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(A) The coin rests on the plate and moves with it. The acceleration of the plate in $S.H.M.$ is given by $a = \omega^{2} x$,where $x$ is the displacem

Vedclass · Accepted Answer

$\frac{g}{\omega^{2}}$

Vedclass · Answer

(A) The coin rests on the plate and moves with it. The acceleration of the plate in $S.H.M.$ is given by $a = \omega^{2} x$,where $x$ is the displacement from the mean position.As the plate moves downward,its acceleration is directed upwards. As the plate moves upward,its acceleration is directed downwards.The coin loses contact with the plate when the downward acceleration of the plate exceeds the acceleration due to gravity $(g)$.At the topmost point of the oscillation,the downward acceleration is maximum,given by $a_{max} = \omega^{2} A$.For the coin to lose contact,the condition is $a_{max} \geq g$.Therefore,the minimum amplitude at which contact is lost is $A = \frac{g}{\omega^{2}}$.

$A$ coin is placed on a horizontal plate. The plate performs $S.H.M.$ vertically with an angular frequency $\omega$. The amplitude $A$ of oscillations is gradually increased. The coin will lose contact with the plate for the first time when the amplitude is ($g =$ acceleration due to gravity).

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