1. Condition for losing contact

A coin placed on a vertically oscillating platform remains in contact as long as the normal reaction $N \ge 0$.

Taking upward as positive, for the coin while in contact: $N - mg = ma$ So, $N = m(g+a)$

The coin will leave the platform when contact just breaks, i.e. $N=0$ Hence, $g+a=0 \Rightarrow a=-g$

So the coin loses contact when the downward acceleration of the platform becomes equal to $g$.

2. Acceleration in SHM

If the platform executes SHM with amplitude $A$ and angular frequency $\omega$, then $a=-\omega^2 y$ where $y$ is displacement from mean position.

Maximum downward acceleration occurs at the topmost position: $a_{\min}=-\omega^2 A$

For the coin to leave contact for the first time, this maximum downward acceleration must just equal $-g$: $-\omega^2 A=-g$ $\omega^2 A=g$ $A=\frac{g}{\omega^2}$

3. Check the instant/position

At the threshold amplitude $A=\dfrac{g}{\omega^2}$, contact is first lost when the platform is at the highest position, because that is where downward acceleration is maximum.

So:

• Option A: false
• Option B: true
• Option C: false
• Option D: false

4. Final answer

The coin leaves contact for the first time when the amplitude reaches $\boxed{\frac{g}{\omega^2}}$ which corresponds to Option B.