Standing Waves & Resonance

Take two waves of equal amplitude and frequency on the same string, one running right and one running left: y₁ = A sin(kx − ωt) and y₂ = A sin(kx + ωt). Superposition adds them, and the identity sin P + sin Q = 2 sin((P + Q)/2) cos((P − Q)/2) collapses the sum to y = 2A sin(kx) cos(ωt). Look at what has happened to the arguments. Neither factor holds x and t together, so nothing in the expression can be written as a function of (x − vt) — there is no travelling phase left to follow. The x-factor assigns a fixed amplitude to each point on the string; the t-factor makes every point rise and fall in unison at the same ω. Two waves that each carried energy along the string have combined into a pattern that carries none. You never have to arrange this by hand: reflection produces the counter-propagating pair automatically.

The point at x swings with amplitude 2A|sin kx|, which vanishes wherever kx = mπ, that is at x = mλ/2. Those are the nodes, and they are motionless for all time, not merely at some instant. Halfway between consecutive nodes |sin kx| = 1 and the point swings with amplitude 2A, twice the amplitude of either travelling wave on its own. So adjacent nodes are λ/2 apart, adjacent antinodes are λ/2 apart, and a node sits λ/4 from the antinode beside it. This is the cheapest wavelength measurement in physics. Drive a cord until the pattern is clean, lay a rule along it, and if the nodes fall 0.25 m apart then λ = 0.50 m — no timing, no oscilloscope. Driven at 440 Hz, that cord carries waves at v = fλ = 220 m s⁻¹.

A string clamped at x = 0 cannot move there, so that point must be a node; the wave reflects with its displacement inverted, which is precisely what makes incident and reflected waves cancel at the clamp. Clamp the far end at x = L and it must be a node too. Nodes are λ/2 apart, so L has to be a whole number of half-wavelengths: L = n λₙ/2, giving λₙ = 2L/n and fₙ = n v/(2L). Before the second clamp went on, any frequency could run on the string. After it, only a discrete ladder survives. Take L = 0.640 m and v = 200 m s⁻¹: the modes are 156.25 Hz, 312.5 Hz, 468.75 Hz, 625 Hz and upward, mode n showing n antinodes and n + 1 nodes counting both ends. Nothing quantum is involved — the discreteness is what fitting a shape into a fixed length costs.

A free end on a string — a light ring sliding on a frictionless post — cannot carry a transverse force, so the slope there must be zero, the reflection returns uninverted, and the end is an antinode. A pipe's open end arrives at the same condition by a different route: the air at the opening stays at room pressure, so it is a pressure node, and a pressure node is a displacement antinode. Either way the two ends now carry different conditions, a node at one and an antinode at the other, and the shortest node-to-antinode distance is λ/4. The medium must therefore hold an odd number of quarter-wavelengths: L = n λₙ/4 with n odd, so fₙ = n v/(4L) for n = 1, 3, 5, … . A pipe 0.500 m long and stopped at one end, with sound at 343 m s⁻¹, sounds 171.5 Hz, then 514.5 Hz and 857.5 Hz; an open pipe of the same length starts an octave higher at 343 Hz and sounds every harmonic. Same length, same air, different spectrum. Real pipes need a small end correction, which pushes the effective length a little past the opening.

The two travelling waves carry equal power in opposite directions, so the time-averaged power crossing any point is zero: the energy is trapped, sloshing rather than flowing. Twice per cycle the string is straight and moving at its fastest, and the energy is all kinetic; a quarter period later it is fully displaced and momentarily still, and the energy is all potential. That trapping is what makes resonance violent. One round trip takes 2L/v. Drive the end at fₙ and the returning wave comes back in step with the driver, so every cycle adds to what is already stored, and the amplitude climbs until the energy lost per cycle to friction and radiation matches the energy fed in. Drive a little off fₙ and successive round trips arrive with a growing phase error, interfere away, and the response stays small. Losses are also why a real resonator answers with peaks of finite width rather than infinitely sharp lines.

Nothing in the argument used a string. Set two mirrors 0.320 m apart and the light between them fits the same way, L = n λₙ/2, so the allowed frequencies are fₙ = n c/(2L) and neighbouring modes lie Δf = c/(2L) = 468 MHz apart. What changes is the size of n. At 632.8 nm, n = 2L/λ = 0.640 ÷ (632.8 × 10⁻⁹) ≈ 1.01 × 10⁶, so the ladder is astonishingly dense — yet it is still a ladder, and the laser oscillates only on the rungs lying under its gain curve. A waveguide runs the transverse version of the same argument: the field must fit a half-wavelength across the guide's width a, so nothing below the cutoff frequency c/(2a) propagates at all. A guide 22.9 mm wide cuts off at 6.55 GHz, and below that it is simply opaque.