A horizontal platform with a small block of mass 1 kg kept on it performs vertical SHM of amplitude 1 cm. The block does not lose contact with the platform anywhere. The minimum possible time period of the platform

Options

A. (a)0.1 s
B. (b)0.2 s
C. (c)0.3 s
D. (d)0.4 s

Explanation

To solve the problem of finding the minimum possible time period of a platform performing vertical simple harmonic motion (SHM) with a block on it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a block of mass \( m = 1 \, \text{kg} \) on a platform that performs vertical SHM.
- The amplitude of the SHM is \( A = 1 \, \text{cm} = 0.01 \, \text{m} \).
- The block must not lose contact with the platform at any point during the motion.

2. **Condition for Maintaining Contact**:
- For the block to remain in contact with the platform, the acceleration of the platform must be greater than or equal to the acceleration due to gravity \( g \) at the lowest point of the motion.
- The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = \omega^2 A \]
- Here, \( \omega \) is the angular frequency of the SHM.

3. **Setting Up the Equation**:
- For the block to stay in contact, we need:
\[ a_{\text{max}} \geq g \]
- Substituting the expression for maximum acceleration:
\[ \omega^2 A \geq g \]

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 \( T \) of SHM is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting our expression for \( \omega \):
\[ T \geq \frac{2\pi}{\sqrt{\frac{g}{A}}} \]

6. **Calculating the Minimum Time Period**:
- Substituting \( g = 10 \, \text{m/s}^2 \) and \( A = 0.01 \, \text{m} \):
\[ T \geq \frac{2\pi}{\sqrt{\frac{10}{0.01}}} = \frac{2\pi}{\sqrt{1000}} = \frac{2\pi}{31.62} \approx 0.199 \, \text{s} \]

7. **Final Result**:
- The minimum possible time period \( T \) is approximately \( 0.199 \, \text{s} \).