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Damping and Initial Conditions

  1. Oct 3, 2015 #1
    Hello,

    I have a question regarding the solution to a second order 'mass-spring-damper' system. Over the years, I have gotten familiar with the idea of system damping in the sense of under damped, over damped, and critically damped systems.

    However, I've began looking closer at the solution to this set up and am somewhat confused on how the initial conditions influence the solution.

    Say for example I have a critically damped system. I would assume that the solution contain no oscillations. However, I have a phase-plane representation of the system (image attached) which clearly show there will exist overshoot given a certain set of initial conditions (i.e. when the initial velocity does not equal 0).

    So my question is: what exactly is the relationship to the damping coefficient of a system and the oscillations in the solution? How do initial conditions play into this?
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2015 #2

    pasmith

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    It doesn't oscillate; it overshoots at most once. The general solution is [itex]x(t) = e^{-\sqrt{(k/m)}t}(A + Bt)[/itex]. The exponential is always strictly positive. If [itex]A/B < 0[/itex] then the solution changes sign exactly once in [itex]t > 0[/itex]; if [itex]A/B > 0[/itex] the solution does not change sign in [itex]t > 0[/itex], and if [itex]B = 0[/itex] or [itex]A = 0[/itex] then the solution does not change sign in [itex]t > 0[/itex].

    "Oscillation" is characterised by overshooting multiple times.
     
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