= Solution
After transients have decayed, seek the driven response $\mathbf x=\mathbf a\sin\omega t$. At the given $\omega^2=k/m$, substitution gives
$$
\begin{pmatrix}2&-2\\-2&0\end{pmatrix}\mathbf a=\begin{pmatrix}F\\0\end{pmatrix}.
$$
The second row forces $a_1=0$, and the first gives $a_2=-F/2$. Hence, in the negligible-damping steady-response limit,
$$
\boxed{x_1(t)=0,\qquad x_2(t)=-\frac F2\sin\omega t.}
$$
The support forcing on the first mass is balanced by the inter-mass spring force while the second mass moves in antiphase. This is a <support-driven unequal-mass spring antiresonance>, not a resonance: the drive frequency lies between the two normal-mode frequencies. A genuinely finite damping coefficient produces small phase shifts and a small nonzero first-mass response; the displayed formula uses the source's instruction to neglect damping except for removal of transients.
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