The simple harmonic oscillator equation
\[m \frac{d^2x}{dt^2}=- \omega^2 x\]
becomes \[m\frac{d^2x}{dt^2}=- \omega^2 x-r v=- \omega^2 x-r \frac{dx}{dt}\]
with a resistive term proportional to the speed. Write this as\[m \frac{d^2x}{dt^2}+ r \frac{dx}{dt}+ \omega^2 x=0\]
Assume a solution of the form
\[x=Ae^{kt}\]
then \[\frac{dx}{dt}= \lambda Ae^{kt}, \; \frac{d^2x}{dt^2}= \lambda^2Ae^{kt}\]
then the equation becomes\[m \lambda^2Ae^{ \lambda t}+ r \lambda Ae^{\lambda t}+ \omega^2 Ae^{ \lambda t}=0\]
The factor
\[e^{ \lambda t}\]
is never zero so we can divide by it to give\[\lambda^2+ r \lambda + \omega^2 =0\]
Using thew quadratic formula,
\[\lambda = \frac{-r \pm \sqrt{r^2-4 m \omega^2}}{2m}\]
.The nature of the vibration depends on the discriminant
\[\Delta = r^2-4m \omega^2\]
.If
\[\Delta \lt 0\]
then \[x=e^{-\frac{r}{2m}t}(C_1 cos( \frac{\sqrt{4m \omega^2-r^2}}{2m}t)+C_2 sin ( \frac{\sqrt{4m \omega^2-r^2}}{2m}t) \]
and motion is vibratory with decreasing amplitude.If
\[\Delta \gt 0\]
then \[x=C_1 e^{{\frac{-r + \sqrt{r^2-4 m \omega^2}}{2m}}t}+C_2e^{\frac{-r - \sqrt{r^2-4 m \omega^2}}{2m}{t}}\]
the motion is not vibratory and the system move to an equilibrium.If
\[\Delta = 0\]
then \[x=e^{-\frac{r}{2m}t}(A+Bt)\]
.