Solved: Forced Vibrations Homework: Using Arctan

[vanceEE]

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?

 

[LCKurtz]

Have a look at this link:

http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-rcostheta-alphaetc.pdf

 

[vanceEE]

Have a look at this link:

http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-rcostheta-alphaetc.pdf

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?

 

[vela]

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