A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.4 `m//s^(2)`. The maximum velocity of the particle is (in m/s)

Options

A. 0.01
B. 0.05
C. 0.5
D. 0.25

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between acceleration and displacement in SHM
In simple harmonic motion (SHM), the acceleration \( a \) of the particle is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( a \) is the acceleration,
- \( \omega \) is the angular frequency,
- \( x \) is the displacement from the mean position.

### Step 2: Substitute the known values into the formula
From the problem, we know:
- The acceleration \( a = 0.4 \, \text{m/s}^2 \) (note that the negative sign indicates direction, but we will use the magnitude for calculation),
- The displacement \( x = 0.02 \, \text{m} \).

Substituting these values into the formula gives:
\[ 0.4 = \omega^2 \cdot 0.02 \]

### Step 3: Solve for \( \omega^2 \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{0.4}{0.02} \]
Calculating this gives:
\[ \omega^2 = 20 \]

### Step 4: Find \( \omega \)
Taking the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{20} = 4.47 \, \text{rad/s} \]

### Step 5: Use the maximum velocity formula
The maximum velocity \( V_{\text{max}} \) in SHM is given by the formula:
\[ V_{\text{max}} = A \omega \]
where \( A \) is the amplitude. Given that the amplitude \( A = 0.1 \, \text{m} \), we can substitute the values:
\[ V_{\text{max}} = 0.1 \cdot 4.47 \]

### Step 6: Calculate \( V_{\text{max}} \)
Calculating this gives:
\[ V_{\text{max}} = 0.447 \, \text{m/s} \]

### Final Answer
Thus, the maximum velocity of the particle is approximately:
\[ \boxed{0.447 \, \text{m/s}} \]

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