Question

A coin is placed on a horizontal platform, which undergoes vertical $\;SHM$ of angular frequency $\omega$ . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time
(A) At the highest position of the platform
(B) At the mean position of the platform
(C) For an amplitude $\dfrac{g}{{{\omega }^{2}}}$
(D) For an amplitude $\sqrt{\dfrac{g}{\omega }}$

Answer 1

Hint : For the coin to leave the horizontal platform, the acceleration due to the $\;SHM$ must be enough to overcome the force exerted by the gravity. Hence, the acceleration of the platform should be greater than gravitational acceleration $g=9.8m{{s}^{-1}}$

Complete Step By Step Answer:
Here, the coin is placed on a horizontal platform. This platform is performing $\;SHM$ with angular frequency $\omega$ .
Hence, the coin can also be said to be performing $\;SHM$ with angular frequency $\omega$
Now, the gravitational force is acting on the coin, due to which it constantly presses the coin to the platform.
In addition to it, when the platform moves up, its acceleration acts in the upward direction. As the gravitational force is always acting in the downward direction, both the acceleration oppose each other and presses the coin more towards the platform.
Hence we can conclude, for the coin to leave the platform, both the accelerations should be acting in the same direction.
As the gravitational acceleration is always acting in the downward direction, the acceleration of the platform should by in downward direction.
The acceleration of the platform is in downward direction when it reaches the maximum height and starts moving downward.
Also, if the acceleration of the platform is less than or equal to the gravitational acceleration, the coin will not leave the platform, as already the gravitational acceleration is acting on it which keeps the coin in contact with the platform.
Hence, the maximum acceleration of the platform should be greater than the gravitational acceleration, by which the platform moves downward leaving the coin.
The maximum magnitude of the acceleration in $\;SHM$ is
${{a}_{\max }}={{\omega }^{2}}A$
This value should be greater than the gravitational acceleration.
$\therefore {{\omega }^{2}}A\ge g$
$\therefore A\ge \dfrac{g}{{{\omega }^{2}}}$
Hence, the correct answer is Option $(C)$ .

Note :
Here, we must understand that even though the direction of both accelerations is the same when the platform is moving downward, an acceleration equal to the gravitational acceleration is already acting on the coin. If the platform also moves with the same acceleration, the coin will move along with the platform.