= Solution
The natural <normal modes> belong to the homogeneous system with both end displacements fixed at zero. Seek $\mathbf y=\mathbf a\cos\Omega t$, so
$$
\omega_0^2\begin{pmatrix}3&-1\\-1&3\end{pmatrix}\mathbf a=\Omega^2\mathbf a.
$$
The <characteristic polynomial> is $(3\omega_0^2-\Omega^2)^2-\omega_0^4=0$. The <eigenvectors> $(1,1)$ and $(1,-1)$ therefore give
$$
\boxed{\Omega_+=\sqrt2\,\omega_0,\quad y_1=y_2;\qquad\Omega_-=2\omega_0,\quad y_1=-y_2}.
$$
The symmetric mode leaves the middle segment horizontal, while the antisymmetric mode bends it strongly and has the higher restoring stiffness. Defining $q_+=(y_1+y_2)/2$ and $q_-=(y_1-y_2)/2$ diagonalizes the driven equations too:
$$
\ddot q_++2\omega_0^2q_+=\omega_0^2h(t),\qquad\ddot q_-+4\omega_0^2q_-=-\omega_0^2h(t).
$$
This shows that the boundary drive excites both natural modes; neither natural <resonance> is removed by a zero drive projection.
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