A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then

The position of momentary rest in SHM is the extreme position where velocity of particle is zero.

As the block loses contact with the plank at this position, i.e., normal force becomes zero, it has to be the upper extreme where acceleration of the block will be g downwards.

$\begin{array}{l}{\mathrm{\omega }}^2\mathrm{A}=\mathrm{g}\\ \Rightarrow {\mathrm{\omega }}^2=\frac{10}{0.4}=25\\ \Rightarrow \mathrm{\omega }=5\mathrm{rad}/\mathrm{s}\therefore \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{5}\mathrm{s}\end{array}$

Acceleration in SHM is given as $\mathrm{a}={\mathrm{\omega }}^2\mathrm{x}$.
From the figure, we can see that at lower extreme, acceleration is g upwards

     $\mathrm{N}-\mathrm{mg}=\mathrm{ma}$

or  $\mathrm{N}=\mathrm{m}(\mathrm{a}+\mathrm{g})=2\mathrm{mg}$

At halfway up acceleration is g/2 downwards.

   $\mathrm{mg}-\mathrm{N}=\mathrm{ma}\text{ or }\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{g}/2)=\frac{1}{2}\mathrm{mg}$

At halfway down, acceleration is g/2 upwards.

   $\mathrm{N}-\mathrm{mg}=\mathrm{ma}\text{ or }\mathrm{N}=\mathrm{m}(\mathrm{g}+\mathrm{g}/2)=\frac{3}{2}\mathrm{mg}$

At mean position, velocity is maximum and acceleration is zero.

   $\therefore \mathrm{N}=\mathrm{mg}$