Sound Waves & the Speed of Sound

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.

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.

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.

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.

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.

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.