Wave Function, Wave Equation & Wave Speed

Write the phase as φ = kx − ωt. The wave number k counts radians of phase per metre at a frozen instant, k = 2π/λ; the angular frequency ω counts radians per second at a fixed place, ω = 2π/T. They are the same bookkeeping done in the two variables. Now ask where a chosen feature — a crest, say — has got to. A crest is a point of constant phase, so kx − ωt = constant, and differentiating with respect to t gives k(dx/dt) − ω = 0, hence dx/dt = ω/k. That is the phase speed, and it is positive: the pattern moves towards +x. Change the argument to kx + ωt and the same line returns dx/dt = −ω/k, the other way. Nothing else in the function decides this — not the sign of A, not the phase constant. With λ = 2.00 m and f = 125 Hz, k = 3.14 rad m⁻¹ and ω = 785 rad s⁻¹, so v = 785/3.14 = 250 m s⁻¹, which is λf rewritten.

Isolate the element between x and x + dx; its mass is μ dx. Tension pulls along the string at each end, so the transverse component is F_T sin θ, and for small slopes sin θ ≈ tan θ = ∂y/∂x. The two ends pull in opposite senses, so the net transverse force is F_T[(∂y/∂x) at x + dx − (∂y/∂x) at x], which is exactly F_T (∂²y/∂x²) dx. Meanwhile cos θ ≈ 1, so the longitudinal components cancel and the magnitude of F_T is left unchanged by the disturbance — that is where the small-slope assumption does its real work. Newton's second law for the element reads F_T (∂²y/∂x²) dx = (μ dx)(∂²y/∂t²). The dx cancels, leaving ∂²y/∂x² = (μ/F_T) ∂²y/∂t². Compare that with ∂²y/∂x² = (1/v²) ∂²y/∂t² and the coefficient is forced: v = √(F_T/μ). The speed was never assumed; it fell out of a restoring force divided by a mass.

∂y/∂x is the slope of the snapshot graph at one instant; ∂y/∂t is the velocity of one particle on the history graph at one place. The wave equation compares their second derivatives: wherever the string is bent concave-up, ∂²y/∂x² > 0, the element sitting there accelerates upward, and the more sharply it is bent the harder it is flung. Feed a sinusoid in and the calculus is one line each way. For y = A sin(kx − ωt), ∂²y/∂x² = −k²y and ∂²y/∂t² = −ω²y, so substitution gives −k²y = −(ω²/v²)y and therefore ω = vk, for every amplitude and every phase constant. Two consequences. The amplitude cancels, so the equation is blind to how hard you drive the string. And k and ω are locked by one constant, so every frequency travels at the same speed: the string is non-dispersive, which is why a chord arrives as a chord.

Let u = x − vt and try y = f(u) for an arbitrary twice-differentiable f. Since ∂u/∂x = 1, the chain rule gives ∂y/∂x = f′(u) and ∂²y/∂x² = f″(u). Since ∂u/∂t = −v, it gives ∂y/∂t = −v f′(u) and ∂²y/∂t² = v² f″(u). Substituting, f″ = (1/v²)(v² f″), which holds whatever f is. No shape is privileged: a Gaussian bump, a steep-sided pulse, a single sine, the whole profile of a plucked string. Each slides to the right at v with its form intact, because the equation only ever sees the combination x − vt. Run the argument again with u = x + vt and it works too, since the equation contains v² and cannot tell left from right; initial and boundary conditions do that. And because the equation is linear, f(x − vt) + g(x + vt) solves it as well.

Test √(F_T/μ) before trusting it: [F_T] = M L T⁻² and [μ] = M L⁻¹, so [F_T/μ] = L² T⁻², whose root is L T⁻¹, a speed. The form is no accident of strings. The numerator always measures how strongly the medium pushes back when it is deformed, and the denominator how much inertia has to be shifted. Swap tension for the bulk modulus B and μ for the mass density ρ, and sound in a fluid runs at √(B/ρ): for air, B = γp = 1.40 × 1.013 × 10⁵ Pa = 1.42 × 10⁵ Pa and ρ = 1.20 kg m⁻³, giving √(1.42 × 10⁵ ÷ 1.20) = 344 m s⁻¹. A thin solid rod uses Young's modulus instead: steel with Y = 2.0 × 10¹¹ Pa and ρ = 7.8 × 10³ kg m⁻³ gives 5.1 × 10³ m s⁻¹, fifteen times faster, because steel's stiffness outruns its density by far more than its density costs. The limits behave: let F_T fall to zero and the wave stops; load the string and it slows as 1/√μ.

Every step above assumed |∂y/∂x| ≪ 1, so put a number on it. A wave of amplitude 2.0 mm and wavelength 0.50 m has maximum slope kA = (2π/0.50)(0.0020) = 0.025, at which sin θ and tan θ differ by 0.03% and the wave stretches the string by only (kA)²/4 = 1.6 × 10⁻⁴ of its length, so it rides on the tension you tuned. Drive the same string to a 100 mm amplitude and kA = 1.3: sin θ and tan θ are now nothing like each other, the tension climbs and falls with the local slope, a single sine breeds harmonics of itself, and the model is no longer linear. Three further assumptions carry just as much load. A real wire resists bending, which adds a term in ∂⁴y/∂x⁴ and makes ω rise faster than k, so a stiff piano string is dispersive and its overtones drift sharp. A rope hanging under its own weight has F_T = μgy at height y above the free end, so v = √(gy) and a pulse speeds up as it climbs. And a lossy medium drains energy, so the shape that arrives is not the shape that left.