A plank with a small block on top of it is under going vertical `SHM`. Its period is `2 sec`. The minium amplitude at which the block will separate from plank is :

Options

A. `(10)/(pi^2)`
B. `(pi^2)/(10)`
C. `(20)/(pi^2)`
D. `(20)/(pi^2)`

Explanation

To solve the problem, we need to determine the minimum amplitude at which a block will separate from a plank undergoing vertical simple harmonic motion (SHM). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Condition for Separation
The block will separate from the plank when the normal force acting on it becomes zero. This occurs when the upward acceleration of the plank due to SHM exceeds the acceleration due to gravity (g).

### Step 2: Relate Acceleration in SHM to Gravity
In SHM, the maximum acceleration (a_max) can be expressed as:
\[ a_{\text{max}} = \omega^2 A \]
where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the motion.

For the block to separate, we need:
\[ a_{\text{max}} > g \]
This leads to:
\[ \omega^2 A > g \]

### Step 3: Express Angular Frequency in Terms of Period
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Given that the period \( T \) is 2 seconds, we can calculate \( \omega \):
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

### Step 4: Substitute \( \omega \) into the Acceleration Condition
Substituting \( \omega \) into the inequality:
\[ \pi^2 A > g \]
Thus, we can express the amplitude \( A \) as:
\[ A > \frac{g}{\pi^2} \]

### Step 5: Calculate the Minimum Amplitude
Now, substituting \( g \) (approximately 10 m/s²):
\[ A > \frac{10}{\pi^2} \]

### Step 6: Calculate the Numerical Value
Calculating \( \pi^2 \) (approximately 9.87):
\[ A > \frac{10}{9.87} \approx 1.01 \, \text{m} \]

### Conclusion
The minimum amplitude at which the block will separate from the plank is approximately:
\[ A \approx 1.01 \, \text{m} \]