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Verify forced response.

  1. Dec 1, 2007 #1
    1. The problem statement, all variables and given/known data

    For the differential equation:
    [tex] x'' + cx' + kx = F_{0}cos(\omage t) [/tex]

    verify that the forced response takes the form, [tex] x_{f}(t) = Ccos(\omega t - \delta) [/tex]


    2. Relevant equations

    [tex] C = \frac { F_{0} } {\sqrt {(k- \omega ^{2})^{2} + c^{2} \omega ^{2} } }[/tex]

    [tex] tan(\delta) = \frac {c \omega} {k - \omega ^ {2} }[/tex]

    3. The attempt at a solution


    I have tried to substitute [tex] x_{f}(t) = Ccos(\omega t - \delta) [/tex]
    into the equation then equate the two sides but i am lost on how i would get the [tex] cos(\omega t - \delta) [/tex] into [tex] cos(\omega t ) [/tex]. The only way i saw that can do this is the trig identity [tex] cos(\omega t - \delta) = cos(\omega t)cos(-\delta) + sin(\omega t )sin(\delta)) [/tex] but doing this seems only to over complicate the problem.

    Is the "forced response" just to solution to the system or is it something more specific than that?
     
    Last edited: Dec 1, 2007
  2. jcsd
  3. Dec 1, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    Well, [itex]\cos(\omega t)\cos(\delta)+\sin(\omega t)\sin(\delta)=\cos(\delta)[cos(\omega t)+\sin(\omega t)\tan(\delta)]. [/itex] Now, you know tan(delta) and I presume delta is small so then [itex]\cos(\delta)\approx 1[/itex].
     
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