## Overdamped solution to a damped driven oscillator

A damped, driven oscillator is described by the equation,

$$m\frac{d^2x}{dt^2}+b\frac{dx}{dt} + kx = F_0\cos(\omega t).$$

If $b^2-4km > 0$, the system is overdamped and the solution has the form,

$$x(t) = C_1 \exp \left(\lambda_1 t\right)+ C_2 \exp \left(\lambda_2 t\right)+\frac{F_0}{\rho}\cos (\omega t -\theta),$$

where

$$\rho = \sqrt{(k-m\omega^2)^2+\omega^2b^2},\qquad\tan\theta = \frac{\omega b}{k-m\omega^2},\qquad \lambda_1 = \frac{-b+\sqrt{b^2-4km}}{2m},\qquad\lambda_2 = \frac{-b-\sqrt{b^2-4km}}{2m},$$ $$C_1 = \frac{1}{\lambda_1-\lambda_2}\left(v_{x0} -\lambda_2x_0 +\frac{\omega F_0}{\rho}\sin (-\theta)+\lambda_2\frac{F_0}{\rho}\cos (-\theta)\right),$$ $$C_2 = \frac{1}{\lambda_2-\lambda_1}\left(v_{x0}-\lambda_1 x_0 +\frac{\omega F_0}{\rho}\sin (-\theta)+\lambda_1\frac{F_0}{\rho}\cos (-\theta)\right).$$

Here $x_0$ is the position at $t=0$ and $v_{x0}$ is the velocity at $t=0$.

 $x$ $t$
 $v_x$ $t$
 $m=$ 1 [kg] $F_0=$ 1 [N] $b/\sqrt{4km}=$ 1 $\omega=$ 1 [rad/s] $k=$ 1 [N/m] $x_0=$ 1 [m] $v_{x0}=$ 1 [m/s]