Speed of sound in a gas and of waves on a stretched string
Newton took sound as isothermal and got 280 m s⁻¹, too low. Laplace's adiabatic correction, v = √(γP/ρ), matches the measured speed.
Newton's formula and Laplace's correction
Newton assumed that the compressions and rarefactions of sound happen at constant temperature. Such a change is called isothermal. This gives v = √(P/ρ). For air at STP, P = 1.01 × 10⁵ N m⁻² and ρ = 1.29 kg m⁻³, so v ≈ 280 m s⁻¹. The measured speed at 0 °C is about 331 m s⁻¹, so Newton's value is about 15 percent too low. Laplace said the changes are adiabatic: no heat flows in or out. He gave two reasons. Air is a poor conductor of heat. The compressions and rarefactions are too rapid for heat to flow. For an adiabatic change the elasticity is γP, so v = √(γP/ρ). With γ = 1.4, this gives about 331 m s⁻¹, which matches the measured value. Your NIOS lesson prints 333 m s⁻¹ (and so a 16 percent error), but working out √(1.4 × 1.01 × 10⁵ ÷ 1.29) gives 331 m s⁻¹, the accepted value at 0 °C.
What changes the speed of sound in a gas
| Factor | Effect on speed | Reason |
|---|---|---|
| Temperature | Rises: v ∝ √T | v = √(γRT/M); about 0.61 m s⁻¹ faster for each 1 °C rise |
| Pressure (temperature fixed) | No effect | If P doubles, ρ also doubles, so P/ρ stays the same |
| Density of the gas | Lighter gas, faster sound: v ∝ 1/√ρ | At the same T and P, sound is 4 times faster in hydrogen than in oxygen, since √(32/2) = 4 |
| Humidity | Rises in moist air | Water vapour is lighter than air, so moist air is less dense |
Derivation: how temperature affects the speed of sound
- 1Start with Laplace's formula v = √(γP/ρ). For one mole of gas of molar mass M and volume V, ρ = M/V, so v = √(γPV/M).
- 2For one mole of an ideal gas, PV = RT. So v = √(γRT/M).
- 3γ, R and M are fixed for a given gas, so v ∝ √T.
- 4Let v₀ be the speed at 0 °C (273 K) and v at t °C. Then v/v₀ = √[(273 + t)/273] = (1 + t/273)^½.
- 5For small t, (1 + t/273)^½ ≈ 1 + t/546. So v ≈ v₀ + (v₀/546)t.
- 6★So, with v₀ ≈ 331 m s⁻¹, v₀/546 ≈ 0.61: the speed of sound in air rises by about 0.61 m s⁻¹ for every 1 °C.
★ key step
Waves on a stretched string
A plucked string carries transverse waves. Their speed is v = √(F/m), where F is the tension and m is the mass per unit length. A tighter or lighter string carries waves faster. A string fixed at both ends vibrates in loops. In the simplest mode, the fundamental, there is a node at each end and one loop in the middle, so l = λ/2. Its frequency is ν = (1/2l)√(F/m). With p loops, ν = (p/2l)√(F/m). In a solid rod, longitudinal waves travel at v = √(E/ρ), where E is the modulus of elasticity. Solids are far more elastic, so sound is fastest in solids: v(gas) < v(liquid) < v(solid).
Worked example: sound in air and waves on a string
(a) The speed of sound in air at 0 °C is 331 m s⁻¹. Find it on a hot summer day at 40 °C. (b) A sonometer wire 0.5 m long has a mass per unit length of 1.0 × 10⁻³ kg m⁻¹ and a tension of 40 N. Find the wave speed and the fundamental frequency.
- (a) v ∝ √T, so v = 331 m s⁻¹ × √(313 K ÷ 273 K) = 331 m s⁻¹ × √1.147.
- v = 331 m s⁻¹ × 1.071 ≈ 354 m s⁻¹. Check with the 0.61 rule: 331 + 0.61 × 40 ≈ 355 m s⁻¹, nearly the same.
- (b) v = √(F/m) = √(40 N ÷ 1.0 × 10⁻³ kg m⁻¹) = √(4.0 × 10⁴) = 200 m s⁻¹.
- In the fundamental, λ = 2l = 2 × 0.5 m = 1.0 m.
- ν = v/λ = 200 m s⁻¹ ÷ 1.0 m = 200 Hz.
(a) about 354 m s⁻¹; (b) 200 m s⁻¹ and 200 Hz
Activity: sonometer, tension against resonating length
Find how the resonating length of a wire depends on its tension, and use the graph to find the mass per unit length.
- 1Stretch the sonometer wire over two bridges. Hang a weight hanger from its end, so the tension is F = Mg. Keep your feet clear of the weights.
- 2Place a small V-shaped paper rider on the middle of the wire.
- 3Strike a tuning fork of known frequency ν on a rubber pad, never on a table. Touch its stem to the sonometer box.
- 4Slide a bridge until the wire resonates: the sound is loudest and the rider flutters and falls off. Record the length L between the bridges.
- 5Add 0.5 kg to the hanger and repeat. Take readings for six or more tensions.
- 6Plot √F on the y-axis against L on the x-axis. Observation: the graph is a straight line through the origin, so √F ∝ L.
- 7Result: from ν = (1/2L)√(F/m), the slope √F/L = 2ν√m. So the mass per unit length is m = (slope ÷ 2ν)².
A sitar or veena player turns the pegs to tighten a string, which raises the frequency ν = (1/2l)√(F/m) and so the note. In a monsoon storm you see the lightning before you hear the thunder, because light is nearly a million times faster than sound.
3 marks: “Newton's formula gave too low a speed of sound in air. Write the formula, give the reason for the error, and state Laplace's correction.” v = √(P/ρ), assuming isothermal changes; it gives 280 m s⁻¹ for air at STP, about 15 percent low. Laplace: changes are adiabatic, because air conducts heat poorly and the changes are very fast; elasticity = γP, so v = √(γP/ρ) ≈ 331 m s⁻¹, matching experiment.
2 marks: “Explain why the speed of sound in air (a) does not change with pressure at constant temperature, (b) is greater on a humid day.” (a) When P changes, ρ changes in the same ratio, so P/ρ and v = √(γP/ρ) stay the same. (b) Water vapour is lighter than air, so humid air is less dense; v ∝ 1/√ρ, so v increases.
2 marks: “At what temperature will the speed of sound in air be 1.5 times its value at 27 °C?” v ∝ √T, so T₂ = (1.5)² × 300 K = 2.25 × 300 K = 675 K, that is 402 °C.
Newton: isothermal, too low. Laplace: adiabatic, just right. Put γ under the root.