= Solution
For a fixed-pivot pendulum the angular inertia is $ml^2$, so $ml^2\ddot\Theta=-dV_{\rm eff}/d\Theta$. Integrating the averaged force gives, up to an arbitrary additive constant,
$$
\boxed{V_{\rm eff}(\Theta)=-mgl\cos\Theta+
\frac{ma^2\omega^2}{4}\cos^2\Theta.}
$$
Linearizing about $\Theta=0$ gives
$$
\ddot\Theta+\left(\frac gl-\frac{a^2\omega^2}{2l^2}\right)\Theta=0,
\qquad V_{\rm eff}''(0)=mgl-\frac{ma^2\omega^2}{2}.
$$
Hence the downward state becomes linearly unstable for
$$
\boxed{a>a_c=\frac{\sqrt{2gl}}{\omega}.}
$$
At $a=a_c$ 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 $a_c/l=\sqrt2\sqrt{g/l}/\omega\ll1$, the threshold is compatible with the assumed small support excursion.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1-horizontal-pendulum.png]
{title=Original averaged pendulum potentials below, at and above the horizontal-vibration instability threshold}
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