= Solution
Separating $G=W(q)-\beta t$ gives $[W'(q)]^2/(2m)+m\omega^2q^2/2=\beta$. On the positive-momentum branch choose
$$
G=\int_0^q\sqrt{2m\left(\beta-\frac12m\omega^2s^2\right)}\,ds-\beta t.
$$
With $Q=\beta$ and $A=\sqrt{2\beta/(m\omega^2)}$, differentiation with respect to $Q$ gives
$$
P=-G_\beta=t-\int_0^q\frac{m\,ds}
{\sqrt{2m(\beta-m\omega^2s^2/2)}}
=t-\frac1\omega\arcsin(q/A).
$$
Since $P=\gamma$ is constant,
$$
\boxed{q=A\sin[\omega(t-\gamma)].}
$$
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.
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