A particle of mass 4g performs S.H.M. between `x = -10 cm and x = + 10` cm along x-axis with frequency 60Hz, initially the particle is at x = +5 cm. Find
the initial phase, amplitude and time period of the particle.

Explanation

To solve the problem step by step, we will find the initial phase, amplitude, and time period of the particle performing Simple Harmonic Motion (SHM).

### Step 1: Determine the Amplitude
The amplitude (A) is the maximum displacement from the equilibrium position. In this case, the particle oscillates between \(x = -10 \, \text{cm}\) and \(x = +10 \, \text{cm}\).

Thus, the amplitude is:
\[
A = \frac{\text{Maximum displacement} - \text{Minimum displacement}}{2} = \frac{10 \, \text{cm} - (-10 \, \text{cm})}{2} = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm}
...