Does a wave on a CHAIN of atoms behave like a wave on a continuous string — one fixed speed for every wavelength?
▶ Launch the interactive simulationSet a ring of N equal masses joined by identical springs vibrating in one normal mode at a time (a standing wave of wavenumber k), and measure each mode's frequency ω from the oscillation period of a single mass; sweep k from long wavelength to the shortest the lattice allows.
the dispersion relation ω(k) = 2√(K/m)·|sin(ka/2)|: at long wavelength ω ≈ √(K/m)·a·k — sound, one fixed speed, non-dispersive like a string — but the discreteness makes ω bend over and SATURATE at a maximum ω_max = 2√(K/m) at the Brillouin-zone edge k = π/a (neighbouring masses in antiphase). A continuum string has no such ceiling; the atoms do, and short waves are dispersive