Solution:

To analyze the motion of the positive charge \( +q \) released from point \( (2a, 0) \), let's consider the forces acting on it due to the two fixed negative charges \( -q \) at points \( (0, a) \) and \( (0, -a) \).

Solution Steps:

1. Symmetry of Forces:
- Each negative charge \( -q \) exerts an attractive force on \( +q \) along the line joining them.
- Due to symmetry, the vertical components of these forces (along the y-axis) will cancel out, while the horizontal components (along the x-axis) will add up, pulling \( +q \) toward the y-axis.

2. Resultant Force Direction:
- The resultant force always points toward the y-axis but varies with distance from it.
- The force does not follow a linear restoring force law (i.e., it’s not proportional to displacement), which is required for simple harmonic motion (SHM).

3. Motion of \( +q \):
- The charge \( +q \) will oscillate back and forth across the y-axis due to the attractive forces from the two fixed charges.
- However, since the force is not proportional to displacement, the motion is **not SHM**.

Conclusion:
The positive charge \( +q \) will oscillate but will not execute SHM.