Forced Vibrations

1. Feb 8, 2014

vanceEE

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. Feb 8, 2014

3. Feb 8, 2014

vanceEE

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
4. Feb 8, 2014

vela

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