RLC Damped Oscillator: Why Quicker to Zero in Critically Damped Case?

[Chen]

Hello,

On this page:
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Oscillations.htm
It says (and shows) that in the case of a critically damped oscillation, it moves more quickly to zero than in the overdamped case.

I don't understand why. The solution to this circuit is:

Q(t) = A e^{(i\omega - \alpha )t}

Where \omega is the square root of some expression that depends on R and L. The critically damped case corresponds to \omega = 0, while the overdamped case corresponds to the case in which \omega is imaginary.

So in the critically damped case the solution is:

Q(t) = A e^{-\alpha t}

And in the overdamped case it is:

Q(t) = A e^{(-P - \alpha )t}

Where P is some positive number, assuming \omega = \sqrt{-P^2}. To my best understanding, the solution should move to zero quicker in the overdamped case. However, that's not the case, as I've seen on my webpages and read in many books.

Can someone please explain this?

Thanks,
Chen

 

[vanesch]

So in the critically damped case the solution is:

Q(t) = A e^{-\alpha t}

You are forgetting the B t e^{-\alpha t} part of the solution, no ?

Have a look here:
http://physics.ucsd.edu/neurophysics/courses/physics_2bl/p2bl_experiment_3_notes_sho.pdf

cheers,
Patrick.

 

[Chen]

Yes, indeed. Thank you!

Chen

 

[SGT]

You are forgetting the B t e^{-\alpha t} part of the solution, no ?

Have a look here:
http://physics.ucsd.edu/neurophysics/courses/physics_2bl/p2bl_experiment_3_notes_sho.pdf

cheers,
Patrick.

This term really tends to zero more slowly than the first term. The problem is that in the overdamped case there is a term
Be^{(+P-\alpha)t}
which tends very slowly to zero.