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A coin is placed on the horizontal plate. Plate performs S.H.M. vertically with
angular frequency ' $\omega^{\prime}$. The amplitude (A) of oscillations is gradually increased.
The coin will lose contact with plate for the first time when amplitude is
$(\mathrm{g}=$ acceleration due to gravity $)$
Options:
angular frequency ' $\omega^{\prime}$. The amplitude (A) of oscillations is gradually increased.
The coin will lose contact with plate for the first time when amplitude is
$(\mathrm{g}=$ acceleration due to gravity $)$
Solution:
1677 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{g}}{\omega^{2}}$
(D)
The coin will lose contact with the plate when its acceleration in downward direction just exceeds acceleration due to gravity.
$\begin{array}{l}
\therefore \omega^{2} \mathrm{x}=\mathrm{g} \\
\therefore \mathrm{x} \quad=\frac{\mathrm{g}}{\omega^{2}}
\end{array}$
The coin will lose contact with the plate when its acceleration in downward direction just exceeds acceleration due to gravity.
$\begin{array}{l}
\therefore \omega^{2} \mathrm{x}=\mathrm{g} \\
\therefore \mathrm{x} \quad=\frac{\mathrm{g}}{\omega^{2}}
\end{array}$
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