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(D) The coin will lose contact with the platform when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.In simple

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At the highest position of the platform.

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(D) The coin will lose contact with the platform when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.In simple harmonic motion,the acceleration is given by $a = -\omega^2 y$.The maximum downward acceleration occurs at the highest point of the oscillation,where the displacement $y = A$ (amplitude).Thus,the downward acceleration is $a = \omega^2 A$.The coin loses contact when the downward acceleration of the platform is greater than or equal to $g$.$\omega^2 A \geq g$$A \geq \frac{g}{\omega^2}$Therefore,the coin first loses contact at the highest point when the amplitude reaches $A = \frac{g}{\omega^2}$.

$A$ coin is placed on a horizontal platform. The platform performs vertical simple harmonic motion with an angular frequency $\omega$. The amplitude of oscillation is gradually increased. At what condition will the coin first lose contact with the platform?

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