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Critically Damped Harmonic System

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    John the door-stopper man sells and installs door-stoppers. He prides himself as being the world's best stopper and guarantees your money back if you can install a better stopper than him.

    John's secret to door-stopping is that he remembers from his lectures that the best way to stop doors is by fine tuning the stopper such that it is critically damped.

    2. Relevant equations
    Trick John (and ask for your money back) by showing him that if you open a critically damed door and release it without also pushing it, it will never ever close again.

    3. The attempt at a solution
    The equation of motion is a second order differential equation in the form: mx"+cx'+kx=0; then let x=e^(zt);

    I realise that for a critically damped system, the determinent for the characteristic equation z^2 + (c/m)z + (k/m) = 0 equals to (c/m)^2 - 4(k/m) = 0;

    Then solving for x to find the equation in the form of x(t) = (A + Bt)e^(-t(k/m)^0.5)

    However, the problem is that this solution converges to the equilibrium position under critically damped conditions.

    How can I show that the door will never close again?? (It seems impossible!!)

    Your help is much appreciated! :smile:
  2. jcsd
  3. Oct 6, 2007 #2
    Careful, you are looking at the discriminant, which is different from the determinate. Also, your solution should be

    [tex]x(t) = (A+Bt)e^{-\frac{c}{2m}t}[/tex]

    But you might as well save same writing and just call c/2m beta or gamma. Those are just some notational issues, and stupid mistakes. They don't really pertain all that much to the solution, but should be corrected nonetheless. (Edit: Oh wait, I see what you (or your instructor) did. You can leave your solution how you have it.)

    You have a couple initial conditions, one is that the door is open some amount x_0 at time = 0. The other is that there is no initial velocity so x'(0) = 0. Then you want to know when the door stops moving, when it is closed.

    See if this gets you anywhere.
    Last edited: Oct 6, 2007
  4. Oct 6, 2007 #3
    You have to set the equilibrium position to the open position. In that case the door will stop there.
  5. Oct 7, 2007 #4
    Thanks you guys!! I see, so the trick is to have the initial condition as being opened already.
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