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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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Verified Answer
The correct answer is:
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$
for an amplitude of $\frac{\mathrm{g}}{\omega^2}$
$A \omega^2=g$
$\Rightarrow A=g / \omega^2$
$\Rightarrow A=g / \omega^2$
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