= Solution
For pendulum length $\ell$, small angular displacement gives $x\simeq\ell\theta$ and horizontal restoring force $-mgx/\ell$. Thus $\omega_0^2=g/\ell$ and
$$
\boxed{\ddot x+\omega_0^2x=\frac{F_0}{m}\sin\omega t}.
$$
For $\omega\ne\omega_0$, the <complementary solution> and a sinusoidal <particular solution> give
$$
x=A\cos\omega_0t+B\sin\omega_0t+
\frac{F_0\sin\omega t}{m(\omega_0^2-\omega^2)}.
$$
Initial rest at the origin fixes $A=0$ and $B=-F_0\omega/[m\omega_0(\omega_0^2-\omega^2)]$, giving
$$
\boxed{x=\frac{F_0}{m(\omega_0^2-\omega^2)}
\left(\sin\omega t-\frac{\omega}{\omega_0}\sin\omega_0t\right)}.
$$
At exact <resonance> the corresponding limit is $x=F_0[\sin\omega_0t-\omega_0t\cos\omega_0t]/(2m\omega_0^2)$; unrestricted initial conditions add the <complementary solution>.
For $\omega=2\omega_0$, set $\tau=\omega_0t$. Then $m\omega_0^2x/F_0=(2\sin\tau-\sin2\tau)/3$: the period is $2\pi$ in $\tau$, zeros are at integer multiples of $\pi$, and the first extrema are $(2\pi/3,\sqrt3/2)$ and $(4\pi/3,-\sqrt3/2)$. Both initial displacement and slope vanish.
Substitution of the damped complex ansatz gives $[m(\omega_0^2-\omega^2)+ib\omega]x_0=F_0$. Hence
$$
\boxed{R(\omega)=\frac1{m(\omega_0^2-\omega^2)+ib\omega}}.
$$
For 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>:
$$
\boxed{R(\omega_0)=-\frac{i}{b\omega_0},\qquad
R(\omega)\simeq\frac1{2m\omega_0(-\delta+i\gamma)}}.
$$
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 $\omega_0$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2014/paper-2-driven-response.png]
{title=Original initial-rest response at twice the natural frequency and weak-damping resonance}
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