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!

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


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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]
    [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]:
    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook