= Solution
Measure both vertical displacements upward from static equilibrium, so gravity cancels. The spring force changes by $k(x_1-x)$, while the damper is attached to a fixed surface and supplies $-b\dot x$. <Newton's second law> gives
$$
\boxed{m\ddot x+b\dot x+kx=kx_1.}
$$
For sinusoidal steady motion, substitution of $x=\operatorname{Re}(RAe^{i\omega t})$ gives the <spring-base displacement response with a grounded damper>
$$
\boxed{R(\omega)=\frac{k}{k-m\omega^2+i b\omega}.}
$$
A decaying homogeneous transient must be added unless initial conditions already select steady state. At low frequency, $R\to1$ and the mass follows the slowly moving support; the first correction is $-ib\omega/k$. At high frequency, $R\sim-k/(m\omega^2)$: inertia suppresses motion and reverses its phase relative to the support. There is no $ib\omega$ in the numerator because the damper's other end does not move.
Minimize the squared denominator $D=(k-m\omega^2)^2+b^2\omega^2$. Its derivative is $2\omega[2m^2\omega^2-2mk+b^2]$, giving
$$
\boxed{\omega_{\rm res}^2=k/m-b^2/(2m^2),\qquad b^2<2mk.}
$$
Write $\omega_0=\sqrt{k/m}$ and $\zeta=b/(2m\omega_0)$. Then $|R|=[(1-r^2)^2+4\zeta^2r^2]^{-1/2}$ with $r=\omega/\omega_0$, and the finite-frequency peak is $1/[2\zeta\sqrt{1-\zeta^2}]$. Increasing light damping lowers and broadens the peak and shifts it left; the low-frequency limit remains1 and the high-frequency decay remains proportional to $\omega^{-2}$.
The <resonance versus oscillatory damping thresholds> differ. For $b^2\ge2mk$, the maximum is at zero frequency and the curve decreases monotonically. This threshold is $\zeta=1/\sqrt2$, distinct from the free-oscillation critical damping threshold $\zeta=1$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2010/paper-1-pa2010-base-response.png]
{title=Base displacement response}
{description=Three light-damping levels and one monotone response; dots mark the finite-frequency peaks.}
{source=paper-1-pa2010-base-response.py}
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