Forced Vibrations

  • Thread starter vanceEE
  • Start date
  • #1
109
2

Homework Statement



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


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?
 

Answers and Replies

  • #3
109
2

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:
  • #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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