= Solution
Put $G=g/\ell$ and $s=\kappa/m$. Applying the <Euler-Lagrange equations> to the <quadratic Lagrangian> gives
$$
\boxed{2\ddot\theta+\ddot\phi+(2G-s)\theta+s\phi=0,\qquad
\ddot\theta+\ddot\phi+s\theta+(G-s)\phi=0.}
$$
A nonzero motion satisfying the specified equal-and-opposite initial displacements can have one <frequency> only if $(1,-1)$ is a normal-mode <vector>. Its acceleration terms are proportional to $(1,0)$, while the restoring terms are proportional to $(2G-2s,2s-G)$. Hence
$$
\boxed{s=G/2,\qquad \kappa=\frac{mg}{2\ell},\qquad \omega^2=G,}
$$
and
$$
\boxed{\theta(t)=\xi\cos(\sqrt{g/\ell}\,t),\qquad
\phi(t)=-\xi\cos(\sqrt{g/\ell}\,t).}
$$
The rods swing oppositely. The hinge <mass> moves sideways, while the final <mass> has no first-order displacement: the two horizontal contributions cancel and its vertical change is quadratic in angle. In the corresponding geometric ansatz it moves vertically at twice the small-oscillation <frequency> to leading nonzero order, but the displayed single-frequency solution is a linearized result. The trivial case $\xi=0$ imposes no parameter condition.
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