Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 3 / e / Solution

Separating $G=W(q)-\beta t$ gives $[W^{\prime }(q){]}^2/(2m)+m{\omega }^2{q}^2/2=\beta$. On the positive-momentum branch chooseWith $Q=\beta$ and $A=\sqrt{2\beta /(m{\omega }^2)}$, differentiation with respect to $Q$ givesSince $P=\gamma$ is constant,The displayed integral is a local positive-$p$ branch. Continuing the trajectory through turning points switches momentum branches and extends the sine solution over the complete orbit. $\beta$ is the oscillator energy, and $\gamma$ fixes its time phase.