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Damping factor in critical damping

  1. May 30, 2014 #1
    The position equation for a oscillator undergoing critical damping is given by

    x(t) = Ate^(-γt) + Be^(-γt)
    where γ = c/2m
    and c is from the original force equation
    ma + cv + kx = 0

    γ is called the damping factor
    my book then goes on to say without explanation that
    γ = c/2m = (k/m)^(1/2) = initial angular velocity

    I understand the units of c/2m work out to be s^-1 so that's good, but other than that I'm not understanding the relation c/2m = w_initial as well as I want to.

    LaTex coming soon to a post near you
     
  2. jcsd
  3. May 30, 2014 #2

    vanhees71

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    Let's write the equation of motion in a somewhat more convenient form
    [tex]\ddot{x}+2 \gamma \dot{x}+\omega^2 x=0.[/tex]
    Then
    [tex]\gamma=\frac{c}{2m}, \quad \omega^2=\frac{k}{m}.[/tex]
    now consider the special case of "critical damping", where [itex]\gamma=\omega[/itex].

    To solve the equation, we make the standard ansatz for this linear differential equation with constant coefficients:
    [tex]x(t)=A \exp(\lambda t).[/tex]
    Plugging this into the equation leads after some simple algebra to
    [tex]\lambda^2 + 2 \omega \lambda + \omega^2=(\omega+\lambda)^2=0.[/tex]
    This means there is only one solution for [itex]\lambda[/itex]:
    [tex]\lambda_1=-\omega,[/tex]
    and one solution to the ODE is
    [tex]x(t)=A \exp(-\omega t).[/tex]
    For a complete solution, however we need one other linearly independent solution.

    Also here, a standard ansatz helps, namely the "ansatz of the variation of the constant", i.e.,
    [tex]x(t)=\exp(-\omega t) y(t).[/tex]
    Plugging this into our oscillator equation of motion yields
    [tex]\exp(-\omega t) [\ddot{y}-2 \omega \dot{y}+\omega^2 y-2 \omega^2 y + 2 \omega \dot{y} + \omega^2 y)]=\exp(-\omega t) \ddot{y}=0,[/tex]
    which immediately implies
    [tex]\ddot{y}=0 \; \Rightarrow \; y(t)=(A+B t).[/tex]
    The general solution of our equation is thus
    [tex]x(t)=(A+B t) \exp(-\omega t),[/tex]
    and thus the two linear independent set solutions is given by
    [tex]x_1(t)=\exp(-\omega t), \quad x_2(t)=t \exp(-\omega t).[/tex]
     
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