Measuring the Speed of Sound
Find two resonances half a wavelength apart, multiply by the frequency.
Lesson at a glanceShow the key points
- Resonance column: a tube whose water level sets the length of a closed air column, driven by a tuning fork at the open end.
- First resonance l₁ + d = λ/4, second l₂ + d = 3λ/4 (d = end correction), so λ = 2(l₂ − l₁) and v = 2(l₂ − l₁)ν (16.20).
- Kundt's tube: a vibrating rod excites the gas in a closed tube; powder collects in heaps at displacement nodes, λ/2 apart.
- Comparing heap spacings in a gas and in air gives v_gas = v_air Δl/Δl′.
- For a rod clamped at its middle (fundamental, λ_rod = 2l): v_rod = (l/Δl) v_air.
A tuning fork, a tube and some water are enough to measure the speed of sound to within a percent. The trick is to measure the difference between two resonances, so the awkward end correction cancels.
The resonance-column method
A long glass tube stands upright, with its bottom connected by a rubber pipe to a water reservoir. Raising or lowering the reservoir moves the water level in the tube. The air above the water is a column closed at the bottom (by the water) and open at the top.
Hold a vibrating fork over the top and slowly lower the water. At certain lengths the air column’s natural frequency matches the fork, and the sound swells loudly: resonance.
Using two resonances
The pressure node sits a small distance d (the end correction) above the open end. With the first two resonance lengths l₁ and l₂:
Subtracting removes d: λ = 2(l₂ − l₁). Then
With a 512 Hz fork, resonances occur at air-column lengths of 15.8 cm and 49.0 cm. Find the speed of sound and the end correction.
Kundt’s tube
Kundt’s method compares the speed of sound in two media. A long glass tube is closed at one end by an adjustable plug. A metal rod, clamped at its middle, pokes into the other end with a light disc attached. Fine powder (lycopodium or cork dust) is sprinkled thinly along the tube.
Rub the rod lengthwise with a resin-coated cloth and it rings in its fundamental longitudinal mode, which drives the air in the tube. Adjust the plug until the air column resonates. The powder is then swept away from the displacement antinodes and settles in little heaps at the displacement nodes, spaced λ/2 apart.
Speed in another gas
Fill the tube with a gas, measure the heap spacing Δl, then repeat with air (Δl′). The rod’s frequency is the same in both cases, so v ∝ λ ∝ heap spacing:
Speed in the rod
A rod of length l clamped at its middle has a node there and antinodes at both free ends, so in its fundamental . Its frequency equals that of the air waves, whose wavelength is 2Δl. Equal frequencies give
A 1.0 m rod clamped at its middle sets up heaps 6.5 cm apart in air (v = 340 m/s). Find the speed of sound in the rod.
- 1Resonance column: first two resonances l₁ + d = λ/4, l₂ + d = 3λ/4.
- 2v = 2(l₂ − l₁)ν; end correction d = (l₂ − 3l₁)/2.
- 3Kundt's tube: powder heaps at displacement nodes, λ/2 apart.
- 4v_gas/v_air = Δl/Δl′; for a rod clamped at its middle, v_rod = (l/Δl)v_air.
Check yourself
Keep exploring
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