Question

A coin is placed on a horizontal plate. The plate performs S.H.M. vertically with angular frequency \(\omega\). The amplitude \(A\) of oscillations is gradually increased. The coin will lose contact with the plate for the first time when amplitude is ( \(g\) = acceleration due to gravity )

• A. \( \dfrac{g}{\omega^2} \)
• B. zero
• C. \( \dfrac{\omega^2}{g} \)
• D. \( \dfrac{A}{2} \)

Hint:

Loss of contact occurs when downward acceleration equals gravitational acceleration.

Answer 1

The Correct Option is A

Step 1: Condition for loss of contact.
The coin loses contact with the plate when the normal reaction becomes zero. This happens when the downward acceleration of the plate equals \(g\).
Step 2: Maximum acceleration in SHM.
The maximum acceleration in SHM is given by \[ a_{\max} = \omega^2 A. \]
Step 3: Applying the condition.
For loss of contact, \[ \omega^2 A = g. \] \[ A = \frac{g}{\omega^2}. \]
Step 4: Conclusion.
The coin loses contact for the first time when the amplitude is \( \dfrac{g}{\omega^2} \).