A particle is performing SHM. At what distance from mean position, the velocity of the particle becomes half of the maximum velocity ?

Answer

To solve the problem of finding the distance from the mean position where the velocity of a particle performing Simple Harmonic Motion (SHM) becomes half of its maximum velocity, we can follow these steps:

### Step-by-step Solution:

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

2. **Velocity at Displacement \( x \)**:
The velocity \( v \) of the particle at a displacement \( x \) from the mean position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

3. **Setting Up the Equation**:
We need to find the distance \( x \) where the velocity \( v \) is half of the maximum velocity:
\[ v = \frac{1}{2} V_{\text{max}} = \frac{1}{2} \omega A \]

4. **Substituting into the Velocity Equation**:
Substitute \( v \) into the equation:
\[ \frac{1}{2} \omega A = \omega \sqrt{A^2 - x^2} \]

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

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

7. **Rearranging the Equation**:
Rearranging the equation to isolate \( x^2 \):
\[ x^2 = A^2 - \frac{1}{4} A^2 \]
\[ x^2 = \frac{3}{4} A^2 \]

8. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ x = \pm \frac{\sqrt{3}}{2} A \]

### Final Answer:
The distance from the mean position where the velocity of the particle becomes half of the maximum velocity is:
\[ x = \pm \frac{\sqrt{3}}{2} A \]