A point particle of mass `0.1kg` is executing `SHM` with amplitude of `0.1m`. When the particle passes through the mean position. Its `K.E.` is `8xx10^(-3)J`. Obtain the equation of motion of this particle if the initial phase of oscillation is `45^(@)`.

Answer

To solve the problem step by step, we will derive the equation of motion for the particle executing Simple Harmonic Motion (SHM) given the parameters.

### Step 1: Understand the given parameters
- Mass of the particle, \( m = 0.1 \, \text{kg} \)
- Amplitude of motion, \( A = 0.1 \, \text{m} \)
- Kinetic energy at mean position, \( KE = 8 \times 10^{-3} \, \text{J} \)
- Initial phase, \( \phi = 45^\circ = \frac{\pi}{4} \, \text{radians} \)

### Step 2: Relate kinetic energy to total energy
At the mean position in SHM, all the energy is kinetic. The total mechanical energy \( E \) in SHM can be expressed as:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Since the kinetic energy at the mean position is given, we can set:
\[ KE = E = \frac{1}{2} m \omega^2 A^2 \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ 8 \times 10^{-3} = \frac{1}{2} \times 0.1 \times \omega^2 \times (0.1)^2 \]
This simplifies to:
\[ 8 \times 10^{-3} = \frac{1}{2} \times 0.1 \times \omega^2 \times 0.01 \]
\[ 8 \times 10^{-3} = 5 \times 10^{-4} \omega^2 \]

### Step 4: Solve for \( \omega^2 \)
Rearranging gives:
\[ \omega^2 = \frac{8 \times 10^{-3}}{5 \times 10^{-4}} = \frac{8}{5} \times 10 = 16 \]
Thus, we find:
\[ \omega = \sqrt{16} = 4 \, \text{rad/s} \]

### Step 5: Write the equation of motion
The general equation of motion for SHM is given by:
\[ y(t) = A \sin(\omega t + \phi) \]
Substituting the values of \( A \), \( \omega \), and \( \phi \):
\[ y(t) = 0.1 \sin(4t + \frac{\pi}{4}) \]

### Final Answer
The equation of motion of the particle is:
\[ y(t) = 0.1 \sin(4t + \frac{\pi}{4}) \, \text{m} \]