A particle starts performing simple harmonic motion. Its amplitude is `A`. At one time its speed is half that of the maximum speed. At this moment the displacement is

Options

A. `(a)(sqrt(2)A)/(3)`
B. `(b)(sqrt(3)A)/(2)`
C. `(c)(2 A)/sqrt(3)`
D. `(d)(3A)/sqrt(2)`

Explanation

To solve the problem, we need to find the displacement \( x \) of a particle performing simple harmonic motion (SHM) when its speed \( v \) is half of its maximum speed \( V_{\text{max}} \).

### Step-by-Step Solution:

1. **Understanding the Maximum Speed in SHM**:
The maximum speed \( V_{\text{max}} \) of a particle in SHM is given by the formula:
\[ V_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Finding the Speed at the Given Condition**:
According to the problem, the speed \( v \) is half of the maximum speed:
\[ v = \frac{1}{2} V_{\text{max}} = \frac{1}{2} A \omega \]

3. **Using the Relationship Between Speed and Displacement**:
The speed \( v \) at any displacement \( x \) in SHM can be expressed as:
\[ v = \sqrt{A^2 \omega^2 - x^2 \omega^2} \]
This can be simplified to:
\[ v = \omega \sqrt{A^2 - x^2} \]

4. **Setting Up the Equation**:
Now, we can set the two expressions for speed equal to each other:
\[ \frac{1}{2} A \omega = \omega \sqrt{A^2 - x^2} \]

5. **Dividing by \( \omega \)**:
Since \( \omega \) is not zero, we can divide both sides by \( \omega \):
\[ \frac{1}{2} A = \sqrt{A^2 - x^2} \]

6. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{1}{2} A\right)^2 = A^2 - x^2 \]
This simplifies to:
\[ \frac{1}{4} A^2 = A^2 - x^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ x^2 = A^2 - \frac{1}{4} A^2 \]
\[ x^2 = \frac{3}{4} A^2 \]

8. **Taking the Square Root**:
Finally, we take the square root to find \( x \):
\[ x = \sqrt{\frac{3}{4} A^2} = \frac{\sqrt{3}}{2} A \]

### Final Answer:
The displacement \( x \) when the speed is half of the maximum speed is:
\[ x = \frac{\sqrt{3}}{2} A \]