Resonance Tube and the End Correction
A student uses a resonance tube, a loudspeaker driven by a signal generator, and a microphone to find the speed of sound. The tube is closed at one end by a movable piston. With the frequency set to 512 Hz, the first strong resonance is heard when the air column is 16.0 cm long and the next at 49.5 cm. The room temperature is 20 °C. The student originally tried to find the speed using only the first resonance and the formula v = 4fL, obtaining 328 m/s. A teacher says this is lower than expected. The microphone is moved slowly and the signal is displayed on an oscilloscope. The tube diameter is 3.0 cm. The frequency is checked with a calibrated meter and is accurate to ±1 Hz. The student repeats the second resonance and gets 49.4 cm and 49.6 cm.
1. Calculate the speed of sound from the two resonance lengths and state the uncertainty indicated by the repeat readings.
Read the worked answer
For a tube closed at one end, the distance between successive resonance positions is half a wavelength:
l2 − l1 = 49.5 − 16.0 = 33.5 cm = 0.335 m.
So:
λ = 2 × 0.335 = 0.670 m.
v = fλ = 512 × 0.670 = 343 m/s.
The repeat readings for the second resonance are 49.4 cm and 49.6 cm, a spread of ±0.1 cm. This gives a wavelength uncertainty of about ±0.002 m and a speed uncertainty of about ±1 m/s. The result is therefore 343 ± 1 m/s.
2. Explain why using only the first resonance gives 328 m/s, and calculate the end correction.
Read the worked answer
Using L = λ/4 assumes the displacement antinode is exactly at the open end. In reality it lies a small distance e outside the tube, so the true relation is:
l1 + e = λ/4.
Using v = 4fL with L = 0.160 m gives:
v = 4 × 512 × 0.160 = 328 m/s.
This is too low because it ignores e. From two resonances:
e = (l2 − 3l1)/2 = (49.5 − 3 × 16.0)/2 = (49.5 − 48.0)/2 = 0.75 cm.
Then l1 + e = 16.75 cm, and:
v = 4f(l1 + e) = 4 × 512 × 0.1675 = 343 m/s.
The end correction is about 0.25 times the tube diameter, which is plausible for a 3.0 cm tube.
3. Explain why the microphone and oscilloscope improve the result compared with listening for the loudest sound, and why the room temperature is recorded.
Read the worked answer
The ear judges loudness over a broad range, and the maximum is shallow, so the resonance position is uncertain. A microphone gives a voltage amplitude that can be read on the oscilloscope; the peak is sharper and can be located more reproducibly, reducing random error.
The speed of sound in air depends on temperature, roughly v ≈ 331 + 0.6T m/s. At 20 °C the expected value is about 343 m/s, so the measured value agrees with the expected value at that temperature. Without recording the temperature, comparison with a standard value would be ambiguous.