= Solution
Define a common displacement coordinate $X=(x_1+qx_2)/(1+q)$ and a relative displacement $\delta=x_1-x_2$. Adding and subtracting the equations of motion gives
$$
\ddot X=0,\qquad
\ddot\delta+\omega^2\delta=0.
$$
The general real motions of the two <normal modes> are therefore
$$
\begin{aligned}
\text{zero mode:}\quad &(x_1,x_2)=(C+Dt)(1,1),\\
\text{oscillatory mode:}\quad &(x_1,x_2)=
\bigl(A\cos\omega t+B\sin\omega t\bigr)(1,-1/q).
\end{aligned}
$$
Here $C,D,A,B$ are independent real constants. The <zero mode> includes uniform drift as well as a static common displacement; writing only a constant times $\cos0t$ would miss the independent linear-in-time solution. Their <superposition> is
$$
\boxed{x_1=C+Dt+A\cos\omega t+B\sin\omega t,\qquad
x_2=C+Dt-\frac1q(A\cos\omega t+B\sin\omega t).}
$$
In the oscillatory <normal mode>, $m_1x_1+m_2x_2=0$ and the two <amplitudes> have ratio $1/q$, with opposite <phases>. Equivalently the oscillatory motion has zero total <angular momentum> about the hoop center.
As $q\to\infty$ with $m_1,k$ fixed, $\omega\to\sqrt{2k/m_1}$ and the heavier particle's oscillatory <amplitude> tends to zero for fixed lighter-particle <amplitude>. The lighter particle then oscillates against two effectively anchored <springs> with combined <spring constant> $2k$. A common uniform <rotation> remains an allowed <zero mode>; the heavy particle is stationary in the limit only if the initial common drift is zero or tends to zero, as it does for bounded total <angular momentum>.
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