Sinusoidal Wave Functions & Phase

Drive the end of a long string, x = 0, in simple harmonic motion: y(0, t) = A cos(ωt). The string passes the disturbance along at speed v, so the piece at position x does exactly what the end did a time x/v earlier: y(x, t) = A cos(ω(t − x/v)) = A cos(ωt − kx), with k = ω/v. Cosine is even, so this equals A cos(kx − ωt), and because cos θ = sin(θ + π/2) it is also A sin(kx − ωt + π/2). Sine or cosine, kx first or ωt first: those choices only move the phase constant φ₀, which records what the string at x = 0 was doing at t = 0. Take f = 5.0 Hz and v = 2.5 m s⁻¹, so ω = 31.4 rad s⁻¹ and k = 12.6 rad m⁻¹. The piece at x = 0.75 m runs 0.75/2.5 = 0.30 s behind the source. That is 1.5 periods, so it is always doing the exact opposite of the source.

The sine repeats when its argument grows by 2π. At a fixed instant that takes a distance λ, so kλ = 2π and k = 2π/λ: the wave number is the phase accumulated per metre, in rad m⁻¹. At a fixed place it takes a time T, so ω = 2π/T = 2πf. The two are exact partners, k doing for space what ω does for time, and because the argument of a sine must be dimensionless, kx and ωt are both pure numbers. Reading a given function is then mechanical. For y = (3.0 mm) sin(8.0x − 50t + 0.60), with x in m and t in s: k = 8.0 rad m⁻¹ gives λ = 2π/8.0 = 0.79 m, and ω = 50 rad s⁻¹ gives T = 2π/50 = 0.13 s and f = 8.0 Hz. The two slips to avoid are reading 8.0 as the wavelength itself, and dropping the 2π to write λ = 1/k. This k has nothing to do with a spring constant.

To find which way the pattern moves, follow one feature, say the crest where Φ = π/2. For Φ = kx − ωt + φ₀ to stay fixed while t increases, x must increase too: k dx − ω dt = 0, so dx/dt = +ω/k. The crest moves toward +x at v = ω/k = λ/T = fλ. Change the phase to kx + ωt + φ₀ and the same argument gives dx/dt = −ω/k, travel toward −x. The rule concerns the relative sign: opposite signs on the x and t terms mean +x, equal signs mean −x. So A cos(50t + 8.0x) travels toward −x at 50/8.0 = 6.3 m s⁻¹, while A sin(ωt − kx) travels toward +x, since sin(ωt − kx) = sin(kx − ωt + π) is the standard form with φ₀ shifted by π. Reversing the overall sign of the phase changes φ₀, never the direction. Notice too that the amplitude appears nowhere in ω/k.

A function of two variables is handled here by holding one still. Set t = t₀ and y(x, t₀) is a function of x alone, the snapshot: a sinusoid of wavelength λ, slid sideways by an amount that depends on t₀. Set x = x₀ and y(x₀, t) is a function of t alone, the history of one piece of string: simple harmonic motion with amplitude A, angular frequency ω and its own phase constant kx₀ + φ₀. Every piece of the string performs the same SHM; they differ only in timing. The symbol ∂y/∂t means the ordinary derivative of the history, with x frozen. It is the velocity of that piece of string, v_y = −ωA cos(kx − ωt + φ₀), and a second derivative gives a_y = −ω²y, the signature of SHM. Likewise ∂y/∂x is the slope of the snapshot, kA cos(kx − ωt + φ₀). Dividing one by the other gives v_y = −v ∂y/∂x for travel toward +x: string just ahead of a crest, where the snapshot slopes down, is on its way up.

Two points on the string at the same instant differ in phase by ΔΦ = kΔx = 2πΔx/λ; one point at two instants differs by ΔΦ = ωΔt = 2πΔt/T. With λ = 0.50 m, points 0.125 m apart are π/2 apart, so one sits at zero when the other is at a crest; 0.25 m apart they are π apart and mirror each other; 0.50 m apart they are 2π apart and move as one. Only the remainder after whole cycles matters. Points 1.30 m apart on that wave are 1.30/0.50 = 2.6 cycles apart, which behaves as 0.6 cycle, 3.8 rad, or equally as 0.4 cycle the other way. As for which point leads: on a wave travelling toward +x the downstream point is late, doing now what the upstream point did Δx/v ago. Phase difference is the quantity that decides, in the superposition topic, whether two waves reinforce or cancel.

The phase constant comes from the state of the string at x = 0, t = 0, and one fact is not enough. Suppose a wave travelling toward +x has y(0, 0) = +A/2. Then sin φ₀ = ½, which allows φ₀ = π/6 or 5π/6: same displacement, opposite motion. The velocity settles it. Since v_y(0, 0) = −ωA cos φ₀, a piece that is moving down needs cos φ₀ > 0, so φ₀ = π/6; moving up would select 5π/6. Always use the displacement and the direction of motion together. Finally, remember what was idealised. The function describes one frequency, a constant amplitude and a string with no ends. It represents a real wave train well away from its ends and over times short compared with the damping time. Where ω/k varies with k, each sinusoid still travels at its own ω/k, but a pulse assembled from many of them spreads as it goes.