= Solution
Use signed <arc length> displacements $x_1,x_2$ in the common clockwise sense shown by both arrows in the source diagram. Near the diametrically opposite equilibrium, one <spring> changes length by $x_2-x_1$ and the other by $x_1-x_2$. Their combined <elastic potential energy> is
$$
U=\frac k2(x_2-x_1)^2+\frac k2(x_1-x_2)^2=k(x_1-x_2)^2.
$$
The <kinetic energy> is $K=(m_1\dot x_1^2+qm_1\dot x_2^2)/2$. The tangential components of <Newton's second law>, or the derivatives $-\partial U/\partial x_i$, give
$$
\boxed{m_1\ddot x_1=-2k(x_1-x_2),\qquad
qm_1\ddot x_2=2k(x_1-x_2),\quad q>0.}
$$
Both <springs> contribute to each restoring <force>. The hoop radius drops out because the coordinates are <arc lengths>; angular coordinates $x_i/r$ give the same equations after cancelling the common factor $r^2$ in the energies.
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