Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 14D (Classical Dynamics)

Hamilton's equations give $\dot{q}=p$ and $\dot{p}=-{\omega }^{2}q$, hence $\ddot{q}+{\omega }^{2}q=0$. The phase-space energy curve is an ellipse. Its action variable isThus $\boxed{E=\omega I}$.

Substituting $x=t^{1/2}y(\log t)$ givesThusand the initial conditions selectDirect differentiation verifies both the equation and $x(1)=0$, $\dot{x}(1)=1$.

Put $\theta =k\log t$. Sincethe instantaneous oscillator energy isThe instantaneous frequency is $\omega (t)=1/(ϵt)$, soIts oscillatory part divided by its mean is $O(k^{-1})=O(ϵ)$. It remains bounded and periodic in $\log t$, so the variations do not accumulate. This is the expected behavior of an adiabatic invariant: slow parameter variation produces small bounded oscillations rather than secular drift.