A block of mass `2kg` executes simple harmonic motion under the reading from at a spring .The angular and the time period of motion are `0.2 cm` and `2pi`sec respectively Find the maximum force execute by the spring in the block.

Options

A. `0.05 N`
B. `0.002 N`
C. `0.003 N`
D. `0.004 N`

Explanation

To find the maximum force exerted by the spring on the block executing simple harmonic motion, we can follow these steps:

### Step 1: Identify the given values
- Mass of the block, \( m = 2 \, \text{kg} \)
- Amplitude of motion, \( A = 0.2 \, \text{cm} = 0.002 \, \text{m} \) (convert cm to m)
- Time period of motion, \( T = 2\pi \, \text{s} \)

### Step 2: Calculate the angular frequency (\( \omega \))
The angular frequency is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{2\pi} = 1 \, \text{rad/s} \]

### Step 3: Calculate the maximum force (\( F_{\text{max}} \))
The maximum force exerted by the spring can be calculated using the formula:
\[ F_{\text{max}} = m \omega^2 A \]
Substituting the values we have:
\[ F_{\text{max}} = 2 \, \text{kg} \times (1 \, \text{rad/s})^2 \times 0.002 \, \text{m} \]
\[ F_{\text{max}} = 2 \times 1 \times 0.002 = 0.004 \, \text{N} \]

### Conclusion
The maximum force exerted by the spring on the block is \( 0.004 \, \text{N} \).

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