University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.7
Sound Waves & the Speed of Sound
Sound is air moving back and forth along the line of travel, and the pressure change that motion produces. The two are a quarter cycle out of step, and the speed they share belongs to the medium, not to the source.
Build the model
Connect the measurement to the mechanism.
A sound wave is one disturbance described two ways. Fluid elements oscillate along the direction of travel by s(x, t), and because a gradient in that displacement compresses or rarefies the gas, the pressure change is Δp = −B ∂s/∂x. Differentiating a cosine gives a sine, so pressure and displacement sit a quarter cycle out of step: where the gas has moved furthest, nothing is compressed and Δp = 0.
The same B that turns displacement into pressure fixes the speed, v = √(B/ρ) — stiffness over inertia, the form already met in v = √(F/μ) on a string. In a gas the compressions are far too fast for heat to escape, so the adiabatic B = γp applies and v = √(γRT/M): independent of pressure, independent of frequency, rising as √T. All of it treats a swarm of colliding molecules as a continuous medium with a local pressure and temperature, which holds because even a 20 kHz wavelength spans some 10⁵ mean free paths.
- Simple definition
- A sound wave is a longitudinal disturbance in which fluid elements oscillate along the direction of propagation; the compressions and rarefactions that result travel at v = √(B/ρ), set by the medium's stiffness and density alone.
- Example
- At 20 °C, √(γRT/M) with γ = 1.40 and M = 28.96 g mol⁻¹ gives 343 m s⁻¹ in air. Helium, lighter and monatomic, gives 1009 m s⁻¹ — which is why a breath of it shifts vocal-tract resonances up, leaving the vocal folds' rate alone.
A slab whose neighbours converge has ∂s/∂x < 0, so its pressure rises.
Pa; s(x, t) = s(max) cos(kx − ωt) is measured along the travel direction
Uses B = ρv² and ω = vk: the same motion at higher frequency costs more pressure.
Pa; a quarter cycle out of phase with s in both x and t
Stiffness over inertia — the same shape as v = √(F/μ) for a string.
m s⁻¹; water, B = 2.2 GPa and ρ = 998 kg m⁻³, gives 1.48 km s⁻¹
Pressure cancels — raising p raises ρ in step — so only T and the gas matter.
T in K, M in kg mol⁻¹; γ = 1.40 for air gives 343 m s⁻¹ at 20 °C
The square root flattens it: near 0 °C one kelvin adds 0.61 m s⁻¹, near 40 °C only 0.57.
T in K, θ in °C; the linear form is within 0.6 m s⁻¹ of exact from −20 to 40 °C
Bulk solid is faster still, near 5.9 km s⁻¹, since sideways bulging is blocked.
Rod narrower than λ; steel, Y = 200 GPa and ρ = 7800 kg m⁻³, gives 5.06 km s⁻¹
One wave, two variables
Take a long tube of air and drive a piston back and forth at one end. Each thin slab of gas oscillates along the tube — the same axis the wave travels — and that shared axis is what makes sound longitudinal. Describe the motion by a slab's displacement from its rest position, s(x, t) = s(max) cos(kx − ωt), where x labels the slab and s is measured along x rather than across it. Slabs crowd together where neighbouring displacements converge and thin out where they diverge, so the gas alternates between compressions and rarefactions moving at speed v = ω/k. The second description is the gauge pressure Δp(x, t) = p − p₀ inside each slab. Both describe the same wave and neither is more real, but pressure is what the medium hands to a detector: an eardrum and a microphone diaphragm respond to the pressure difference across them, not to how far the air travelled. Almost every measurement of sound is therefore a measurement of Δp.
Pressure is the displacement gradient
The bulk modulus links the two. A slab between x and x + Δx has its length changed by s(x + Δx) − s(x), so its fractional volume change is ΔV/V = ∂s/∂x, and B = −Δp/(ΔV/V) rearranges to Δp = −B ∂s/∂x. Differentiate the displacement wave: ∂s/∂x = −k s(max) sin(kx − ωt), so Δp = B k s(max) sin(kx − ωt). A cosine has become a sine. Displacement and pressure are a quarter cycle out of step in space and in time, and the reason is physical, not algebraic — where a slab has moved furthest, its neighbours have moved just as far, nothing is squeezed, and Δp = 0. With B = ρv² and ω = vk the amplitude is Δp(max) = ρ v ω s(max), which exposes how little air actually moves. The faintest audible 1 kHz tone has Δp(max) ≈ 2.9 × 10⁻⁵ Pa; with ρv = 413 kg m⁻² s⁻¹ and ω = 6.28 × 10³ s⁻¹, s(max) = 2.9 × 10⁻⁵/(413 × 6.28 × 10³) ≈ 1.1 × 10⁻¹¹ m, about a tenth of an atomic diameter.
