 Solution

The natural normal modes belong to the homogeneous system with both end displacements fixed at zero. Seek $\mathbf{y}=\mathbf{a}\cos \Omega t$, soThe characteristic polynomial is $(3{\omega }_0^2-{\Omega }^2{)}^2-{\omega }_0^4=0$. The eigenvectors $(1,1)$ and $(1,-1)$ therefore giveThe 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:This shows that the boundary drive excites both natural modes; neither natural resonance is removed by a zero drive projection.