Resonance Tube, End Correction and a Beat-Rate Check
A student investigates resonance in a vertical tube closed at the lower end by a water column whose level can be adjusted. A tuning fork of stated frequency 512 Hz is held above the open end and the water level is lowered until a loud sound is heard. The first resonance occurs at air-column length 16.0 cm and the second at 50.0 cm. The room temperature is 24 °C. The student calculates v = fλ using λ = 4L1 and obtains 328 m/s, which is lower than the textbook value of 343 m/s, and concludes that the tuning fork is mislabelled. A second student suggests that the difference is due to an end correction and that the fork frequency should be checked by finding the beat frequency with a 512 Hz signal from a loudspeaker; 4 beats per second are heard.
1. Use the two resonance lengths to find the wavelength, the speed of sound and the end correction. Explain why λ = 4L1 gives a misleading speed.
Read the worked answer
In a tube closed at one end, resonance occurs when L + e = (2n − 1)λ/4. Subtracting successive conditions eliminates e:
L2 − L1 = λ/2 → 0.500 − 0.160 = 0.340 m → λ = 0.680 m
v = fλ = 512 × 0.680 = 348 m s⁻¹
e = λ/4 − L1 = 0.170 − 0.160 = 0.010 m = 1.0 cm
The formula λ = 4L1 ignores the end correction: it gives λ = 0.640 m and v = 328 m s⁻¹. The end correction is 1.0 cm, about 6% of L1, and it is multiplied by four in that calculation, so it cannot be neglected. The two-resonance method is much more reliable because e cancels.
2. Four beats per second are heard against a 512 Hz loudspeaker. Deduce the possible fork frequencies and decide whether the 'mislabelled fork' conclusion is justified. Use v ≈ 331 + 0.6T m/s for the room temperature.
Read the worked answer
Beats are the superposition of two close frequencies, so f_beat = |f_fork − 512| = 4 Hz. This gives f_fork = 508 Hz or 516 Hz; the beat count alone does not distinguish them. Adding wax lowers the fork frequency: if the beat rate falls, the fork was above 512 Hz; if it rises, it was below.
With λ = 0.680 m: 508 Hz → 345 m s⁻¹; 516 Hz → 351 m s⁻¹. At 24 °C the expected speed is about 331 + 0.6 × 24 ≈ 345 m s⁻¹, and the measured 348 m s⁻¹ lies between the two, within a few m s⁻¹. The fork is therefore consistent with 512 Hz within experimental uncertainty. The 328 m s⁻¹ result came from the flawed λ = 4L1 method, so the label is not shown to be wrong.
3. Predict how the first and second resonance lengths change if the room warms to 30 °C. State whether the end correction changes.
Read the worked answer
For sound, v ∝ √T (approximately v ≈ 331 + 0.6T), so warming from 24 °C to 30 °C raises v by roughly 1%: at 30 °C, v ≈ 349 m s⁻¹, so λ = v/f ≈ 0.683 m compared with 0.680 m.
L1 = λ/4 − e increases by about 0.0008 m (≈1 mm), and L2 = 3λ/4 − e increases by about 0.002 m (≈2 mm). These shifts are small but measurable with a millimetre scale if the water level is adjusted carefully.
The end correction is a geometric effect of the open end (roughly e ≈ 0.6r, where r is the tube radius), so it is essentially unchanged by temperature; thermal expansion of the tube is negligible on this scale. The resonances move upwards mainly because the wavelength grows.