Speed is stiffness over inertia
Apply Newton's second law to that slab — net force from the pressure difference across its faces, mass ρA Δx — and a wave equation drops out with v² = B/ρ. The shape is the one already met on a stretched string, v = √(F/μ): a restoring stiffness above the line, an inertia below it. Nothing about the source survives, so the speed belongs to the medium. Water at 20 °C has B ≈ 2.2 GPa and ρ ≈ 998 kg m⁻³, giving v = √(2.2 × 10⁹/998) = 1.48 × 10³ m s⁻¹, over four times the speed in air. A solid needs one adjustment: in a rod narrower than a wavelength the material can bulge sideways as it is compressed, so Young modulus replaces the bulk modulus and v = √(Y/ρ). Steel with Y = 200 GPa and ρ = 7800 kg m⁻³ gives √(200 × 10⁹/7800) = 5.06 × 10³ m s⁻¹. In bulk steel the sideways bulge is blocked by the surrounding material, the effective stiffness is larger, and the longitudinal wave runs nearer 5.9 km s⁻¹.
Adiabatic, not isothermal
Which bulk modulus belongs in √(B/ρ)? Newton assumed the compressions held ambient temperature. An isothermal ideal gas has B = p, so v = √(1.013 × 10⁵/1.204) = 290 m s⁻¹ at 20 °C — about 15 % below the measured 343 m s⁻¹, a gap that stood for a century until Laplace closed it. Compression warms a slab, and the question is whether that heat can reach the cooler rarefaction half a wavelength away before the slab expands again. Air has thermal diffusivity α ≈ 2.1 × 10⁻⁵ m² s⁻¹, so during one period of a 1 kHz tone heat diffuses roughly √(α/f) = √(2.1 × 10⁻⁸) ≈ 1.4 × 10⁻⁴ m, against a wavelength of 0.34 m. It covers about a two-thousandth of the distance required. No heat is exchanged, the correct modulus is the adiabatic B = γp, and v = √(γp/ρ) = √(1.40 × 1.013 × 10⁵/1.204) = 343 m s⁻¹ — agreement to three figures.
What sets the speed, and what does not
Substituting p/ρ = RT/M for an ideal gas turns v = √(γp/ρ) into v = √(γRT/M), and the list of what matters shortens sharply. Pressure disappears: raise it at fixed temperature and the density rises in exact step. Sound is not slower on a mountain because the air is thin, only because it is cold. What remains is temperature, molar mass and γ. Helium at 20 °C, with M = 4.00 g mol⁻¹ and γ = 1.67, gives √(1.67 × 8.314 × 293.15/0.00400) = 1.01 × 10³ m s⁻¹, nearly three times air — breathe it and the vocal-tract resonances shift up by that factor while the folds keep vibrating at the same rate. Temperature enters under a square root, v = v₀√(T/T₀), which near room temperature is straight enough to write as v ≈ (331 + 0.60 θ) m s⁻¹ with θ in °C: 343 m s⁻¹ at 20 °C. Frequency never appears. A 20 Hz and a 20 kHz component leave together and arrive together, which is why a chord is still a chord 100 m away.
The assumptions underneath
Every step treated the gas as a continuum carrying a local pressure, density and temperature, and every step assumed the disturbance was small. Both deserve a test. The continuum holds when a fluid element contains many molecules yet stays small against a wavelength: air at room conditions has a mean free path near 6.8 × 10⁻⁸ m, while even a 20 kHz wave has λ = 343/(2.0 × 10⁴) = 1.7 × 10⁻² m, some 2.5 × 10⁵ mean free paths. Only in rarefied gas, or at gigahertz frequencies, does that margin close and the model fail outright. Linearity is comfortable too: a very loud 1 Pa amplitude is 10⁻⁵ of atmospheric pressure, so Δp stays proportional to strain and the wave keeps its shape. Raise the amplitude to a blast wave and it does not — a compression is hotter, so locally faster, crests overtake troughs, and the front steepens into a shock. Finally, nothing here dissipates. Viscosity and conduction absorb sound roughly as f², which is why distant thunder rumbles with the crack stripped out.
