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(C) For the object to remain on the platform, the normal force $N$ must be greater than or equal to zero. The equation of motion for the object of mas

Vedclass · Accepted Answer

$0.4$

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(C) For the object to remain on the platform, the normal force $N$ must be greater than or equal to zero. The equation of motion for the object of mass $m$ is given by $mg - N = ma$, where $a$ is the acceleration of the platform.For the object to just lose contact with the surface, the normal force $N$ becomes $0$.Thus, $mg = ma$, which implies $a = g$.The acceleration of a particle in $S.H.M.$ is given by $a = A\omega^2$, where $A$ is the amplitude and $\omega$ is the angular frequency.To avoid detachment, the maximum downward acceleration of the platform must not exceed $g$.Therefore, $A\omega^2 \leq g$.Substituting the given values, $A = 40 	ext{ cm} = 0.4 	ext{ m}$ and $g = 10 	ext{ m/s}^2$:$0.4 \omega^2 = 10$$\omega^2 = \frac{10}{0.4} = 25$$\omega = 5 	ext{ rad/s}$.Since $\omega = \frac{2\pi}{T}$, we have:$T = \frac{2\pi}{\omega} = \frac{2\pi}{5} = 0.4\pi 	ext{ s}$.Thus, the least period of oscillation is $0.4\pi 	ext{ s}$.

$A \ (mm \ s^{-2})$	$16$	$8$	$0$	$-8$	$-16$
$x \ (mm)$	$-4$	$-2$	$0$	$2$	$4$

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