Past exam of the mathematical methods II course of the University of Oxford / 2013 resit / 9 / c / Solution

After transients have decayed, seek the driven response $\mathbf{x}=\mathbf{a}\sin \omega t$. At the given ${\omega }^2=k/m$, substitution givesThe second row forces $a_1=0$, and the first gives $a_2=-F/2$. Hence, in the negligible-damping steady-response limit,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.