Resonance Tube: Is the 500 Hz Fork Mislabelled?
A tuning fork marked 500 Hz is held over a resonance tube whose lower end is closed by a movable water surface. The first strong resonance is heard when the air column is 17.0 cm long and the next when it is 52.0 cm long. A thermometer in the laboratory reads 30 °C and the barometer reads normal atmospheric pressure. The student must report a value for the speed of sound in air, with an uncertainty, and say whether the tuning fork is likely to be accurately marked. Take the speed of sound in dry air at 0 °C as 331 m/s and assume the speed varies as the square root of the absolute temperature.
The teacher warns that the tube is narrow, so the displacement antinode lies a little outside the open end, and that a 1 mm error in the 17.0 cm reading matters much more than a 1 mm error in the 52.0 cm reading. The student also wonders whether the air is 'too warm' for the fork to be trusted.
1. From the two resonance lengths find the wavelength, the speed of sound in the laboratory air and the end correction of the tube. State the uncertainty in the speed if each length is read to ±1 mm.
Read the worked answer
For a tube closed at one end, successive resonances are separated by λ/2: λ = 2(l₂ − l₁) = 2(0.520 − 0.170) = 0.700 m. v = fλ = 500 × 0.700 = 350 m/s.
End correction. l₁ + e = λ/4 and l₂ + e = 3λ/4, so e = (l₂ − 3l₁)/2 = (0.520 − 0.510)/2 = 0.005 m = 5 mm. Check: 0.170 + 0.005 = 0.175 m = λ/4 ✓ and 0.520 + 0.005 = 0.525 m = 3λ/4 ✓. (A 5 mm correction corresponds to about 0.3 × the tube radius, the expected size.)
Uncertainty. A ±1 mm error in either length changes (l₂ − l₁) by up to ±2 mm, so λ changes by ±0.004 m and v by ±2 m/s, about 0.6 %. Report v = (3.50 ± 0.02) × 10² m/s.
2. Predict the speed of sound at 30 °C using v ∝ √T and decide whether the fork is likely to be mislabelled. What would change at 35 °C?
Read the worked answer
v = 331 √(T/273) with T = 303 K: v = 331 √(303/273) = 331 × 1.0534 = 348.7 m/s ≈ 349 m/s.
The measured 350 m/s is only 1.3 m/s (0.4 %) above this - smaller than the ±2 m/s reading uncertainty. Moist laboratory air is less dense than dry air and adds a further ~0.2 %, so the measurement sits comfortably on the prediction.
The frequency implied by the measurement and the theoretical speed is f = 348.7/0.700 = 498 Hz, only 0.3 % below the mark, well inside normal tuning-fork tolerance. So there is no evidence of mislabelling; a claim of mislabelling would need a discrepancy clearly outside the uncertainty, or a beat-count against a calibrated source.
At 35 °C (308 K): v = 331 √(308/273) = 331 × 1.0622 = 351.6 m/s, so λ = v/f = 0.7032 m and l₁ = λ/4 − e = 0.1758 − 0.005 = 0.1708 m, l₂ = 3λ/4 − e = 0.5274 − 0.005 = 0.5224 m. Both resonances move outward by only about 2 mm, so a 5 K temperature change is not a serious error source here - but it must be recorded if a precision better than 1 % is claimed.
3. Explain why the resonance lengths are combined as (l₂ − l₁) rather than using l₁ alone, and why the second resonance is the better one to use.
Read the worked answer
The difference cancels the end correction: (l₂ + e) − (l₁ + e) = λ/2, so λ = 2(l₂ − l₁) contains no unknown e. Using l₁ alone forces you to assume a value for e (about 0.3 × tube radius), and that assumed value carries its own error because the antinode is not at a sharply defined plane - the open end radiates, so the effective length is slightly longer than the tube.
Relative error: 1 mm in 170 mm is 0.6 %, whereas 1 mm in 520 mm is 0.2 %, and the difference of 350 mm gives 0.3 % - the subtraction uses the well-measured long column and rejects the badly-measured short one.
Physical disturbance: the first resonance at λ/4 places the antinode right at the mouth, where the fork, the hand holding it and the surrounding air disturb the field; the second at 3λ/4 is less affected and gives a louder, sharper maximum.
A useful check is to look for a third resonance near 5λ/4 − e = 0.870 m; agreement with 87 cm confirms both λ and e.