Resonance tube: end correction and temperature shift
A student measures the speed of sound using a resonance tube closed at one end by a water surface. A tuning fork of unknown frequency is held above the open end. The first strong resonance is found when the air column is 17.0 cm long, the second at 52.5 cm, and the third at 88.0 cm. The tube has internal radius 1.2 cm. The room temperature is 20 °C, at which the speed of sound is 343 m/s. The student first calculates frequency using only the first length as λ/4, then repeats the calculation using the difference between the first and second resonance lengths. The two answers differ. The student also notices that the antinode is not exactly at the open end of the tube. Explain the discrepancy and decide which method is more reliable. Use the end correction to check consistency with the tube radius.
1. Calculate the frequency using the first length alone and using the difference between the first and second resonance lengths. Determine the end correction and compare it with the tube radius.
Read the worked answer
Using L1 alone: if the first resonance were exactly λ/4,
λ = 4 × 0.170 = 0.680 m
f = v/λ = 343 / 0.680 = 504 Hz
Using the difference: for a tube closed at one end, successive resonances differ by λ/2:
L2 − L1 = 0.525 − 0.170 = 0.355 m
λ = 2 × 0.355 = 0.710 m
f = 343 / 0.710 = 483 Hz
The difference method gives the more reliable frequency, about 4.8 × 10² Hz. With this wavelength, λ/4 = 0.1775 m. The first resonance is at 0.170 m, so the end correction is
e = λ/4 − L1 = 0.1775 − 0.170 = 0.0075 m = 0.75 cm
For a tube of radius 1.2 cm, the usual estimate e ≈ 0.6r gives about 0.72 cm, which is consistent with the measured correction.
2. Explain why using the difference between resonance lengths removes the end correction, and why using L1 alone gives a frequency that is too high.
Read the worked answer
The open end is not a true displacement antinode; it lies a small distance e outside the tube. The resonance condition is therefore
L_n = (2n − 1)λ/4 − e
For the first two resonances:
L1 = λ/4 − e
L2 = 3λ/4 − e
Subtracting gives
L2 − L1 = λ/2
so the unknown e cancels. This is why the difference method is more reliable.
Using L1 alone assumes e = 0. Then the calculated wavelength is 4L1, which is shorter than the true wavelength by 4e. A shorter wavelength gives a higher frequency, so the L1-only result of 504 Hz is too high compared with the difference result of 483 Hz. The discrepancy is not random: it is a systematic error caused by the end correction.
3. The room temperature rises to 25 °C, so the speed of sound increases slightly to about 346 m/s. The tuning fork frequency is unchanged. Predict whether the resonance lengths become longer or shorter, and whether recalculating the frequency from the new lengths using 343 m/s would give a value too high or too low.
Read the worked answer
The tuning fork fixes the frequency. Since v = fλ, an increase in v at fixed f means λ increases. The resonance lengths are proportional to λ, so all three resonance positions move slightly farther from the open end: they become longer.
For example, with the reliable frequency 483 Hz and v = 346 m/s,
λ_new = 346 / 483 = 0.716 m
compared with 0.710 m at 20 °C. The first resonance would move from about 17.0 cm to about 17.2 cm, a small but measurable shift.
If the student uses the new, longer lengths but still assumes v = 343 m/s, the calculated frequency is
f_calc = 343 / 0.716 = 479 Hz
which is lower than the true 483 Hz. The error arises because the calculation uses too small a speed for the actual air temperature; the measured lengths correctly reflect the higher speed.