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Heavily Damped Simple Harmonic System - How To Start?

  • Thread starter phyzmatix
  • Start date
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

[tex]x=e^{-pt}(F\cosh qt + G\sinh qt)[/tex]

where

[tex]p=\frac{r}{2m}[/tex]

and

[tex]q=(\frac{r^2}{4m^2}-\frac{s}{m})^{\frac{1}{2}}[/tex]

that the ratio

[tex]\frac{G}{F}=\frac{r}{(r^2-4ms)^{\frac{1}{2}}}[/tex]


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

Homework Helper
3,304
2
hi phyz,

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

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

[tex]m\ddot{x}+r\dot{x}+sx=0[/tex]

???

Let me give it a go :smile:
 
Last edited:

HallsofIvy

Science Advisor
41,625
823
[tex]x'(t)= -pe^{-pt}(F\cosh qt+ G\sinh qt)+ e^{-pt}(qF\sinh qt+ qG\cosh qt)[/tex]

Since it is released from rest, we have
[tex]x'(0)=-pF+ qG= 0[/tex]
That should be enough.
 
[tex]x'(t)= -pe^{-pt}(F\cosh qt+ G\sinh qt)+ e^{-pt}(qF\sinh qt+ qG\cosh qt)[/tex]

Since it is released from rest, we have
[tex]x'(0)=-pF+ qG= 0[/tex]
That should be enough.
Thank you! That did the trick :smile:
 

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