Past exam of the physics 1 paper of the University of Oxford / 2015 / 10 / e / Solution

Take the natural release interpretation: both block and bob are initially at rest, the block starts at $x_0$, and the bob has small horizontal displacement $\delta =l{\theta }_0$ relative to the downward line through the pivot. Then $P=0$, $\dot{\theta }(0)=0$ andThe bob's horizontal position isThe block and bob therefore oscillate in opposite phase about the fixed horizontal centre of mass $x_0+m\delta /(M+m)$. Their horizontal oscillation amplitudes are $m\delta /(M+m)$ and $M\delta /(M+m)$ respectively. They pass the centre-of-mass line together when the rod is vertical. If the bob were also given an unspecified initial velocity, the additional sine and translational terms in part (d) would be needed; the source's displacement-release case selects the cosine solution.