resonances and not a mis-identified minimum.
2. Predict the first resonance length for a 440 Hz fork, using the same apparatus. Explain why a student who ignored the end correction would wrongly blame the fork.
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Use the measured speed, since the same tube and same air are involved: λ = 348 / 440 = 0.791 m, so λ/4 = 0.198 m. With the end correction: l1 = λ/4 - e = 0.198 - 0.015 = 0.183 m.
The trap: using λ/4 = l1 directly gives 0.198 m, a 15 mm error. That is far larger than the ±1 mm reading precision, so the resonance would appear at the 'wrong' place and the student would suspect the 440 Hz fork or the speed of sound. The discrepancy is entirely geometrical: the air column behaves as if it is 15 mm longer than the water level suggests. Any prediction made with a tube of a different diameter must use that tube's end correction, since e depends on the bore, not on the frequency.
3. The theoretical speed of sound in dry air at 21 °C is about 344 m/s. Decide whether the measured 348 m/s is acceptable or indicates a systematic error, and state what you would do next.
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Uncertainty budget: the wavelength rests on a difference of two readings, giving about ±2 mm on 0.340 m, i.e. ±0.6%, plus the assumed exactness of the fork. That is roughly ±2 m/s. The discrepancy is 4 m/s (about 1.2%), which is about twice the estimated uncertainty - large enough to be worth investigating, small enough not to be dismissed.
Possible causes: humidity raises the speed in air by up to about 0.3% at these temperatures, worth roughly 1 m/s. Temperature uncertainty of 1 °C is worth only 0.6 m/s, so the whole 4 m/s cannot be blamed on the thermometer. The end correction being slightly different from 0.3d affects each resonance separately but cancels from the difference.
Verdict: 348 m/s is a defensible value, not an outright error, but the fair conclusion is 'consistent within about 1%'. Next: repeat with several forks, plot the resonance lengths against 1/f, and take the gradient, which averages the reading errors and gives a value with a proper uncertainty.