A block is placed on a horizontal plank. The | vedclass

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(D) The block loses contact with the plank when the normal force becomes zero. For a block on a plank performing $SHM$ vertically,the normal force $N

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all of the above

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(D) The block loses contact with the plank when the normal force becomes zero. For a block on a plank performing $SHM$ vertically,the normal force $N = m(g - a)$,where $a$ is the acceleration of the plank. At the topmost point,the acceleration is $a = -\omega^2 A$ (downward). The block loses contact when $N = 0$,which implies $g - a = 0$,so $a = g$. Since $a = \omega^2 A$ at the extreme position,we have $\omega^2 A = g$. Given $A = 0.4 \, m$ and $g = 10 \, m/s^2$,$\omega^2 = 10/0.4 = 25$,so $\omega = 5 \, rad/s$. The period $T = 2\pi/\omega = 2\pi/5 \, s$. Thus,option $A$ is correct.At the bottommost point,the acceleration is $a = +\omega^2 A = g$ (upward). The normal force $N = m(g + a) = m(g + g) = 2mg$. Thus,the block weighs double its weight at the bottom position. Option $B$ is correct.Halfway down from the mean position,the displacement is $y = A/2$. The acceleration is $a = -\omega^2 y = -\omega^2 (A/2) = -g/2$. The normal force $N = m(g - a) = m(g - (-g/2)) = 1.5mg$. Thus,option $C$ is correct.Since $A, B,$ and $C$ are correct,option $D$ is the correct answer.

$A$ block is placed on a horizontal plank. The plank is performing $SHM$ along a vertical line with an amplitude of $40 \, cm$. The block just loses contact with the plank when the plank is momentarily at rest. Then:

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