 Solution

Both masses in the source's absorber diagram are free: there is no additional spring attaching $M$ to a wall. Let their displacements be $X$ and $x$. The equations areFor a periodic particular response $X=A\cos \omega t$, $x=B\cos \omega t$, the amplitude equations are $(k-M{\omega }^2)A-kB=F_0$ and $-kA+(k-m{\omega }^2)B=0$. Their solution isAt $\boxed{k=m{\omega }^2}$, $A=0$ and $B=-F_0/k$: the spring force on the larger mass cancels the applied force at every instant. This is antiresonance, a frequency-selective cancellation, rather than dissipation of energy.

There is a nearby undamped resonance at $k_{\mathrm{res}}=Mm{\omega }^2/(M+m)$. For $m\ll M$ it lies close to the cancellation setting. More quantitatively, write $k=m{\omega }^2(1+ϵ)$. ThenFor $|ϵ|\ll m/(M+m)$, the amplitude relative to the unconnected mass's amplitude $F_0/(M{\omega }^2)$ is approximately $(M/m)|ϵ|$. A light absorber therefore needs very accurate tuning and loses suppression when the driving frequency changes. Its own travel $|B|=F_0/(m{\omega }^2)$ can also become large, limiting linearity and available stroke.

Without damping, resonance growth and persistent free oscillations remain; special initial conditions select the displayed periodic particular response. The free pair also has rigid-translation motion. These limitations prevent interpreting the cancellation as broadband damping or as a unique attracting late-time state.