= Solution
Keep the same fixed laboratory equilibrium reference, and let the pivot move downward by $z(t)=z_0\cos\omega_0t$. The dynamic upper-spring extension is now $x_1-z$, so
$$
m\ddot{\mathbf x}+k\begin{pmatrix}2&-1\\-1&1\end{pmatrix}\mathbf x=k\begin{pmatrix}z(t)\\0\end{pmatrix}.
$$
At $\omega_0^2=k/m$, seek a particular solution $\mathbf x_p=\mathbf b\cos\omega_0t$. The amplitude equation is
$$
\begin{pmatrix}1&-1\\-1&0\end{pmatrix}\mathbf b=\begin{pmatrix}z_0\\0\end{pmatrix},\qquad \mathbf b=\begin{pmatrix}0\\-z_0\end{pmatrix}.
$$
Let $s_\pm=(3\pm\sqrt5)/2$, $\omega_\pm=\sqrt{ks_\pm/m}$ and $\mathbf u_\pm=(1-s_\pm,1)^{\mathsf T}$. The unrestricted general motion is
$$
\boxed{\mathbf x(t)=\sum_{\sigma=\pm}\mathbf u_\sigma[C_\sigma\cos\omega_\sigma t+D_\sigma\sin\omega_\sigma t]+\begin{pmatrix}0\\-z_0\cos\omega_0t\end{pmatrix}.}
$$
Four initial-condition constants are retained because none are specified. The driven particular response is not a resonance: $\omega_0$ differs from both eigenfrequencies. Its stationary upper mass is the <pivot-driven two-mass spring antiresonance>. Relative to the moving pivot, subtract $z(t)$ from both components.
Back to article page
