 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 isThe kinetic energy is $K=(m_1{\dot{x}}_1^2+q{m}_1{\dot{x}}_2^2)/2$. The tangential components of Newton's second law, or the derivatives $-\partial U/\partial x_i$, giveBoth 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.