Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S7 / 2

Use rotating basis vectors with ${\dot{\mathbf{e}}}_1=\Omega {\mathbf{e}}_2$, ${\dot{\mathbf{e}}}_2=-\Omega {\mathbf{e}}_1$. The inertial velocity is $({\dot{x}}_1-\Omega x_2){\mathbf{e}}_1+({\dot{x}}_2+\Omega x_1){\mathbf{e}}_2$. The two fixed rotating anchors are $(0,\pm a)$, so the spring lengths are ${\ell }_{\pm }=\sqrt{x_1^2+(x_2\mp a{)}^2}$. Combining kinetic and elastic energies givesA constant elastic offset has no dynamical effect.

Differentiate the quadratic Lagrangian with respect to coordinates and velocities. The equations areThe velocity-cross terms are Coriolis accelerations; moving $-{\Omega }^2{x}_i$ to the force side identifies the outward centrifugal acceleration. The remaining terms are elastic restoring forces. The sign pair is fixed by the rotating-basis convention in part a.

Substitution of the exponential ansatz gives the homogeneous amplitude systemA nonzero mode requires zero determinant. Thus $\boxed{({\omega }_1^2-{\omega }^2)({\omega }_2^2-{\omega }^2)=4{\Omega }^2{\omega }^2}$ with the stated two effective stiffnesses. The off-diagonal product is positive, which determines the sign on the right.

Put $X={\omega }^2$ and $S={\omega }_1^2+{\omega }_2^2+4{\Omega }^2$. Its equation is $X^2-SX+{\omega }_1^2{\omega }_2^2=0$. If both effective stiffnesses are positive, the sum and product of roots are positive and $S^2-4{\omega }_1^2{\omega }_2^2\ge 0$, so both roots are positive. They can coincide in a degenerate zero-rotation isotropic case; “two” need not mean distinct.

If ${\omega }_1^2<0<{\omega }_2^2$, the product is negative, so one root is negative. Its imaginary frequency gives a growing exponential displacement: centrifugal softening has overwhelmed one elastic restoring direction, making the central configuration unstable. Coriolis coupling mixes the two coordinates but does not remove this negative-product instability.