Overdamped condition

  • #1
648
15
How to get overdamping condition of equation
[tex]m\ddot{x}+\dot{x}+kx=0,[/tex]
Taking ##x=\mbox{e}^{\lambda t}##, we got
[tex]\lambda_{1/2}=\frac{-1 \pm \sqrt{1-4mk}}{2m}.[/tex]
Is it possible from this ##\lambda## values to got overdamped condition?
I found that if we have equation
[tex]m \ddot{x}+\gamma \dot{x}=f(x),[/tex]
then ##-4m\frac{\partial f}{\partial x} \leq \gamma^2## is overdamped condition. How to find it? Any help?
 

Answers and Replies

  • #2
Orodruin
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Hint: The overdamped condition is just that your eigenvalues are real and different. What does that translate to?

Also, you can easily rewrite your equation on this form:
[tex]m \ddot{x}+\gamma \dot{x}=f(x),[/tex]
What do you get for ##f(x)## and ##\gamma## when doing so?
 
  • #3
2,298
647
Hint: The overdamped condition is just that your eigenvalues are real and different.

This is precisely the correct answer. There is really nothing more to be said.
 
  • #4
648
15
Thanks a lot. Do you have some references for that?
 
  • #5
648
15
And one more question. What if equation is nonlinear
[tex]\frac{d^2y}{dx^2}+\sin y=0[/tex]
Should I linearized equation first or what?
 
  • #6
2,298
647
What you should do first depends upon what you want to accomplish. If you want the solution to the linearized equation, by all means proceed to linearize it. If, on the other hand, you want the solution to the nonlinear equation, linearization is simply wasted effort.
 

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