= Solution
A <mass> $M$ on a <spring> of stiffness $K$, subject to linear <viscous damping> and a sinusoidal drive, obeys the <forced harmonic oscillator> equation
$$
M\ddot x+2M\gamma\dot x+Kx=F_0\cos\Omega t,\qquad
\omega_0=\sqrt{K/M},\quad\gamma>0.
$$
Here $x$ is <displacement>, $2M\gamma$ has units $\mathrm{kg\,s^{-1}}$, and $F_0$ is a <force> in <newtons>. A <series RLC circuit> is an electrical analogue, with $M\leftrightarrow L$, $2M\gamma\leftrightarrow R$, $K\leftrightarrow1/C$ and $x\leftrightarrow q$.
For free motion set $F_0=0$ and solve the <characteristic equation> $s^2+2\gamma s+\omega_0^2=0$. The roots $s=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}$ distinguish three cases:
* For <underdamping>, $\gamma<\omega_0$, the motion is $x=Ae^{-\gamma t}\cos(\omega_dt-\phi)$, where $\omega_d=\sqrt{\omega_0^2-\gamma^2}$. The <amplitude> decays on time $1/\gamma$ and the <frequency> is reduced from its undamped value.
* At <critical damping>, $\gamma=\omega_0$, the repeated root gives $x=(A+Bt)e^{-\omega_0t}$. This marks the boundary between oscillatory and nonoscillatory relaxation; it is the usual fastest settling without oscillation for a release from rest.
* For <overdamping>, $\gamma>\omega_0$, $x=Ae^{s_+t}+Be^{s_-t}$ with both roots real and negative. At large <damping> the slow root approaches $-\omega_0^2/(2\gamma)$, so increasing the resistance to motion can slow equilibration.
The constants are set by initial <position> and <velocity>. The mechanical <energy> and its rate of change in free motion are
$$
E=\frac12M\dot x^2+\frac12Kx^2,\qquad
\frac{dE}{dt}=-2M\gamma\dot x^2\leq0.
$$
For weak <damping>, the cycle-averaged <energy> decays approximately as $e^{-2\gamma t}$. The <quality factor> is $Q=\omega_0/(2\gamma)$; for $Q\gg1$ it equals $2\pi$ times stored <energy> divided by the loss per cycle. <Logarithmic decrement> of successive free-decay peaks determines $\gamma$ experimentally.
For driven motion the complete answer is the decaying free transient plus a steady response at the driving <angular frequency>:
$$
x_{\rm ss}=X(\Omega)\cos(\Omega t-\delta),\qquad
\boxed{X(\Omega)=\frac{F_0/M}{\sqrt{(\omega_0^2-\Omega^2)^2+4\gamma^2\Omega^2}},\quad
\delta=\operatorname{atan2}(2\gamma\Omega,\omega_0^2-\Omega^2).}
$$
At low <frequency>, $X\to F_0/K$ and the <displacement> is nearly in phase with the drive. At $\Omega=\omega_0$, the lag is $\pi/2$ and $X=F_0/(2M\gamma\omega_0)$. At high <frequency>, $X\simeq F_0/(M\Omega^2)$ and the lag tends to $\pi$. The <displacement> <resonance> maximum lies at
$$
\Omega_r=\sqrt{\omega_0^2-2\gamma^2}
$$
only when $\gamma<\omega_0/\sqrt2$. Thus an underdamped oscillator need not have a nonzero-frequency displacement resonance peak.
In the steady state the mean input <power> equals the mean dissipated <power>:
$$
\boxed{\overline P=M\gamma\Omega^2X^2
=\frac{F_0^2\gamma\Omega^2/M}{(\omega_0^2-\Omega^2)^2+4\gamma^2\Omega^2}.}
$$
This absorption peaks exactly at $\Omega=\omega_0$, rather than at $\Omega_r$. Its half-power points have bandwidth $\Delta\Omega=2\gamma$, so $Q=\omega_0/\Delta\Omega$ for this linear model. For weak <damping>, a high <quality factor> therefore means a narrow strong <resonance> and a long free-decay time. Exactly zero <damping> at resonant driving has no finite steady response: a secular <amplitude> grows with time.
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