Taking a driving force:
$$ f(t)=F_0\cos\omega t $$
and adding it to the forced damped oscillations we get:
$$ \ddot{x} + \gamma \dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos \omega t $$
To solve the above equation we will change it to the complex form by subbing in $z=x+iy$
$$ \ddot{z} + \gamma \dot{z} + \omega_0^2 z = \frac{F_0}{m} \exp (i \omega t) $$
Trying the following answer as a solution:
$$ z(t) = A(\omega) \exp^{i(\omega t - \delta)} $$
we find that the following need to be satisfied for it to be an answer:
$$ \begin{aligned} \mathrm{Real:} \quad \quad (\omega_0^2- \omega^2) A(\omega) &= \frac{F_0}{m} \cos \delta \\ \mathrm{Im:} \;\; \quad \quad \quad \quad \quad \gamma \omega A(\omega) &= \frac{F_0}{m} \sin \delta \end{aligned} \qquad $$
$$ \tan \delta = \frac{ \gamma \omega}{\omega_0^2 - \omega^2} $$
Phase angle: $\delta(\omega)$
Phase angle graph
Amplitude response function graph
$$ \begin{aligned} A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + \omega^2 \gamma^2}} \end{aligned} $$
Amplitude response function: $A(\omega)$
$$ \begin{aligned} \lim_{\omega\to0}{ A(\omega)} &= \frac{F_0}{m \omega_0^2} \left( =\frac{F_0}{k} \right) \\ \lim_{\omega\to\infin}{A(\omega)}&=0 \\ \lim_{\omega\to\omega_0}{A(\omega)}&=\frac{F_0}{m\omega_0\gamma} \end{aligned} $$
$$ A_\text{max} = \frac{F_0}{m \omega_0 \gamma} \frac{1}{\sqrt{1 - \frac{\gamma^2}{4 \omega_0^2}}}\qquad \text{for} \qquad \omega= \omega_0 \left ( 1- \frac{\gamma^2}{2 \omega_0^2} \right )^{\frac1 2} $$
$$ \lim_{\begin{aligned} \omega&\to\omega_0 \\ \gamma&\to0 \end{aligned}}{A(\omega)}=+\infin $$