A particle moving with SHM in a straight line has a speed of `6m//s` with when 4m from the centre of oscillation and a speed of `8m//s` when 3m from the centre of oscillation . Find the amplitude of oscillation and the shortest time taken by the particle in moving from the extremen position to a point mid way between the extreme position and the centre.

Explanation

To solve the problem step by step, we will use the principles of Simple Harmonic Motion (SHM) and the relationships between displacement, velocity, and amplitude.

### Step 1: Understanding the given information
We are given:
- Speed \( v_1 = 6 \, \text{m/s} \) when the particle is \( y_1 = 4 \, \text{m} \) from the center of oscillation.
- Speed \( v_2 = 8 \, \text{m/s} \) when the particle is \( y_2 = 3 \, \text{m} \) from the center of oscillation.

### Step 2: Using the velocity formula in SHM
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