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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 a particle in $SHM$ 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 $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.Substituting the values into the equation $A = \frac{g}{\omega^2}$,we use $g = 10 \, m/s^2$.$A = \frac{10}{\pi^2} \, m$.Thus,the minimum amplitude is $\frac{10}{\pi^2} \, m$.

$A$ 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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