Past exam of the theoretical physics I course of the University of Cambridge / 2015 / Paper 1 / 1 / d / Solution

For a fixed-pivot pendulum the angular inertia is $m{l}^2$, so $m{l}^2\ddot{\Theta }=-d{V}_{\mathrm{eff}}/d\Theta$. Integrating the averaged force gives, up to an arbitrary additive constant,Linearizing about $\Theta =0$ givesHence the downward state becomes linearly unstable forAt $a=a_c$ its quadratic restoring term vanishes: the leading quartic coefficient is positive, so it is marginal at linear order, rather than already a local maximum. Above threshold the two stable tilted equilibria satisfy $\cos \Theta =2gl/(a^2{\omega }^2)$. Horizontal vibration therefore destabilizes the downward orientation and produces tilted minima; it does not stabilize the upward orientation in this model. Because $a_c/l=\sqrt{2}\sqrt{g/l}/\omega \ll 1$, the threshold is compatible with the assumed small support excursion.