In terms of time period of oscillations T, find the shortest time in moving a particle from `+ (A)/(2)` to `-(sqrt(3))/(2)`.

Explanation

To solve the problem of finding the shortest time for a particle in simple harmonic motion (SHM) to move from \( +\frac{A}{2} \) to \( -\frac{\sqrt{3}}{2}A \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Position of the Particle**:
- The particle is oscillating between \( -A \) and \( +A \) with an amplitude \( A \).
- The starting position is \( x_1 = +\frac{A}{2} \).
- The ending position is \( x_2 = -\frac{\sqrt{3}}{2}A \).

2. **Identifying the Angles**:
- In SHM, the position can be related to angles on a reference circle.
- For \( x_1 = +\frac{A}{2} \), we can find the corresponding angle \( \theta_1 \):
\[ \cos(\theta_1) = \frac{x_1}{A} = \frac{1}{2} \implies \theta_1 = 60^\circ \]
- For \( x_2 = -\frac{\sqrt{3}}{2}A \), we can find the corresponding angle \( \theta_2 \):
\[ \cos(\theta_2) = \frac{x_2}{A} = -\frac{\sqrt{3}}{2} \implies \theta_2 = 150^\circ \]

3. **Calculating the Angle Difference**:
- The angle difference \( \Delta \theta \) between the two positions is:
\[ \Delta \theta = \theta_2 - \theta_1 = 150^\circ - 60^\circ = 90^\circ \]

4. **Relating the Angle to Time**:
- The total time period \( T \) for a complete oscillation is associated with \( 360^\circ \).
- The time taken to move through an angle of \( 90^\circ \) is:
\[ t = \frac{90^\circ}{360^\circ} \cdot T = \frac{1}{4} T \]

5. **Final Result**:
- Therefore, the shortest time to move from \( +\frac{A}{2} \) to \( -\frac{\sqrt{3}}{2}A \) is:
\[ t = \frac{T}{4} \]

### Summary
The shortest time for the particle to move from \( +\frac{A}{2} \) to \( -\frac{\sqrt{3}}{2}A \) is \( \frac{T}{4} \).