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(A) For the block to separate from the plank,the downward acceleration of the plank must exceed the acceleration due to gravity $g$.At the point of se

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$\frac{10}{\pi^2} \ m$

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(A) For the block to separate from the plank,the downward acceleration of the plank must exceed the acceleration due to gravity $g$.At the point of separation,the downward acceleration of the $SHM$ is equal to $g$.The maximum acceleration of an $SHM$ is given by $a_{max} = A \omega^2$.Setting $a_{max} = g$,we get $A \omega^2 = g$.Given the time period $T = 2 \ s$,the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2} = \pi \ rad/s$.Substituting the values,we have $A (\pi)^2 = g$.Taking $g = 10 \ m/s^2$,we get $A \pi^2 = 10$.Therefore,the minimum amplitude is $A = \frac{10}{\pi^2} \ m$.

$A$ plank with a small block on top of it is undergoing vertical $SHM$. Its period is $2 \ s$. The minimum amplitude at which the block will separate from the plank is:

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