= Solution
\b[Response and <resonance>.] The terms $m\ddot x$, $b\dot x$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring <force>; the right side is the applied harmonic <force>. Substituting the complex trial motion and taking <real parts> gives
$$
\boxed{A=R(\omega)F_0,\qquad R(\omega)=\frac1{k-m\omega^2+ib\omega}}.
$$
The steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m\omega^2)^2+b^2\omega^2]^{-1/2}$. Differentiating the denominator gives $2\omega[-2m(k-m\omega^2)+b^2]=0$, so
$$
\boxed{\omega_{\rm res}=\sqrt{k/m-b^2/(2m^2)}}.
$$
A nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero <frequency>. Increasing <damping> lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m\omega^2)$ at high <frequency>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-2-response-resonance.png]
{title=Original normalized oscillator response for several damping ratios, distinguishing amplitude and current resonance}
\b[Series circuit.] The PDF circuit has the <resistor>, <inductor> and <capacitor> in one loop. With current $I=\dot q$, <Kirchhoff's voltage law> gives $V=L\dot I+RI+q/C$, hence
$$
\boxed{L\ddot q+R\dot q+q/C=V}.
$$
Thus $m,b,k$ correspond to $L,R,1/C$. The <resistor>'s rms voltage is
$$
V_{R,\rm rms}=\frac{|V_0|}{\sqrt2}\frac{R}{\sqrt{R^2+(\omega L-1/(\omega C))^2}}.
$$
It is maximized exactly at $\omega_{\max}=\omega_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude <resonance> is lower. Expanding its square root for $R^2C/L\ll1$ gives
$$
\boxed{\omega_{\max}-\omega_{\rm res}
=\omega_0-\sqrt{\omega_0^2-R^2/(2L^2)}
\simeq\frac{R^2}{4L}\sqrt{\frac CL}}.
$$
The two maxima refer to different observables: <resistor> voltage follows current, whereas charge follows displacement.
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