PHY053 · Waves & Acoustics · Classes 11–12
Resonance in air column
Objective: To find the speed of sound in air using resonance in a tube closed at one end by water.
🧰 Apparatus
- Resonance tube (glass tube in a tall jar of water)
- Tuning forks (256, 384, 512 Hz)
- Rubber pad
- Metre scale
- Thermometer
📐 Formulas and SI units
λ = 4L (first resonance)
v = 4 L f (m/s)
with end correction: v = 4f(L + 0.3d)
💡 Theory
A vibrating fork over a tube sends sound down the air column. When the column length is a quarter of the wavelength, the reflected wave reinforces the incoming one and the sound becomes loud: resonance. Then λ = 4L (ignoring end correction) and v = fλ = 4Lf.
🪜 Procedure
- Lower the tube until almost all of it is under water.
- Strike the fork on the rubber pad and hold it over the mouth.
- Slowly raise the tube until the sound is loudest; mark the water level.
- Measure the air column length L.
- Calculate v = 4Lf.
- Repeat with another fork and note the room temperature.
📋 Observations Simulated values
✅ Result and conclusion
A 512 Hz fork resonates with a 16.6 cm air column, giving v = 340 m/s.
🔬 In a real lab
The antinode is a little above the mouth (end correction ≈ 0.3 × diameter), so the simple formula gives a value slightly low; the second resonance removes this.
⚠️ Safety
Strike tuning forks only on a rubber pad (never on the glass); clamp the jar so it cannot tip.
❓ Quick check
1. At first resonance the air column is…
2. A higher-frequency fork resonates with a…
3. Sound travels faster in warmer air because…
🏋️ Practice
Easyf = 256 Hz, L = 33 cm. v?
Answer: ≈ 338 m/s
MediumExpected L for 384 Hz at 340 m/s?
Answer: ≈ 22.1 cm
ChallengeSecond resonance length for 512 Hz?
Answer: ≈ 49.8 cm (3λ/4)