A horizontal platform is undergoing vertical S.H>M of time period T. A block of mass M is placed on the platform . The maximum mplitude of the platform so that the block is not detached form it is

Options

A. `(gT^(2))/(2pi^*2)`
B. `(gT^(2))/(4pi^(2)`
C. `(2gT^(2))/(2pi^(2))`
D. `(4gT^(2))/(pi^(2)`

Explanation

To solve the problem, we need to determine the maximum amplitude \( A \) of the platform undergoing vertical simple harmonic motion (SHM) such that the block of mass \( M \) does not detach from the platform.

### Step-by-Step Solution:

1. **Understanding Forces Acting on the Block**:
The block experiences two forces: the gravitational force acting downwards, \( Mg \), and the normal force \( N \) acting upwards from the platform. The block will not detach from the platform as long as the normal force \( N \) is greater than or equal to zero.

2. **Applying Newton's Second Law**:
According to Newton's second law, the net force acting on the block in the vertical direction can be expressed as:
\[ N - Mg = Ma \]
where \( a \) is the acceleration of the block due to the SHM of the platform.

3. **Acceleration of the Platform**:
The acceleration \( a \) of the platform can be expressed in terms of the amplitude \( A \) and angular frequency \( \omega \):
\[ a = -\omega^2 A \sin(\omega t) \]
Here, \( \omega = \frac{2\pi}{T} \), where \( T \) is the time period of the SHM.

4. **Substituting Acceleration into the Force Equation**:
Substituting the expression for \( a \) into the force equation gives:
\[ N - Mg = -M\omega^2 A \sin(\omega t) \]
Rearranging this, we find:
\[ N = Mg - M\omega^2 A \sin(\omega t) \]

5. **Condition for the Block to Remain in Contact**:
For the block to remain in contact with the platform, the normal force \( N \) must be greater than or equal to zero:
\[ Mg - M\omega^2 A \sin(\omega t) \geq 0 \]
This simplifies to:
\[ g \geq \omega^2 A \sin(\omega t) \]

6. **Finding Maximum Value of Sine Function**:
The maximum value of \( \sin(\omega t) \) is 1. Therefore, we can write:
\[ g \geq \omega^2 A \]
Rearranging this gives:
\[ A \leq \frac{g}{\omega^2} \]

7. **Substituting for Angular Frequency**:
Since \( \omega = \frac{2\pi}{T} \), we have:
\[ \omega^2 = \left(\frac{2\pi}{T}\right)^2 = \frac{4\pi^2}{T^2} \]
Substituting this into the inequality gives:
\[ A \leq \frac{g T^2}{4\pi^2} \]

8. **Conclusion**:
Therefore, the maximum amplitude \( A \) of the platform so that the block does not detach from it is:
\[ A = \frac{g T^2}{4\pi^2} \]

### Final Answer:
The maximum amplitude of the platform so that the block is not detached from it is \( A = \frac{g T^2}{4\pi^2} \).