Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), there is a relationship between the velocity of the particle and its displacement from the mean position. The velocity is maximum at the mean position ($x=0$) and zero at the extreme positions ($x=\pm A$).
Step 2: Key Formula or Approach:
The velocity $v$ of a particle in SHM at a displacement $x$ from the mean position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where $\omega$ is the angular frequency and A is the amplitude.
The maximum velocity ($v_{max}$) occurs at $x=0$:
\[ v_{max} = \omega \sqrt{A^2 - 0^2} = \omega A \]
We are given the condition $v = \frac{1}{2}v_{max}$.
Step 3: Detailed Explanation:
We are given the condition:
\[ v = \frac{v_{max}}{2} \]
Substitute the formulas for $v$ and $v_{max}$:
\[ \omega \sqrt{A^2 - x^2} = \frac{\omega A}{2} \]
The angular frequency $\omega$ cancels out from both sides:
\[ \sqrt{A^2 - x^2} = \frac{A}{2} \]
Square both sides to solve for $x$:
\[ A^2 - x^2 = \left(\frac{A}{2}\right)^2 = \frac{A^2}{4} \]
Rearrange the equation to isolate $x^2$:
\[ x^2 = A^2 - \frac{A^2}{4} = \frac{4A^2 - A^2}{4} = \frac{3A^2}{4} \]
Take the square root of both sides:
\[ x = \pm \sqrt{\frac{3A^2}{4}} = \pm \frac{\sqrt{3}A}{2} \]
The position is at $\frac{\sqrt{3}A}{2}$ from the mean position.
Step 4: Final Answer:
The position of the particle is at $\frac{\sqrt{3}A}{2}$. Therefore, option (D) is correct.