Question
A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency $\omega $. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
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Answer
For block $A$ to move in $SHM.$
$\mathrm{mg}-\mathrm{N}=\mathrm{m} \omega^{2} \mathrm{x}$
where $x$ is the distance from mean position
For block to leave contact $\mathrm{N}=0$
$\Rightarrow m g=m \omega^{2} x \Rightarrow x=\frac{g}{\omega^{2}}$
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