A mass m = 1.0,kg is placed on a flat pan att | vedclass

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(C) For the mass $m$ to just detach from the pan,the normal force $N$ between the mass and the pan must become zero at the highest point of the oscill

Vedclass · Accepted Answer

$A > 2.0\,cm$

Vedclass · Answer

(C) For the mass $m$ to just detach from the pan,the normal force $N$ between the mass and the pan must become zero at the highest point of the oscillation.At the highest point,the acceleration of the mass is directed downwards and is equal to $a = \omega^2 A$.The equation of motion for the mass at the highest point is $mg - N = ma$.Setting $N = 0$ for the condition of detachment,we get $mg = m\omega^2 A$,which simplifies to $g = \omega^2 A$.Since $\omega^2 = k/m$,we have $g = (k/m) A$.Substituting the given values: $10 = (500 / 1.0) 	imes A$.$A = 10 / 500 = 0.02\,m = 2.0\,cm$.Therefore,the mass will tend to detach if the amplitude $A$ is greater than or equal to $2.0\,cm$.

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