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^(@)`.

Options

A. `y=0.1 "cos" (3t+(pi)/(4))`
B. `y=0.1 "sin"(6t+(pi)/(4))`
C. `y=0.1 "sin"(4t+(pi)/(4))`
D. `y=0.1 "cos" (4t+(pi)/(4))`

Explanation

To solve the problem step by step, we will follow these instructions:

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

### Step 2: Write the equation for kinetic energy at mean position
At the mean position (where displacement \( x = 0 \)), the kinetic energy is given by:
\[ KE = \frac{1}{2} m \omega^2 A^2 \]
Substituting the known values:
\[ 8 \times 10^{-3} = \frac{1}{2} \times 0.1 \times \omega^2 \times (0.1)^2 \]

### Step 3: Simplify the equation
First, calculate \( (0.1)^2 = 0.01 \):
\[ 8 \times 10^{-3} = \frac{1}{2} \times 0.1 \times \omega^2 \times 0.01 \]
This simplifies to:
\[ 8 \times 10^{-3} = \frac{1}{200} \omega^2 \]

### Step 4: Solve for \( \omega^2 \)
Multiply both sides by 200 to isolate \( \omega^2 \):
\[ \omega^2 = 8 \times 10^{-3} \times 200 \]
Calculating the right side:
\[ \omega^2 = 1.6 \, \text{(since } 8 \times 200 = 1600 \text{ and } 10^{-3} \text{ gives } 1.6\text{)} \]

### Step 5: Find \( \omega \)
Taking the square root of both sides:
\[ \omega = \sqrt{1.6} \approx 4 \, \text{(taking the positive value)} \]

### Step 6: Write the equation of motion
The standard equation of motion for SHM is:
\[ y(t) = A \sin(\omega t + \phi) \]
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:
\[ y(t) = 0.1 \sin(4t + \frac{\pi}{4}) \]

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