a
The spring has a length \(l .\) Whem \(m\) is placed over it, the equilibrium position becomes \(O^{\prime}\)

If it is pressed from \(O^{\prime}\) \((the\, equilibrium\, position)\) to \(O^{\prime \prime}, O^{\prime} O^{\prime \prime}\) is the amplitude.

\(O O^{\prime} =\frac{m g}{k}=\frac{2 \times 10}{200}=0.10 \mathrm{m} \)

\(m g =k x_{0}\)

If the restoring force \(m A \omega^{2}>m g,\) then the mass will move up with acceleration, detached from the pan.

i.e. \(A>\frac{g}{k / m} \Rightarrow A>\frac{20}{200}>0.10 \mathrm{m}\)

\(The\, amplitude >10 \mathrm{cm}\)

i.e. the minimum is just greater than \(10 \mathrm{cm}\). (The actual compression will include \(x_{0}\) also. But when talking of amplitude, it is always from the equilibrium position with respect to which the mass is oscillating.