Consider a forced, damped harmonic oscillator
$$\ddot{\phi} + 2\beta \dot{\phi} + \omega_0^2 \phi = j(t) \, .\tag{1}$$
If I pick a sinusoidal driving force $j(t) = A \cos(\Omega t)$, I find
$$\phi(t) = \text{Re} \left[ e^{-i \Omega t} \frac{-A}{\Omega^2 – \omega_0^2 + 2i\beta \Omega} \right] \, .\tag{2}$$
From here, how do I define the "resonance"?
Is it the point where $\langle \phi(t)^2 \rangle$ is maximized?
Things I do know:
The frequency at which $\langle \phi(t)^2 \rangle$ is maximized is $$\omega_r ~:=~ \omega_0 \sqrt{1 – 2(\beta/\omega_0)^2},\tag{3}$$ but I thought I read/heard that the resonance frequency of a damped oscillator is just $\omega_0$.
I also calculated that the free oscillation frequency is $$\omega_{\text{free}} ~:=~ \omega_0 \sqrt{1 – (\beta / \omega_0)^2},\tag{4}$$ but I don't think that's the same thing as the resonance frequency under steady driving.
Best Answer
At resonance, the energy flow from the driving source is unidirectional, i.e., the system absorbs power over the entire cycle.
When $\Omega = \omega_0$, we have
$$\phi(t) = \frac{A}{2\beta \omega_0}\sin\omega_0 t$$
thus
$$\dot \phi(t) = \frac{A}{2\beta}\cos\omega_0 t$$
The power $P$ per unit mass delivered by the driving force is then
$$\frac{P}{m} = j(t) \cdot \dot \phi(t) = \frac{A^2}{2\beta}\cos^2\omega_0 t = \frac{A^2}{4\beta}\left[1 + \cos 2\omega_0 t \right] \ge 0$$
When $\Omega \ne \omega_0$ the power will be negative over a part of the cycle when the system does work on the source.
What you've labelled as $\omega_r$ is the damped resonance frequency or resonance peak frequency.
Unqualified, the term resonance frequency usually refers to $\omega_0$, the undamped resonance frequency or undamped natural frequency.