 Solution

Put $G=g/\ell$ and $s=\kappa /m$. Applying the Euler-Lagrange equations to the quadratic Lagrangian givesA 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 $(2G-2s,2s-G)$. HenceandThe 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.