A particle is executing simple harmonic motion. The velocity of the particle is 10 m `s^(-1)`. What at a distance of 5 cm from mean position and 8 cm `s^(-1)` when at a distance of 12 cm from mean position. What will be the amplitude of motion ?

Explanation

To solve the problem of finding the amplitude of a particle executing simple harmonic motion (SHM) given its velocities at specific displacements, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship in SHM**:
The velocity (v) of a particle in SHM can be expressed as:
\[
v = \omega \sqrt{A^2 - y^2}
...