# Heavily Damped Simple Harmonic System - How To Start?

#### phyzmatix

1. The problem statement, all variables and given/known data

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

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#### lanedance

Homework Helper
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

Last edited:

#### 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

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