• For undamped oscillator $P=0$
• For damped oscillators the power absorbed to sustain the motion is equal to the rate of energy dissipation. Hence:

$$ P=\frac{W}{\Delta t} \; {\color{#979A9B}\text{ and }} \; W=F_D \Delta x \; {\color{#979A9B}\text{ so }} \Rightarrow \; P=F_D\frac{\Delta x}{\Delta t} $$

Since $F_D$ is the damping force and ${\Delta x}/\Delta t$ is the infinitesimal velocity then:

$$ P=\mathrm{Damping \, force} \, \times \, \mathrm{Velocity} $$

Velocity can be found by using $x(t) =A(\omega) \cos(\omega t - \delta)$

$$ \begin{aligned} v(t)= \frac{dx}{dt} &= - A(\omega) \omega \sin (\omega t - \delta)\\ &= - v_0 ( \omega) \sin ( \omega t - \delta) \end{aligned} $$

• Hence instantaneous power is

$$ P(t)= b v(t) \cdot v(t) = b v_0^2 (\omega) \sin^2 (\omega t - \delta) $$

• The average power absorbed per cycle $\bar{P}(\omega)$ is

$$ \bar{P}(\omega) = \frac{1}{T} \int_{t_0}^{t_0 +T} b v_0^2 (\omega)\, \sin^2(\omega t - \delta)\, dt $$

• $1/T$ ensures average per cycle
• $b v^2_0 (\omega)$ is independent of $t$
• $\int \sin^2(\omega t-\delta)=(1/2)\, T$

$$ \begin{aligned} \bar{P}(\omega) &= \frac{1}{2} b v_0^2(\omega)\\ &= \frac{\omega^2 F_0^2 \gamma}{2 m [(\omega^2 - \omega_0^2)^2 + \omega^2 \gamma^2]} \end{aligned} $$

Plot of the power resonance curve

• The width of the curve around $\omega=\omega_0$ defines the sharpness of the resonance.
• Close to the peak $\omega \sim \omega_0$ thus we can simplify the above equation to:

$$ \hat{P}(\omega) = \frac{F_0^2}{2 m \gamma \left(\frac{4 \Delta \omega^2}{\gamma^2} +1 \right )} $$

where $\Delta\omega=\omega-\omega_0$

• The maximum value is when $\Delta\omega=0$ and $\bar{P}_\text{max}=F_0^2/4m\gamma$
• $\gamma=2\Delta\omega=\omega_\text{fwhh}$ and thus $Q=\omega_0/\gamma=\omega_0/\omega_\text{fwhh}$

Solving second order linear ODE