## 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\[\lambda^2+ r \lambda + \omega^2 =0\]

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 \[x=e^{-\frac{r}{2m}t}(A+Bt)\]

.