$\textbf{Given the two equations}$
$$ (1) \text{ } m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt}+kx=F$$ , where x is the distance from the "resting position", k is the spring constant, m is the mass, $\gamma$ is the reduction factor and F is the outer force $$(2) \text{ } \ddot{x}+2 \xi \Omega \dot{x}+ \Omega^2 x=fcos(\omega t) $$ , where $(\ddot{})$ means differentiation 2 times with respect to the time variable t. Here $\omega >0, \Omega > 0, \xi \geq 0$ and f is a given size
How can one show that Eq(1) can be rewritten to Eq(2) as well as determining $\Omega$ and $\xi$ in terms of physical sizes?
Simply divide by $m$ and set
$\Omega=\sqrt{\frac{k}{m}}$
$2 \xi \Omega=\frac{\gamma}{m}\implies \xi =\sqrt{\frac{m}{k}}\frac{\gamma}{2m}=\frac{\gamma}{2\sqrt{km}}$
$f=\frac F m$