= 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 are
$$
M\ddot X=F_0\cos\omega t+k(x-X),\qquad m\ddot x=-k(x-X).
$$
For 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 is
$$
\boxed{A=\frac{F_0(k-m\omega^2)}{\omega^2[Mm\omega^2-k(M+m)]},\qquad B=\frac{F_0k}{\omega^2[Mm\omega^2-k(M+m)]}}.
$$
At $\boxed{k=m\omega^2}$, $A=0$ 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 $k_{\rm res}=Mm\omega^2/(M+m)$. For $m\ll M$ it lies close to the cancellation setting. More quantitatively, write $k=m\omega^2(1+\epsilon)$. Then
$$
A=-\frac{F_0\epsilon}{\omega^2[m+\epsilon(M+m)]}.
$$
For $|\epsilon|\ll m/(M+m)$, the <amplitude> relative to the unconnected mass's <amplitude> $F_0/(M\omega^2)$ is approximately $(M/m)|\epsilon|$. 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.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-coupled-spring-modes-and-absorber.png]
{title=Symmetric spring modes and narrow tuning of a free-pair dynamic absorber}
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.
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