Past exam of the theoretical physics I course of the University of Cambridge / 2013 / Paper 1 / 1 / c / Solution

Removing the hinge mass gives the exact LagrangianThe stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomesConsequentlyFor unequal squared frequencies, $\nu =0$: the rods remain aligned to first order and the end mass swings like a pendulum of length $2\ell$. This degenerate Lagrangian gives the difference coordinate no quadratic kinetic term, so its equation is an algebraic constraint, not a second oscillator. If ${\omega }_0={\omega }_1$, the quadratic action loses all dependence on $\nu$. Its motion is then undetermined at this order; nonlinear terms or a small hinge mass must resolve the degeneracy. For larger spring stiffness the straight constrained branch is not a minimum of the bending potential, despite the absence of a finite-inertia growing mode in this strictly massless quadratic description.