## Simple Harmonic Oscillator With a Resistive Term rv

In general a particle in motion is subject to many forces. There is a frictional or resistive force in most mechanical systems. Typically the resistive force depends on the speed of the vibration, and is always opposed to motion.
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
$\lambda^2+ r \lambda + \omega^2 =0$

$\lambda = \frac{-r \pm \sqrt{r^2-4 m \omega^2}}{2m}$
.
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
$x=e^{-\frac{r}{2m}t}(A+Bt)$
.