A coin is placed on the horizontal plate. Plate performs S.H.M. vertically with angular frequency '…
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 $)$
$\frac{\mathrm{g}}{\omega^{2}}$
zero
$\frac{\omega^{2}}{\mathrm{~g}}$
$\frac{\mathrm{A}}{2}$
Solution
As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesn't get separated) increases.
If we draw the FBD for coin at one of the extreme positions as shown
then from Newton's law, $m g-N=m \omega^2 A$
For loosing contact with the platform, $N=0$
So, $\quad A=g / \omega^2$
