 Solution

Use a translating accelerated reference frame; its axes do not rotate. If $\mathbf{R}=(\xi ,\eta ,\zeta )$ is the block's position relative to the table, then its inertial position is $\mathbf{r}=\mathbf{R}+\mathbf{A}\sin \omega t$. Its relative velocity is $\dot{\mathbf{R}}$ and its inertial acceleration is $\ddot{\mathbf{R}}-{\omega }^2\mathbf{A}\sin \omega t$. Let $N$ be the upward normal force. During contact $\zeta =\dot{\zeta }=\ddot{\zeta }=0$, and the linear tangential resistance givesThese are Newton's second law transformed to the table's reference frame; the extra terms arise from the table's acceleration. The last equation gives $N=m(g-A_z{\omega }^2\sin \omega t)$ while contact persists. If contact is lost, both the contact resistance and $N$ vanish, and free flight must be described by the corresponding gravity-only equations.

For $A_z=0$, assume $\lambda >0$ and put $\gamma =\lambda /m$. Seek a periodic particular solution ${\xi }_p=b\sin \omega t+c\cos \omega t$. Substitution into the first differential equation givesSolving yieldsThe homogeneous solutions are a constant and a multiple of $e^{-\gamma t}$. Choose the phase $0<\theta <\pi /2$ with $\tan \theta =\omega /\gamma =m\omega /\lambda$. Then $b=-A_x{\sin }^2\theta$ and $c=-A_x\sin \theta \cos \theta$, soAlso $\eta (t)={\eta }_{\infty }+C_{\eta }{e}^{-\lambda t/m}$. The decaying transients leave the stated steady orbit. Positive damping is essential to this late-time assertion; for $\lambda =0$ an additional constant drift may persist.

No attractive contact force is available, so $N$ cannot be negative. Over a whole oscillation its minimum is $m(g-{\omega }^2|A_z|)$. Thus strictly positive contact loading throughout requires $\boxed{{\omega }^2|A_z|<g}$, or the printed form with $A_z$ taken as a nonnegative vertical amplitude. If ${\omega }^2|A_z|>g$, contact cannot persist: it would require $N<0$ during part of the cycle. At equality, $N$ vanishes only at isolated instants and the ideal constrained motion can still remain in contact. Hence the strict inequality characterizes positive loading; mere nonnegative contact permits the marginal equality case.