A horizontal platform with an object placed on it is executing SHM in the vertical direction . The amplitude of oscillation is 2.5 cm what must be the least period of these oscillations so that the object is not detached ?

Answer

To solve the problem, we need to determine the least period of oscillation for an object placed on a platform executing simple harmonic motion (SHM) in the vertical direction. The object should not detach from the platform, which means the maximum upward acceleration of the platform must be greater than or equal to the acceleration due to gravity (g).

### Step-by-Step Solution:

1. **Understanding the Problem:**
- The platform is executing SHM vertically with an amplitude \( A = 2.5 \, \text{cm} = 2.5 \times 10^{-2} \, \text{m} \).
- The acceleration due to gravity \( g \) is approximately \( 10 \, \text{m/s}^2 \).
- The condition for the object not to detach is that the maximum acceleration of the platform must be at least equal to \( g \).

2. **Maximum Acceleration in SHM:**
- The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = \omega^2 A \]
- Here, \( \omega \) is the angular frequency.

3. **Setting Up the Equation:**
- For the object to remain attached to the platform:
\[ a_{\text{max}} \geq g \]
- Therefore, we have:
\[ \omega^2 A \geq g \]

4. **Solving for Angular Frequency:**
- Rearranging the inequality gives:
\[ \omega^2 \geq \frac{g}{A} \]
- Taking the square root:
\[ \omega \geq \sqrt{\frac{g}{A}} \]

5. **Finding the Time Period:**
- The time period \( T \) is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting for \( \omega \):
\[ T \leq \frac{2\pi}{\sqrt{\frac{g}{A}}} \]
- This simplifies to:
\[ T \leq 2\pi \sqrt{\frac{A}{g}} \]

6. **Substituting Values:**
- Substitute \( A = 2.5 \times 10^{-2} \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \):
\[ T \leq 2\pi \sqrt{\frac{2.5 \times 10^{-2}}{10}} \]
- Calculate:
\[ T \leq 2\pi \sqrt{0.0025} = 2\pi \times 0.05 = 0.1\pi \, \text{s} \]

7. **Final Calculation:**
- Thus, the least period \( T \) is:
\[ T \leq \frac{\pi}{10} \, \text{s} \]

### Conclusion:
The least period of oscillation for the object to remain attached to the platform is \( \frac{\pi}{10} \, \text{s} \).