Velocity & Acceleration in SHM question from JEE Main 2006 (AIEEE)

Answer: (b)

1. At the highest point the platform's acceleration is $\omega^2 A$ downward.
2. The normal force $N = m(g - \omega^2 A)$ becomes zero first when $\omega^2 A = g$.
3. So contact is first lost for amplitude $A = \dfrac{g}{\omega^2}$.
4. So the answer is (b) $\dfrac{g}{\omega^2}$.