Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 1 / b / iii / Solution

For the $n$th normal mode, the denominator just obtained isIts forcing coefficient is nonzero for $c\ne 0$ and $\omega >0$. Therefore the periodic response grows like $({\omega }_n^2-{\omega }^2{)}^{-1}$ as $\omega$ approaches ${\omega }_n$: this is resonance. At exact resonance there is no bounded periodic particular solution in the linear inviscid model; the resonant amplitude instead contains a secular term proportional to $t\sin ({\omega }_{n}t)$ or $t\cos ({\omega }_{n}t)$. Physical damping and nonlinear effects eventually matter. Away from resonance, undamped free oscillations can persist alongside the periodic particular solution.