C

Step 1: In S.H.M., the velocity can be determined using the equation: \(v_x = \sqrt{v^2 - \left(\frac{x}{A}\right)^2 \cdot v_{max}^2}\) where \(v\) is the velocity at the mean position, \(v_{max}\) is the maximum velocity, \(A\) is the amplitude, and \(x\) is the displacement from the mean position.
Step 2: Given that \(x = \frac{A}{2}\), the equation becomes: \(v_x = \sqrt{v^2 - \left(\frac{A/2}{A}\right)^2 \cdot v_{max}^2} = \sqrt{v^2 - \frac{1}{4} v_{max}^2}
Step 3: Since \(v_{max} = v\), we have: \(v_x = \sqrt{v^2 - \frac{1}{4} v^2} = \sqrt{\frac{3}{4} v^2} = \frac{\sqrt{3}}{2} v\).
Step 4: Thus, the velocity of the particle at a distance equal to half of the amplitude is \(\frac{\sqrt{3}}{2} v\),
which matches Option C: \(\frac{\sqrt{3}}{2} v\).