1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

RLC damped oscillator

  1. Feb 6, 2006 #1
    Hello,

    On this page:
    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Oscillations.htm [Broken]
    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:

    [tex]Q(t) = A e^{(i\omega - \alpha )t}[/tex]

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

    So in the critically damped case the solution is:

    [tex]Q(t) = A e^{-\alpha t}[/tex]

    And in the overdamped case it is:

    [tex]Q(t) = A e^{(-P - \alpha )t}[/tex]

    Where P is some positive number, assuming [tex]\omega = \sqrt{-P^2}[/tex]. 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
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 6, 2006 #2

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  4. Feb 6, 2006 #3
    Yes, indeed. Thank you!

    Chen
     
  5. Feb 8, 2006 #4

    SGT

    User Avatar

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: RLC damped oscillator
  1. Damped Oscillation (Replies: 5)

Loading...