= Solution
For $q=2$, the common coordinate has $X(0)=(x_1(0)+2x_2(0))/3=0$ and $\dot X(0)=\pi\omega r/20$. The relative coordinate has $\delta(0)=3\pi r/20$ and $\dot\delta(0)=0$. Consequently
$$
X(t)=\frac{\pi\omega r}{20}t,\qquad
\delta(t)=\frac{3\pi r}{20}\cos\omega t.
$$
Since $x_1=X+2\delta/3$ and $x_2=X-\delta/3$, the full <superposition> is
$$
x_1(t)=\frac{\pi\omega r}{20}t+\frac{\pi r}{10}\cos\omega t,\qquad
x_2(t)=\frac{\pi\omega r}{20}t-\frac{\pi r}{20}\cos\omega t.
$$
The heavier particle's <velocity> is
$$
\dot x_2(t)=\frac{\pi\omega r}{20}(1+\sin\omega t).
$$
It first vanishes when $\sin\omega t=-1$, so
$$
\boxed{t_{\rm first}=\frac{3\pi}{2\omega}.}
$$
This <velocity> is nonnegative and only touches zero: the heavier particle does not reverse its direction at that instant. Both the uniform <zero mode> and the oscillatory <normal mode> are needed to obtain this result.
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