Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S7 / 1 / 4 / Solution

For $\varphi =-\Omega$, $p_1=m({\dot{x}}_1-R\Omega \sin \Omega t)$ and $p_2=m({\dot{y}}_1+R\Omega \cos \Omega t)$. The equations are ${\ddot{x}}_1+{\omega }^2{x}_1=R{\Omega }^2\cos \Omega t$, ${\ddot{y}}_1+{\omega }^2{y}_1=R{\Omega }^2\sin \Omega t$. For $\Omega \ne \omega$,At resonance the forced terms are $R\omega t\sin \omega t/2$ and $-R\omega t\cos \omega t/2$. Constants depend on the marble's initial position and velocity. A general cup orientation is obtained by multiplying the vector solution by $\mathcal{R}(-\alpha t)$; the physical inertial solution is unchanged by relabeling the cup axes. Resonant growth eventually invalidates the small-horizontal-motion/neglected-vertical-speed approximation.