Driven oscillator

A driven oscillator has some forcing (external force dependent only on $t$ ) applied.

\ddot x + \Gamma \dot x + \omega_0^2 x = f_0 \cos(\omega_d t).

Here the force has amplitude $f_0$ and driving frequency $\omega_d$ .

We solve this by guessing $x=\mathrm{Re}[z]$ where $z=Ae^{i (\omega_d t - \delta)}$ . We also rewrite the forcing term as $f_0 \cos(\omega_d t) = \mathrm{Re}\left[ f_0 e^{i \omega_d t} \right]$ .

\begin{align*} \left( -\omega_d^2 + i \omega_d \Gamma + \omega_0^2 \right) A e^{i (\omega_d t - \delta)} &= f_0 e^{i \omega_d t} \\ \left( -\omega_d^2 + i \omega_d \Gamma + \omega_0^2 \right) A &= f_0 e^{i \delta}. \end{align*}

Let us take the real and imaginary parts of the equation.

\begin{align*} \mathrm{Re}: &&\left(-\omega_d^2 + \omega_0^2\right) A &= f_0 \cos \delta \\ \mathrm{Im}: && \omega_d \Gamma A &= f_0 \sin \delta. \end{align*}

We solve for $\delta$ by dividing $\frac{\mathrm{Im} }{\mathrm{Re} }$ .

\tan \delta = \frac{\omega_d \Gamma}{\omega_0^2 - \omega_d^2}.

We solve for $A$ by squaring and adding $\mathrm{Im}^2 + \mathrm{Re}^2$ .

A = \frac{f_0}{\sqrt{ \left( \omega_0^2 - \omega_d^2 \right)^2 + \omega_d^2 \Gamma^2} }.

Substituting these values we find a solution to our differential equation.

x(t) = A \cos(\omega_d t - \delta).

Notice however that there are no free parameters. We expect the general solution to have two free parameters defined by the initial conditions, and indeed, this is not the general solution. What we have is called the “steady-state” solution, or the inhomogenous part of the general solution.

We find the general solution is a superposition of the “transient” (homogenous) solution and the “steady-state” solution.

x(t) = B e^{-\Gamma t/2}\cos(\omega t - \phi) + A \cos(\omega_d t - \delta).

Where $\omega = \sqrt{\omega_0^2 - \Gamma^2/4}$ .

The transient solution just represents a underdamped undriven oscillator while the steady-state solution represents the body moving at the same frequency as the driving force, with some phase shift.