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. The least period of these oscillations, so that the object is not detached from the platform is `pi/(5x)` second. Find value of x. Take g=10m/`s^(2)`

Answer

To solve the problem, we need to determine the value of \( x \) given the conditions of the simple harmonic motion (SHM) of an object on a platform.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The platform is executing SHM in the vertical direction with an amplitude \( A = 4 \times 10^{-3} \) m.
- The least period of oscillation such that the object does not detach from the platform is given as \( T = \frac{\pi}{5x} \) seconds.
- We need to find the value of \( x \).

2. **Condition for Detachment**:
- The object will not detach from the platform if the maximum acceleration of the object is equal to or less than the acceleration due to gravity \( g \).
- The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = \omega^2 A \]
- Here, \( \omega \) is the angular frequency, and \( A \) is the amplitude.

3. **Relating Angular Frequency and Period**:
- The angular frequency \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
- Substituting this into the equation for maximum acceleration gives:
\[ a_{\text{max}} = \left(\frac{2\pi}{T}\right)^2 A \]

4. **Setting Up the Equation**:
- For the object to remain on the platform, we need:
\[ a_{\text{max}} \leq g \]
- Therefore:
\[ \left(\frac{2\pi}{T}\right)^2 A \leq g \]

5. **Substituting Values**:
- Substituting \( A = 4 \times 10^{-3} \) m and \( g = 10 \, \text{m/s}^2 \):
\[ \left(\frac{2\pi}{T}\right)^2 (4 \times 10^{-3}) \leq 10 \]

6. **Solving for Period \( T \)**:
- Rearranging gives:
\[ \left(\frac{2\pi}{T}\right)^2 \leq \frac{10}{4 \times 10^{-3}} = 2500 \]
- Taking the square root:
\[ \frac{2\pi}{T} \leq 50 \]
- Thus:
\[ T \geq \frac{2\pi}{50} = \frac{\pi}{25} \]

7. **Equating to Given Period**:
- We know \( T = \frac{\pi}{5x} \), so we set:
\[ \frac{\pi}{5x} = \frac{\pi}{25} \]

8. **Solving for \( x \)**:
- Cancelling \( \pi \) from both sides:
\[ \frac{1}{5x} = \frac{1}{25} \]
- Cross-multiplying gives:
\[ 25 = 5x \implies x = 5 \]

### Final Answer:
The value of \( x \) is \( 5 \).