Solution:

In Simple Harmonic Motion (SHM), the relationship between displacement, velocity, and amplitude can be described using the following equations.

1. The maximum speed \( V_0 \) is given by:
\[
V_0 = \omega A
\]
where \( \omega \) is the angular frequency.

2. The speed \( v \) at a displacement \( x \) in SHM is given by the formula:
\[
v = \sqrt{V_0^2 - \left(\frac{\omega x}{\omega A}\right)^2}
\]
Simplifying this using \( \omega = \frac{V_0}{A} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot x\right)^2}
\]
Substituting \( x = \frac{A}{2} \):
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{A} \cdot \frac{A}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \left(\frac{V_0}{2}\right)^2}
\]
\[
v = \sqrt{V_0^2 - \frac{V_0^2}{4}} = \sqrt{\frac{3V_0^2}{4}} = \frac{\sqrt{3}}{2} V_0
\]

Thus, the speed of the particle at displacement \( \frac{A}{2} \) is:
\[{\frac{\sqrt{3}}{2} V_0}
\]