A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency `omega`. The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time

Options

A. at the mean position of platform
B. for an amplitude `g/omega^(2)`
C. for an amplitude `g^(2)/omega^(2)`
D. at the lowest position of the platform

Explanation

To solve the problem of when a coin will leave contact with a platform undergoing vertical simple harmonic motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a coin placed on a horizontal platform that is oscillating vertically with an angular frequency \( \omega \) and an increasing amplitude \( A \).
- The forces acting on the coin are its weight \( mg \) (acting downwards) and the normal force \( N \) (acting upwards).

2. **Condition for Losing Contact**:
- The coin will lose contact with the platform when the normal force \( N \) becomes zero. This occurs when the downward acceleration of the platform equals the acceleration due to gravity \( g \).

3. **Acceleration of the Platform**:
- For a simple harmonic motion, the maximum acceleration \( a \) of the platform can be expressed as:
\[ a = \omega^2 A \]
- Here, \( A \) is the amplitude of the oscillation.

4. **Setting Up the Equation**:
- At the lowest point of the oscillation, the forces acting on the coin can be described by the equation:
\[ mg - N = ma \]
- When the coin just loses contact, \( N = 0 \), thus the equation simplifies to:
\[ mg = ma \]

5. **Substituting for Acceleration**:
- Substituting \( a = \omega^2 A \) into the equation gives:
\[ mg = m(\omega^2 A) \]

6. **Solving for Amplitude**:
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g = \omega^2 A \]
- Rearranging this equation to solve for the amplitude \( A \):
\[ A = \frac{g}{\omega^2} \]

7. **Conclusion**:
- The coin will leave contact with the platform for the first time when the amplitude of oscillation reaches \( A = \frac{g}{\omega^2} \).

### Final Answer:
The amplitude at which the coin will leave contact with the platform for the first time is given by:
\[ A = \frac{g}{\omega^2} \]