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Damped vs Undamped Driven Springs and superposition?

  1. Jan 20, 2016 #1
    Modeling driven undamped spring systems in my diff eqs text at the moment.

    So I've just worked through the derivation of

    [tex]x(t) = C\cos{(\omega_0t - \alpha)} + \frac{F_0/m}{\omega_0^2-\omega^2}\cos{\omega t}[/tex]

    And it's clear that this describes the superposition of two different oscillations.

    I was wondering, just off the top of my head, if a damped system would consist of the superposition of three different oscillations?

    My immediate guess would be no, because you'd be solving the diff eq which included the addition of the damping constant multiplied by the first differential of motion, which would just give you a different complementary solution depending on whether it was underdamped, critically damped or overdamped. In the over and critically damped cases, the complementary solution is not sinusoidal, and so the nonhomogeneous driving force would have its own frequency, even though the particular solution would be solved by running a linear combination of sinusoids through the diff eq, this will not change the drivers frequency, just its phase shift...and solving the complementary function for the underdamped case also just produces sinusoids with the natural frequency and a phase shift, but at no point will a third frequency be introduced...is that about right? Is there anything else I should be considering?
     
  2. jcsd
  3. Jan 20, 2016 #2

    jambaugh

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    I would first remind you of Euler's formula [itex]e^{i\theta}=\cos(\theta)+i\sin(\theta)[/itex]. I bring this up because you will find that the dissipative aspect of dampening will introduce an exponential decay as a multiplicative factor. But this can simply be understood as a real component to the imaginary exponential we write a trig.

    Another point is that you get a superposition of solutions because the (homogeneous) differential equation is linear so that linear combinations of solutions are also solutions. You get two independent solutions (before applying boundary/initial conditions) because you have a second order equation (acceleration = second time derivative of position and force is a function of position).

    So you will get solutions of the form [itex] x(t)=e^{(-r+i\omega)t}=e^{-r t}(\cos(\omega t)+i\sin(\omega t))[/itex], representing a sinusoidal solution with decaying amplitude.
    There are some additional particulars and cases based on the amount of dampening.

    Haven fun.
     
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