Past exam of the physics A course of the University of Cambridge / 2014 / Paper 2 / Section B / B6

For pendulum length $\ell$, small angular displacement gives $x\simeq \ell \theta$ and horizontal restoring force $-mgx/\ell$. Thus ${\omega }_0^2=g/\ell$ andFor $\omega \ne {\omega }_0$, the complementary solution and a sinusoidal particular solution giveInitial rest at the origin fixes $A=0$ and $B=-F_0\omega /[m{\omega }_0({\omega }_0^2-{\omega }^2)]$, givingAt exact resonance the corresponding limit is $x=F_0[\sin {\omega }_{0}t-{\omega }_{0}t\cos {\omega }_{0}t]/(2m{\omega }_0^2)$; unrestricted initial conditions add the complementary solution.

For $\omega =2{\omega }_0$, set $\tau ={\omega }_{0}t$. Then $m{\omega }_0^{2}x/F_0=(2\sin \tau -\sin 2\tau )/3$: the period is $2\pi$ in $\tau$, zeros are at integer multiples of $\pi$, and the first extrema are $(2\pi /3,\sqrt{3}/2)$ and $(4\pi /3,-\sqrt{3}/2)$. Both initial displacement and slope vanish.

Substitution of the damped complex ansatz gives $[m({\omega }_0^2-{\omega }^2)+ib\omega ]x_0=F_0$. HenceFor the near-resonant mechanical susceptibility, put $\gamma =b/(2m)\ll {\omega }_0$ and $\delta =\omega -{\omega }_0$, and neglect ${\delta }^2$ and $b\delta$ near resonance:Thus $|R{|}^2\simeq [4m^2{\omega }_0^2({\delta }^2+{\gamma }^2){]}^{-1}$, with angular-frequency full width $2\gamma =b/m$. The displacement lags the force by $\pi /2$ at ${\omega }_0$.