A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when the initial phase of oscillation is `45^(@)`.

Explanation

To solve the problem step by step, we will follow the outlined process:

### Step 1: Understand the given data
- Mass of the particle, \( m = 0.1 \, \text{kg} \)
- Amplitude of SHM, \( A = 0.1 \, \text{m} \)
- Kinetic energy at mean position, \( KE = 8 \times 10^{-3} \, \text{J} \)

### Step 2: Relate kinetic energy to velocity
At the mean position in SHM, the kinetic energy is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( v \) is the velocity at the mean position.

### Step 3: Substitute the known values
Substituting the known values into the kinetic energy formula:
\[ 8 \times 10^{-3} = \frac{1}{2} \times 0.1 \times v^2 \]

### Step 4: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{8 \times 10^{-3} \times 2}{0.1} = \frac{16 \times 10^{-3}}{0.1} = 0.16 \]

### Step 5: Find \( v \)
Taking the square root to find \( v \):
\[ v = \sqrt{0.16} = 0.4 \, \text{m/s} \]

### Step 6: Relate velocity to angular frequency and amplitude
The velocity at the mean position can also be expressed in terms of angular frequency \( \omega \) and amplitude \( A \):
\[ v = \omega A \]
Substituting the known values:
\[ 0.4 = \omega \times 0.1 \]

### Step 7: Solve for \( \omega \)
Rearranging gives:
\[ \omega = \frac{0.4}{0.1} = 4 \, \text{rad/s} \]

### Step 8: Write the equation of motion
The general equation of motion for SHM is given by:
\[ y(t) = A \sin(\omega t + \phi) \]
where \( \phi \) is the initial phase. Given that the initial phase is \( 45^\circ \) (or \( \frac{\pi}{4} \) radians), we can substitute the values:
\[ y(t) = 0.1 \sin(4t + \frac{\pi}{4}) \]

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