A horizontal platform with an object placed on it is executing SHM in the vertical direction . The amplitude of oscillation is `4 xx 10^(-3)` m. What must be least period of these oscillations , so that the object is not detached from the platform ? (take , `g=10 ms^(-2)`)

Options

A. `(pi)/(25) s`
B. `(pi)/(5) s`
C. `(pi)/(10) s`
D. `(pi)/(50) s`

Explanation

To solve the problem, we need to determine the least period of oscillation for the object on the platform executing simple harmonic motion (SHM) such that the object does not detach from the platform.

### Step-by-Step Solution:

1. **Understanding the Forces**:
The object will remain in contact with the platform as long as the upward acceleration of the platform (due to SHM) is greater than or equal to the acceleration due to gravity (g).

2. **Acceleration in SHM**:
The maximum acceleration (a_max) of an object in SHM is given by the formula:
\[ a_{max} = \frac{A \cdot \omega^2}{g} \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

3. **Angular Frequency**:
The angular frequency \( \omega \) is related to the period \( T \) of the oscillation by the formula:
\[ \omega = \frac{2\pi}{T} \]

4. **Substituting Angular Frequency**:
Substituting \( \omega \) into the acceleration formula gives:
\[ a_{max} = A \cdot \left(\frac{2\pi}{T}\right)^2 \]

5. **Condition for Contact**:
For the object to remain in contact with the platform, we need:
\[ a_{max} \geq g \]
Thus, we can set up the inequality:
\[ A \cdot \left(\frac{2\pi}{T}\right)^2 \geq g \]

6. **Rearranging the Inequality**:
Rearranging gives:
\[ \frac{4\pi^2 A}{T^2} \geq g \]

7. **Solving for T**:
Rearranging for \( T^2 \):
\[ T^2 \leq \frac{4\pi^2 A}{g} \]
Taking the square root:
\[ T \leq 2\pi \sqrt{\frac{A}{g}} \]

8. **Substituting Values**:
Now we can substitute the given values:
- \( A = 4 \times 10^{-3} \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Thus:
\[ T \leq 2\pi \sqrt{\frac{4 \times 10^{-3}}{10}} \]

9. **Calculating T**:
\[ T \leq 2\pi \sqrt{4 \times 10^{-4}} = 2\pi \cdot 0.02 = 0.1256 \, \text{s} \]

10. **Final Result**:
Therefore, the least period of oscillation \( T \) is approximately:
\[ T \approx 0.1256 \, \text{s} \]

### Summary:
The least period of oscillation for the object to remain in contact with the platform is approximately \( 0.1256 \, \text{s} \).