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

Let the pivot's prescribed horizontal position be $x(t)$ and measure $\theta$ from the downward vertical. The bob's coordinates are $X=x+l\sin \theta$ and $Y=-l\cos \theta$. ThereforeWith zero gravitational potential energy at pivot height, the bob's Lagrangian isHere $x(t)$ is a prescribed input, not yet a free coordinate. Terms depending only on that prescribed time function can be dropped without changing the $\theta$ equation. The moving block's own kinetic energy becomes relevant when its motion is free in part (d).

The Euler-Lagrange equation for $\theta$ isFor small angles, set $\cos \theta \simeq 1$ and $\sin \theta \simeq \theta$. With $x=A\cos \omega t$ and ${\omega }_0^2=g/l$, the harmonic forcing of a horizontally driven pendulum obeysFor $\omega \ne {\omega }_0$, its general solution isThe constants $C,D$ set the initial angle and angular velocity. If the bob starts vertical with zero angular velocity at $t=0$, this reduces toThe solution remains an approximation only while the resulting angle is small.

As $\omega$ approaches ${\omega }_0=\sqrt{g/l}$, the driving motion couples resonantly to the pendulum. The particular-solution amplitude has denominator ${\omega }_0^2-{\omega }^2$, so a steady harmonic-response formula alone is singular. For a specified finite initial state, the near-resonant motion instead develops beats, and its exact resonant limit has a growing envelope.

For example, with the vertical zero-angular-velocity initial condition above, taking the limit givesThe ideal undamped linear theory predicts amplitude growing linearly with time, not an infinite displacement at the first instant. In reality the motion leaves the small-angle regime, and damping and nonlinear changes of the pendulum's oscillation period alter the response. Below and above resonance, the sign of the harmonic response coefficient changes, corresponding to a phase reversal.

When the block is free, $x$ becomes a dynamical generalized coordinate and its kinetic energy $M{\dot{x}}^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant givesThe coordinate $x$ is cyclic. Its conserved canonical momentum is the total horizontal momentum,The two linear Euler-Lagrange equations areEliminating $\ddot{x}$ givesThus the pendulum with a freely sliding support has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution isHere $V=P/(M+m)$ and $X_c+Vt$ is the horizontal centre of mass coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to \infty$ restores $\Omega \to \sqrt{g/l}$.

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.