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Critical Damping: T/4?

  1. Aug 28, 2008 #1
    Critical Damping: T/4???

    I read a few Textbooks on SHM. It happens to be the ones that show no mathematical derivations, that claim that Critical Damping has the period (i.e. time from 0 to equilibrium) of T/4

    So I went back to my derivations of SHM considering second order differential equations. The solutions I got were similar to that stated in Wikipedia:
    http://en.wikipedia.org/wiki/Harmonic_oscillator#Simple_harmonic_oscillator

    I have tried numerous approaches to ecen try and show how in a Damped Oscillator the period is T/4 compared to Overdamping or Under Damping. Can anyone explain or guide me to a way of showing this result
     
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  3. Aug 29, 2008 #2

    Redbelly98

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    Re: Critical Damping: T/4???

    I'm assuming T is the natural period of an undamped system.

    I don't see how you can say "Critical Damping has the period (i.e. time from 0 to equilibrium) of T/4".

    First, if you mean "period" in the sense of an oscillation, it doesn't apply since critically damped systems do not oscillate. Also, they never actually reach equilibrium, but approach closer and closer to it as time increases.

    Edit:
    The 2nd of the 3 figures here illustrates what I am saying:
    http://hyperphysics.phy-astr.gsu.edu/Hbase/oscda.html
     
  4. Aug 29, 2008 #3
    Re: Critical Damping: T/4???

    Yes. But I did however, mention that by "period" the books referred to the time they reached the Equilibrium position. THey claim that this time is 1/4 of the time for overdamping to reach its equilibrium.

    Of course they never reach the equilibrium position. However, critical damping seems to go towards the point 4 times faster than that of overdamping.

    I just want a mathematical proof of this phenomena...
     
  5. Aug 30, 2008 #4

    uart

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    Re: Critical Damping: T/4???

    Yes. But Redbelly98 did however mention that critically damped systems never actually reach equilibrium. So your question makes no sense.

    Now if you want to modify your question and ask something sensible, say for example the time taken for it to get to within 10% of it's final value following a step change, then I'm sure someone could help you.
     
  6. Aug 30, 2008 #5

    Redbelly98

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    Re: Critical Damping: T/4???

    Well, this will be true only for a specific value of overdamping. Overdamping just means the damping parameter is greater than the critical value, and the ratio could really be any number you want, not just 4.

    Here is a graph showing an example where overdamping is about 2.5 times slower in approaching equilibrium:
    http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html
     
  7. Aug 30, 2008 #6

    uart

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    Re: Critical Damping: T/4???

    You wont get a mathematical proof becasue it's just not true.

    BTW, the time for the critically damp osc to settle to with 10% of it's final value (after a step change) is approx 60% of the period of the undampled osc.


    Equations for a step change from x0 to zero, given zero initial derivative (equiv to system in steady state prior to step).

    undamped : [itex]y(t) = x_0 cos (w_0 t)[/itex]

    critially damped : [itex]y(t) = x_0 (1+\alpha \, t) \, e^{-\alpha t} [/itex], where [itex]\alpha = w_0[/itex] in the critical damped case.

    So just plug the numbers in, [itex]T=(2 \pi)/w_0[/itex] for the undamped case.

    62% of T is approx 3.9/w_0, so in the critically damped case alpha*t = w_0*t = 3.9.

    And (1+3.9) * exp(-3.9) is approx 0.1
     
  8. Aug 30, 2008 #7
    Re: Critical Damping: T/4???

    Well that is what I thought. My equations were:

    I solved:

    m*d^2x/dt^2+b*dy/dx+ky = 0

    k=spring constant
    m=mass
    p=drag constant (F=pv)
    x=displacement
    t=time
    A,B = Constants of Integration

    n1 = -p/2m + SQRT( (p/2m)^2 - k/m )
    n2 = -p/2m + SQRT( (p/2m)^2 - k/m )

    OverDamping: x = Ae^(n1*t) + Be^(n1*t)

    Critical Damping: x = (At+B)*e^(-p/2m *t)

    Under Damping: x = e^(-p/2m * t) * ( SQRT(A^2 + B^2) * sin(t/2*SQRT(k/m - (p/2m)^2 ) ) )

    So I could not see Initially why they referred to "T/4"
     
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