= Response to the added load
Measure downward displacement relative to the former mean equilibrium. The added $M=20\,\mathrm{kg}$ changes the equilibrium by
$$
\boxed{\Delta=\frac{Mg}{k}=4.905\,\mathrm{mm}}.
$$
Assume the load is deposited without a significant impact impulse and then moves with the table. The moving <mass> becomes $m+M$. Ignoring the tiny pre-existing $10\,\mathrm{Hz}$ vibration in the sketch, the <added-load step response of a supported table> is
$$
x(t)=\Delta\left[1-e^{-\beta t}\left(\cos\omega_dt+\frac\beta{\omega_d}\sin\omega_dt\right)\right],\quad
\beta=\frac b{2(m+M)},\quad\omega_d=\sqrt{\frac{k}{m+M}-\beta^2}.
$$
For the estimated <mass>, $\beta=0.194\,\mathrm{s^{-1}}$, the envelope time is $1/\beta=5.15\,\mathrm s$, and the period is about $1.13\,\mathrm s$. The table overshoots and rings about the new equilibrium, with the displacement starting continuously at zero and with zero initial slope. A real impact adds an initial-velocity transient; the problem gives no impact history. The residual driven vibration, of order a micrometre, can be superposed.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2-b6-table-response.png]
{title=Lightly damped response amplitude and phase, followed by the table's load-step displacement and a critically damped comparison}
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