WAVES
Two travelers passing forever — their sum stands still.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
Setup
A stretched string, nailed down at both ends. Pluck it and waves race away, reflect, and pile onto each other. Out of that traffic, only a few patterns survive — the string is about to hand you a menu.
Constraint
The walls are non-negotiable: y = 0 at both ends, forever. Any wave that fails to place nodes there destroys itself in reflection. What survives is the discrete family λ = 2L/n — the boundary condition writes the menu, and the menu is countable.
Solve
The string sets its own tempo: c = √(F/μ), tension against heft. Pair that speed with the allowed wavelengths and the frequencies fall out in a ladder — fₙ = nc/2L, every rung an integer multiple of the first. Tighten the peg and the whole ladder rises.
Oscillate
Three periods, honestly slowed. The faint amber and cyan travelers run opposite ways; their violet sum is the standing wave — nodes nailed still while antinodes trade places. Nothing travels, yet everything moves.
Audit
The books: boundary values zero to the last bit, interior nodes pinned at kL/n, fλ = c to machine precision, and the superposition identity verified at every sampled point — a standing wave really is two travelers passing forever.