At particle is executing `SHM` with amplitude `A` and has maximum velocity `v_(0)`. Find its speed when it is located at distance of `(A)/(2)` from mean position.

Explanation

To find the speed of a particle executing Simple Harmonic Motion (SHM) when it is located at a distance of \( \frac{A}{2} \) from the mean position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between maximum velocity and amplitude**:
In SHM, the maximum velocity \( v_0 \) is given by the formula:
\[ v_0 = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Express angular frequency**:
From the equation above, we can express \( \omega \) as:
\[ \omega = \frac{v_0}{A} \]

3. **Use the formula for velocity in SHM**:
The velocity \( v \) of a particle at a distance \( x \) from the mean position is given by:
\[ v = \sqrt{v_0^2 - \left(\omega x\right)^2} \]

4. **Substitute \( x = \frac{A}{2} \)**:
We need to find the speed when \( x = \frac{A}{2} \). First, we calculate \( \omega x \):
\[ \omega x = \left(\frac{v_0}{A}\right) \left(\frac{A}{2}\right) = \frac{v_0}{2} \]

5. **Substitute into the velocity formula**:
Now, substitute \( \omega x \) back into the velocity formula:
\[ v = \sqrt{v_0^2 - \left(\frac{v_0}{2}\right)^2} \]

6. **Simplify the expression**:
Calculate \( \left(\frac{v_0}{2}\right)^2 \):
\[ \left(\frac{v_0}{2}\right)^2 = \frac{v_0^2}{4} \]
So we have:
\[ v = \sqrt{v_0^2 - \frac{v_0^2}{4}} = \sqrt{\frac{4v_0^2}{4} - \frac{v_0^2}{4}} = \sqrt{\frac{3v_0^2}{4}} = \frac{v_0 \sqrt{3}}{2} \]

7. **Final Result**:
Therefore, the speed of the particle when it is located at a distance of \( \frac{A}{2} \) from the mean position is:
\[ v = \frac{v_0 \sqrt{3}}{2} \]