= Solution
For horizontal forcing the exact angular equation is
$$
\ddot\theta=-\frac gl\sin\theta+\frac{a\omega^2}{l}\cos\theta\cos\omega t.
$$
Write $\theta=\Theta+\xi$ with $\Theta=\theta_1$ slowly varying and $\xi=-(a/l)\cos\Theta\cos\omega t$. To leading order in the slow-to-fast frequency ratio, $\ddot\xi=(a\omega^2/l)\cos\Theta\cos\omega t$, which balances the rapid forcing. Expand the right-hand side to first order in $\xi$:
$$
\ddot\Theta+\ddot\xi
=-\frac gl(\sin\Theta+\xi\cos\Theta)
+\frac{a\omega^2}{l}(\cos\Theta-\xi\sin\Theta)\cos\omega t.
$$
During one fast period treat $\Theta$ as fixed. The averages $\langle\xi\rangle$ and $\langle\cos\omega t\rangle$ vanish, whereas
$$
\langle\xi\cos\omega t\rangle=-\frac{a}{2l}\cos\Theta.
$$
The leading averaged equation is therefore
$$
\boxed{\ddot\Theta=-\frac gl\sin\Theta+
\frac{a^2\omega^2}{2l^2}\sin\Theta\cos\Theta.}
$$
The effective force comes from correlation between the rapid displacement and the rapidly varying driving <torque>; simply averaging the driving cosine to zero misses it. This keeps the leading nonzero fast-motion contribution, proportional to $a^2\omega^2$, while neglecting terms suppressed by additional powers of $a/l$ or of the slow rate divided by $\omega$. The gravitational expansion at order $g(a/l)^2$ is smaller in the stated high-frequency regime. The resulting slow frequencies must also remain small compared with $\omega$ for averaging to be self-consistent.
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