Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Section C / 11

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, isThe loaded equilibrium follows from $k{z}_{\mathrm{eq}}=mg$, 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: $dE/dt=-b{\dot{z}}^2\le 0$. Differentiating the energy gives $\dot{z}(m\ddot{z}+kz-mg)=-b{\dot{z}}^2$, which yields, with continuity through instants of zero velocity,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 ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$: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.

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 isAt 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 $B-\gamma A=v_0$, henceSubtracting 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 requireswith $v_0$ 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.