1.5 The driven oscillator

Lets consider a case with a driven or forced oscillator. We now have $$m\ddot{x}+kx=F_0{e}^{i\omega t}$$ or $$\ddot{x}+{\omega }_0^{2}x=\frac{F_0}{m}{e}^{i\omega t}\text{.}$$ Try $$x=A{e}^{i(\omega t+\alpha )}$$ then $$\dot{x}=i\omega A{e}^{i(\omega t+\alpha )}$$ and $$\ddot{x}=-{\omega }^{2}A{e}^{i(\omega t+\alpha )}\text{.}$$ So now get $$-{\omega }^{2}A{e}^{i(\omega t+\alpha )}+{\omega }_0^{2}A{e}^{i(\omega t+\alpha )}=\frac{F_0}{m}{e}^{i\omega t}$$ $$({\omega }_0^2-{\omega }^2)A{e}^{i(\omega t+\alpha )}=\frac{F_0}{m}{e}^{i\omega t}$$ $$({\omega }_0^2-{\omega }^2)A=\frac{F_0}{m}{e}^{-i\alpha }$$ $$({\omega }_0^2-{\omega }^2)A=\frac{F_0}{m}\cos \alpha -i\frac{F_0}{m}\sin \alpha$$ Next you spearate the Real and Imaginary parts. The Imaginary part gives $$0=-\frac{F_0}{m}\sin \alpha$$ so $\alpha =0,\pi \dots$ implies $\cos \alpha =\pm 1$ The Real part gives $$({\omega }_0^2-{\omega }^2)A=\frac{F_0}{m}\cos \alpha$$ $$({\omega }_0^2-{\omega }^2)A=\pm \frac{F_0}{m}$$ $$A=\frac{\pm \frac{F_0}{m}}{({\omega }_0^2-{\omega }^2)}$$ So a solution is $$x=\frac{\frac{F_0}{m}}{({\omega }_0^2-{\omega }^2)}{e}^{i(\omega t+\alpha )}$$ where $\alpha =0,\pi \dots$

This is an extremely important result, this is the phenomenum of resonance. When you drive an oscillator at its resonant frequency then the amplitude ofthe oscillation will become huge. In the equation above, it becomes infinite, but in practice there will be some damping that prevents that. You have knownthis since your childhood, this is how you swing on a swing. If you live in a snowy climate, you know (or at least should know) that a trick to get your carout of a snow bank is is to rock it back and forth - if you get the frequency right you will make the car oscillate with a large amplitude and dislodge it.The electrical analogue is used to tune a radio.

A driven damped oscillator will be given as a homework problem.