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

Take laboratory displacements from the equilibrium positions with the support at $f=0$. The left spring's dynamic extension is $x_1-f(t)$ and the spring between the masses has dynamic extension $x_2-x_1$. Hooke's law and Newton's second law giveEquivalently,The natural length sets the equilibrium separations but disappears from these displacement equations. The support is externally prescribed, so no additional support equation is required.

Set the support displacement to zero and seek a normal mode $\mathbf{x}=(A,B{)}^{\mathsf{T}}\cos (\omega t+\phi )$. With $s=m{\omega }^2/k$, the generalized eigenvalue problem requiresThusThe second row gives $A=(1-s)B$. The signed displacement ratios are thereforeThe slow mode is in phase and the fast mode out of phase. Keeping the unequal mass matrix is essential: replacing it by the identity changes the eigenfrequencies.

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.