= Solution
Use the diagram's downward displacement $z$, measured from the unloaded plate position, and take $z(0)=0$. The <mechanical energy>, with gravitational potential zero at that reference, is
$$
\boxed{E=\frac12m\dot z^2+\frac12kz^2-mgz.}
$$
The loaded equilibrium follows from $kz_{\mathrm{eq}}=mg$, so its downward shift is $\boxed{z_{\mathrm{eq}}=mg/k}$.
The shock absorber's <force> is $-b\dot z$, hence its power is <force> times <velocity>: $dE/dt=-b\dot z^2\leq0$. Differentiating the <energy> gives $\dot z(m\ddot z+kz-mg)=-b\dot z^2$, which yields, with continuity through instants of zero <velocity>,
$$
\boxed{m\ddot z+b\dot z+kz=mg.}
$$
Set $\omega_0=\sqrt{k/m}$ and $\gamma=b/(2m)$.
For release from rest, the three <step response of a damped oscillator> regimes are easy to compare after shifting to $z-z_{\mathrm{eq}}$. With light damping $0<\gamma<\omega_0$, write $\omega_d=\sqrt{\omega_0^2-\gamma^2}$:
$$
\frac z{z_{\mathrm{eq}}}=1-e^{-\gamma t}\left(\cos\omega_dt+\frac\gamma{\omega_d}\sin\omega_dt\right).
$$
It overshoots and oscillates around the loaded position with a decaying envelope. At <critical damping>, $\gamma=\omega_0$, it becomes $z/z_{\mathrm{eq}}=1-(1+\gamma t)e^{-\gamma t}$, approaching monotonically from below. For heavy damping $\gamma>\omega_0$, set $\eta=\sqrt{\gamma^2-\omega_0^2}$ and replace the trigonometric bracket by $\cosh\eta t+(\gamma/\eta)\sinh\eta t$. This is also monotonic, but its slow exponential delays settling.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017-damped-scale.png]
{title=Damping controls the scale's settling, while initial impact velocity can still cause overshoot}
<Critical damping> gives the fastest nonoscillatory settling from the unloaded rest condition, avoiding a repeatedly swinging reading without the sluggishness of <overdamping>. Its required coefficient is
$$
\boxed{b_{\mathrm{crit}}=2\sqrt{km}.}
$$
At <critical damping> the repeated characteristic root is $-\gamma$. Thus the general displacement under constant load is $(A+Bt)e^{-\gamma t}+C$. The equilibrium and initial conditions give $C=z_{\mathrm{eq}}$, $A=-z_{\mathrm{eq}}$ and $B-\gamma A=v_0$, hence
$$
\boxed{A=-\frac{mg}{k},\qquad B=v_0-\gamma\frac{mg}{k},\qquad C=\frac{mg}{k},\qquad \gamma=\sqrt{\frac km}.}
$$
Subtracting the final position gives $z-z_{\mathrm{eq}}=[-z_{\mathrm{eq}}+(v_0-\gamma z_{\mathrm{eq}})t]e^{-\gamma t}$. The exponential is positive and the bracket starts negative. It becomes positive at some $t>0$ if and only if its slope is positive. Therefore <critical-damping overshoot from initial velocity> requires
$$
\boxed{v_0>\gamma z_{\mathrm{eq}}=g\sqrt{\frac mk}=\frac g\gamma,}
$$
with $v_0$ positive downwards. At equality the scale still approaches from below. <Critical damping> prevents oscillatory modes, but does not guarantee a non-overshooting response to an arbitrarily large impact <velocity>.
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