A particle in S.H.M. crosses its equilibrium position at time t = 2 sec. When the motion advances by one second more, its velocity is found to be `pi//3 m//s`. If the frequency of the motion is `1//6s^(10)`, find the amplitude of the motion in meter.

Explanation

To solve the problem, we need to find the amplitude of a particle in simple harmonic motion (SHM) given certain conditions.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The particle crosses its equilibrium position at \( t = 2 \) seconds.
- After 1 second (at \( t = 3 \) seconds), the velocity is \( v = \frac{\pi}{3} \, \text{m/s} \).
- The frequency \( f = \frac{1}{6} \, \text{s}^{-1} \).
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