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$[b]$: See this SO question which I posted specifically to help generate this answer.

$[b]$: See this SO question which I posted specifically to help generate this answer.

$[a]$: In fact because of the way the quantities are set up here, if $\phi$ is a displacement of a mass on a spring, then what I'm calling "work" is actually "work divided by mass".

$[a]$: In fact because of the way the quantities are set up here, if $\phi$ is a displacement of a mass on a spring, then what I'm calling "power" is actually "power divided by mass".

The work done by the driving force is $\text{force}\times\text{velocity}^{[a]}$: $$W(t) = j(t)\dot{\phi}(t) = \frac{A^2}{2\beta}\cos(\omega_0 t)^2 = \frac{A^2}{4 \beta}[1 + \cos(2\omega_0 t)] \, .$$ Note that $W$ is always positive. This is actually the definition of resonance: the resonance frequency is the one such that the work done by the drive is always positive. In an electrical circuit this is the same as saying that the resonance frequency is the one where the impedance of the damped oscillator is purely real. No other frequency has this property, so we've just shown that $\omega_0$ is the resonance frequency of the damped system$^{[b]}$. Since we're in steady state, this work must also be precisely the work lost by the system to the damping. We can therefore compute the average power loss in one cycle:

$$\langle P_{\text{loss}}\rangle = \frac{\omega_0}{2\pi} \int_{0}^{2\pi / \omega_0} W(t)\,dt = \frac{A^2}{4\beta}\,.$$

The power exerted by the driving force is $\text{force}\times\text{velocity}^{[a]}$: $$P(t) = j(t)\dot{\phi}(t) = \frac{A^2}{2\beta}\cos(\omega_0 t)^2 = \frac{A^2}{4 \beta}[1 + \cos(2\omega_0 t)] \, .$$ Note that $P(t)$ is always positive. This is actually the definition of resonance: the resonance frequency is the one such that the work done by the drive is always positive. In an electrical circuit this is the same as saying that the resonance frequency is the one where the impedance of the damped oscillator is purely real. No other frequency has this property, so we've just shown that $\omega_0$ is the resonance frequency of the damped system$^{[b]}$. Since we're in steady state, this work must also be precisely the work lost by the system to the damping. We can therefore compute the average power loss in one cycle:

$$\langle P_{\text{loss}}\rangle = \frac{\omega_0}{2\pi} \int_{0}^{2\pi / \omega_0} P(t)\,dt = \frac{A^2}{4\beta}\,.$$