= Solution
For the steady response of a <forced harmonic oscillator>, write $x=\operatorname{Re}(Ae^{i\omega t})$ and use the same <complex amplitude> convention for the driving <force>. Substitution gives
$$
[-\omega^2+i\gamma\omega+\omega_0^2]A=F_0/m,\qquad
\boxed{A=R(\omega)F_0,\quad R(\omega)=\frac1{m(\omega_0^2-\omega^2+i\gamma\omega)}}.
$$
Thus the response magnitude and displacement <phase> relative to the <force> are
$$
|R|=\frac1{m\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad
\phi=-\operatorname{atan2}(\gamma\omega,\omega_0^2-\omega^2).
$$
The <phase> goes from $0$ at low <frequency>, through $-\pi/2$ at $\omega_0$, towards $-\pi$ at high <frequency>. The static magnitude is $1/(m\omega_0^2)=1/k$, while the high-frequency magnitude is approximately $1/(m\omega^2)$.
Define the <quality factor> by $Q=\omega_0/\gamma$. If $Q>1/\sqrt2$, the displacement peak occurs at $\omega_r=\sqrt{\omega_0^2-\gamma^2/2}$; for $Q\leq1/\sqrt2$ the magnitude decreases monotonically from its static value. For $Q\gg1$ there is a tall narrow peak close to $\omega_0$, of height approximately $Q/k$ and width approximately $\gamma$. At very small $Q$ the resonant peak disappears and the response is broad. The plotted functions are normalized as $k|R|$, at fixed $m,k$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-b6-driven-response.png]
{title=Normalized amplitude and phase of a driven oscillator at three quality factors}
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