Change one variable at a time
Make the relationship visible.
Push f up with s(max) held fixed and the top curve keeps its height while the bottom one grows — Δp(max) = ρvω s(max) — then warm the air instead and the whole pattern stretches, because v, and so λ, belongs to the medium. Wherever the dashed quarter-wavelength line lands, s is zero and Δp is at its peak.
SPEED v343.0 m s⁻¹
WAVELENGTH λ0.686 m
IMPEDANCE ρv413 Pa s m⁻¹
PRESSURE Δp(max)12.97 Pa
Live interpretationSPEED v: 343.0 m s⁻¹. WAVELENGTH λ: 0.686 m. IMPEDANCE ρv: 413 Pa s m⁻¹. PRESSURE Δp(max): 12.97 Pa
Catch the common trap
Explain before calculating.
A balloon carries a microphone to an altitude where the air pressure is half its sea-level value but the temperature is still 15 °C, the same as at the ground. At sea level the speed of sound is 340 m s⁻¹. What is it at that altitude?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA tuning fork sounds at 512 Hz in a room at 25 °C. Find the speed of sound there and the wavelength of the note.
- Speed belongs to the medium, so start from temperature: v ≈ (331 + 0.60θ) m s⁻¹ with θ = 25 °C gives v = 331 + 0.60 × 25 = 331 + 15 = 346 m s⁻¹.
- Check it against the exact form: v = 331 × √(298.15/273.15) = 331 × 1.0448 = 346 m s⁻¹ — the linear rule is good to well under a metre per second here.
- Only now bring in the source: λ = v/f = 346 ÷ 512 = 0.676 m.
Answerv = 346 m s⁻¹; λ = 0.676 m
MediumA 1.00 kHz tone in air at 20 °C (ρ = 1.20 kg m⁻³, v = 343 m s⁻¹) has pressure amplitude Δp(max) = 28 Pa — a painfully loud 120 dB. Find the displacement amplitude, and say where in the cycle the pressure is zero.
- Δp(max) = ρvω s(max), so s(max) = Δp(max)/(ρvω).
- Acoustic impedance first: ρv = 1.20 × 343 = 412 kg m⁻² s⁻¹. Then ω = 2πf = 2π × 1.00 × 10³ = 6.28 × 10³ s⁻¹.
- s(max) = 28 ÷ (412 × 6.28 × 10³) = 28 ÷ (2.59 × 10⁶) = 1.1 × 10⁻⁵ m.
- Set that against λ = 343 ÷ 1000 = 0.343 m: even at 120 dB the air shifts 11 μm, about 3 × 10⁻⁵ of a wavelength.
- Δp = −B ∂s/∂x, and the gradient of a cosine vanishes at its peak, so Δp = 0 exactly where the displacement is largest — a quarter cycle from the pressure maxima.
Answers(max) = 1.1 × 10⁻⁵ m (11 μm); the pressure vanishes at the displacement antinodes, a quarter cycle from the pressure maxima
HardSound travels at 319 m s⁻¹ through argon at 20 °C. Argon's molar mass is 39.95 g mol⁻¹. Find γ for argon, and say what that value reveals about the molecule.
- Rearrange v = √(γRT/M): square it to v² = γRT/M, so γ = v²M/(RT).
- SI units first: M = 39.95 g mol⁻¹ = 3.995 × 10⁻² kg mol⁻¹, and T = 20 + 273.15 = 293.15 K.
- Numerator: v²M = (319)² × 3.995 × 10⁻² = 1.0176 × 10⁵ × 3.995 × 10⁻² = 4065 J mol⁻¹.
- Denominator: RT = 8.314 × 293.15 = 2437 J mol⁻¹.
- γ = 4065 ÷ 2437 = 1.67, which is 5/3 — the monatomic value. Argon stores energy in three translational degrees of freedom and nothing else; air's diatomic molecules also rotate, which adds two more and drops γ to 1.40.
Answerγ = 1.67, i.e. 5/3 — argon is monatomic, with no rotational modes to absorb energy