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Forced Vibrations

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data

    $$ M \frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = F_{0}cos \omega t $$


    3. The attempt at a solution
    $$x(t) = asin\omega t + bcos \omega t $$
    $$ a = \frac{\omega c F_{0}}{(k-\omega^2M)^2 + \omega^2c^2} $$
    $$ b = \frac{(k-\omega^2M)F_{0}}{(k-\omega^2M)^2 + \omega^2c^2} $$
    $$ x_{0}(t) = \frac{F_{0}}{(k-\omega^2M)^2 + \omega^2c^2} [\omega c sin \omega t + (k-\omega^2M)cos \omega t] $$

    How can I use arctan to write this particular solution in a more useful form?
     
  2. jcsd
  3. Feb 8, 2014 #2

    LCKurtz

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  4. Feb 8, 2014 #3
    Thank's for the link, it was really helpful!

    So ##[\omega sin \omega t + (k-\omega ^2 M) cos \omega t] \equiv [(Rcos\phi)cos\omega t + (Rsin\phi)sin \omega t] ## by the sum-difference formulas
    Therefore,
    $$\frac{F_{0}}{(k-\omega^2M)^2 + \omega^2c^2} [\omega c sin \omega t + (k-\omega^2M)cos \omega t] $$ $$\equiv \frac{F_{0} R cos (\omega t -\phi)}{R^2} $$ $$\equiv \frac{F_{0}}{√((k-\omega^2M)^2 + \omega^2c^2)} [cos \omega t - \phi] $$ where $$ \phi = \arctan (\frac{\omega}{k-\omega ^2M}) $$
    Right?
     
    Last edited: Feb 8, 2014
  5. Feb 8, 2014 #4

    vela

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    You forgot a factor of ##c##. The phase angle should be
    $$\phi = \arctan\left(\frac{\omega c}{k-\omega^2 M}\right).$$
     
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