= Solution
Removing the hinge <mass> gives the exact <Lagrangian>
$$
L=\frac{m\ell^2}{2}[\dot\theta^2+\dot\phi^2+2\dot\theta\dot\phi\cos(\theta-\phi)]
+mg\ell(\cos\theta+\cos\phi)-\kappa\ell^2[1+\cos(\theta-\phi)].
$$
The stated equations again concern its small-angle quadratic approximation. In the sum and difference coordinates this becomes
$$
\boxed{L_2=\frac{m\ell^2}{2}\left[\dot\eta^2-\omega_0^2\eta^2
-(\omega_0^2-\omega_1^2)\nu^2\right],\quad
\omega_0^2=\frac g{2\ell},\quad \omega_1^2=\frac\kappa m.}
$$
Consequently
$$
\boxed{\ddot\eta+\omega_0^2\eta=0,\qquad
(\omega_0^2-\omega_1^2)\nu=0.}
$$
For 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 $\nu$. \b[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.
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