Heavily Damped Simple Harmonic System - How To Start?

[phyzmatix]

Homework Statement

A heavily damped simple harmonic system is displaced a distance F from its equilibrium positio and released from rest. Show that in the expression for the displacement

$$x=e^{-pt}(F\cosh qt + G\sinh qt)$$

where

$$p=\frac{r}{2m}$$

and

$$q=(\frac{r^2}{4m^2}-\frac{s}{m})^{\frac{1}{2}}$$

that the ratio

$$\frac{G}{F}=\frac{r}{(r^2-4ms)^{\frac{1}{2}}}$$


2. The attempt at a solution

I've been thinking and thinking and thinking, but no luck. I'd really appreciate it if someone could just tell me where to start.

Thanks!
phyz

 

[lanedance]

hi phyz,

the differential equation of the damped harmonic oscillator is probably a good place to start...

 

[phyzmatix]

Hi lanedance!

You mean I should simply calculate xdot and xddot and plug them into

$$m\ddot{x}+r\dot{x}+sx=0$$

?

Let me give it a go

 

[HallsofIvy]

$$x'(t)= -pe^{-pt}(F\cosh qt+ G\sinh qt)+ e^{-pt}(qF\sinh qt+ qG\cosh qt)$$

Since it is released from rest, we have
$$x'(0)=-pF+ qG= 0$$
That should be enough.

 

[phyzmatix]

$$x'(t)= -pe^{-pt}(F\cosh qt+ G\sinh qt)+ e^{-pt}(qF\sinh qt+ qG\cosh qt)$$

Since it is released from rest, we have
$$x'(0)=-pF+ qG= 0$$
That should be enough.

Thank you! That did the trick