Describing travelling waves

A travelling wave carries a disturbance, not matter. Pluck a stretched cord and the bump runs along it, yet no piece of cord ends up further from the peg than it began: each element rises, falls and returns. How an element moves relative to the direction the disturbance travels splits waves into two families. In a transverse wave the elements move perpendicular to the travel direction — a wave on a string, a seismic S wave. In a longitudinal wave they move along it, so the medium alternates between compression and rarefaction; sound is the standard case. The distinction is not cosmetic. A transverse mechanical wave needs a restoring force against shear, which a fluid cannot supply, and that is why S waves stop dead at the Earth's liquid outer core while longitudinal P waves pass through. The displacements involved can be startlingly small: ordinary speech heard at a metre moves air molecules back and forth through about 1 × 10⁻⁸ m, while the disturbance itself advances at 343 m s⁻¹.

Freeze a pulse at t = 0 and call its shape f(x), a function of position alone. If the medium carries it at speed v without distortion, the snapshot at any later time is that same curve slid bodily to the right by vt, so y(x, t) = f(x − vt). Nothing more is needed. Test it by following one feature, say the peak. The peak sits wherever the argument takes the value that makes f largest, so x − vt = constant along it, and differentiating gives dx/dt = v: the feature moves at exactly the wave speed. Replace the minus with a plus and the same step returns dx/dt = −v, a wave travelling towards −x. Numerically, a pulse whose peak is at x = 0.40 m at t = 0 on a cord carrying waves at 12 m s⁻¹ has that peak at 0.40 + 12 × 0.050 = 1.00 m at t = 0.050 s. Shape preservation is an assumption, not a theorem: it holds for a linear, non-dispersive medium, and what fixes the value of v is the next topic's business.

Drive the source periodically and the wave acquires two repeats, one in time and one in space. Hold position and watch: the element returns to the same state every period T, and f = 1/T cycles pass each second. Hold time and look along the wave: the pattern repeats every wavelength λ. The travel links them, because in exactly one period the whole pattern advances exactly one wavelength, so v = λ/T = fλ. The dimensions agree at once: Hz × m = s⁻¹ m = m s⁻¹. The relation hides which quantity is actually free, though. The medium fixes v — for a cord, the tension and the mass per unit length; for air, the temperature alone, because v = √(γp/ρ) becomes v = √(γRT/M) once p/ρ is replaced by RT/M, and the pressure cancels. The source fixes f, and no passive medium can change it, because the elements at a boundary are shaken at whatever rate arrives. So λ = v/f is the dependent one. A 686 Hz tone in 20 °C air has λ = 343/686 = 0.500 m; the same 686 Hz in water, where v = 1480 m s⁻¹, has λ = 2.16 m.

Two elements at different positions do the same thing at different times, and the cleanest way to say how different is an angle. One whole wavelength of separation is one whole cycle of lag, 2π rad, so a separation Δx measured along the direction of travel corresponds to Δφ = 2π Δx/λ = k Δx, where k = 2π/λ is the wave number in rad m⁻¹. Worked: two microphones 0.25 m apart on the line from a 686 Hz source in air, where λ = 0.500 m, differ by Δφ = 2π × 0.25/0.500 = π rad — exactly out of step, one rising while the other falls. Only the separation modulo λ matters, so microphones 1.25 m apart give 2π × 2.5 = 5π rad, which is the same π once whole cycles are discarded. Two cautions: Δx must be resolved along the direction of travel, not measured as a straight line to a point off to one side, and the same bookkeeping works in time at a single place, where waiting Δt advances the phase by 2π Δt/T = ω Δt.

Both are sinusoids, and both are routinely mislabelled. A snapshot graph plots y against x at one instant: it is a photograph of the whole cord, its horizontal axis is a length in metres, and the repeat distance measured on it is the wavelength. A history graph plots y against t for one chosen element: it is what a single detector records, its horizontal axis is a time in seconds, and the repeat measured on it is the period. Read a wavelength off a history graph and you have quoted a time as a length. Each graph also carries something the other cannot. The snapshot gives the slope ∂y/∂x at every point, which is the shape the cord actually has; the history gives that element's velocity ∂y/∂t. And a single snapshot cannot say which way the wave travels, because left-going and right-going look identical. Settling that needs a second snapshot, a stated direction, or the sign of one element's velocity.

Because the whole wave is y(x, t) = f(x − vt), its two partial derivatives are not independent. The chain rule gives ∂y/∂x = f′(x − vt) and ∂y/∂t = −v f′(x − vt), so ∂y/∂t = −v ∂y/∂x: the transverse velocity of an element is minus the wave speed times the local slope of the snapshot. Two consequences follow. Where the snapshot is flat, at a crest or a trough, the element is momentarily at rest; where the snapshot is steepest, at the zero crossings, it moves fastest. And the two speeds are very different numbers. For a sinusoid of amplitude A the largest slope is kA, so uₘₐₓ = v k A = ωA = 2πfA, and uₘₐₓ/v = 2πA/λ. On a cord with A = 3.0 mm, f = 120 Hz and v = 80 m s⁻¹, uₘₐₓ = 2π × 120 × 0.0030 = 2.3 m s⁻¹ against a wave speed of 80 m s⁻¹, a ratio of 0.028. That smallness is exactly the condition the linear description needs.