A particle stars SHM at time t=0. Its amplitude is A and angular frequency is `omega.` At time t=0 its kinetic energy is `E/4`. Assuming potential energy to be zero and the particle can be written as (E=total mechanical energy of oscillation).

Options

A. x=A `cos(omegat+pi/6)`
B. x = A`sin(omegat+pi/3)`
C. X = A`sin(omegat-(2pi)/3)`
D. x = A `cos(omegat-(pi)/6)`

Answer

To solve the problem step by step, we need to analyze the given information about the simple harmonic motion (SHM) of the particle.

### Step 1: Understand the Given Information
The particle starts SHM at time \( t = 0 \) with:
- Amplitude \( A \)
- Angular frequency \( \omega \)
- Kinetic energy at \( t = 0 \) is \( \frac{E}{4} \)
- Potential energy is assumed to be zero at this time.

### Step 2: Write the Expression for Kinetic Energy
The kinetic energy \( K \) of a particle in SHM can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 3: Write the Expression for Total Mechanical Energy
The total mechanical energy \( E \) in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant. We can also express \( k \) in terms of \( \omega \) and \( m \):
\[ k = m \omega^2 \]
Thus, the total energy can also be expressed as:
\[ E = \frac{1}{2} m \omega^2 A^2 \]

### Step 4: Relate Kinetic Energy to Total Energy
Since we know that at \( t = 0 \), the kinetic energy is \( \frac{E}{4} \):
\[ \frac{1}{2} mv^2 = \frac{E}{4} \]
Substituting \( E \):
\[ \frac{1}{2} mv^2 = \frac{1}{4} \left(\frac{1}{2} m \omega^2 A^2\right) \]
This simplifies to:
\[ mv^2 = \frac{1}{4} m \omega^2 A^2 \]
Dividing both sides by \( m \):
\[ v^2 = \frac{1}{4} \omega^2 A^2 \]

### Step 5: Find the Position of the Particle
The velocity \( v \) in SHM can also be expressed in terms of displacement \( x \):
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting for \( v^2 \):
\[ \frac{1}{4} \omega^2 A^2 = \omega^2 (A^2 - x^2) \]
Dividing both sides by \( \omega^2 \):
\[ \frac{1}{4} A^2 = A^2 - x^2 \]
Rearranging gives:
\[ x^2 = A^2 - \frac{1}{4} A^2 = \frac{3}{4} A^2 \]
Taking the square root:
\[ x = \pm \frac{\sqrt{3}}{2} A \]

### Step 6: Determine the Phase Constant
To find the phase constant, we can express the position in terms of sine or cosine functions. The general equation for SHM can be written as:
\[ x = A \sin(\omega t + \phi) \]
At \( t = 0 \):
\[ x = A \sin(\phi) \]
Substituting \( x = \frac{\sqrt{3}}{2} A \):
\[ \frac{\sqrt{3}}{2} A = A \sin(\phi) \]
Dividing by \( A \):
\[ \sin(\phi) = \frac{\sqrt{3}}{2} \]
Thus, \( \phi = \frac{\pi}{3} \) or \( \phi = \frac{2\pi}{3} \).

### Final Result
The position of the particle can be expressed as:
1. \( x = A \sin(\omega t + \frac{\pi}{3}) \)
2. \( x = A \sin(\omega t + \frac{2\pi}{3}) \)
3. \( x = A \cos(\omega t + \frac{\pi}{6}) \)
4. \( x = A \cos(\omega t - \frac{\pi}{6}) \)

Thus, the correct options are 1, 2, 3, and 4.