When a particle is performing `SHM` of time period `T_(1)` the force acting on it is `F_(1)` for a certain displacement. When the same particle is performing `SHM` of time period `T_(2)` the force acting is `F_(2)` for the same displacement. What will be the time period of the particle when a combined force of `F_(1)` and `F_(2)` produce the same displacement in `SHM` is different cases?

Explanation

To solve the problem, we will analyze the situation in three different cases based on the direction of the forces \( F_1 \) and \( F_2 \) acting on the particle performing simple harmonic motion (SHM).

### Step-by-Step Solution:

**Step 1: Understand the forces in SHM**
- In SHM, the force acting on a particle is given by Hooke's law, which states that the force is proportional to the displacement from the equilibrium position. Mathematically, it can be expressed as:
\[
F = -m \omega^2 x
...