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 a particle undergoing simple harmonic motion (SHM).

### Step 1: Understand the given parameters
- The particle starts SHM at time \( t = 0 \).
- The amplitude of the motion is \( A \).
- The angular frequency is \( \omega \).
- At \( t = 0 \), the kinetic energy \( KE = \frac{E}{4} \), where \( E \) is the total mechanical energy.

### Step 2: Write the expression for total mechanical energy
The total mechanical energy \( E \) in SHM is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
This is because at the maximum displacement (amplitude), all the energy is potential, and at the mean position, all the energy is kinetic.

### Step 3: Write the expression for kinetic energy
The kinetic energy \( KE \) at any point in SHM is given by:
\[ KE = \frac{1}{2} m v^2 \]
At \( t = 0 \), we know that the kinetic energy is \( \frac{E}{4} \). Therefore, we can write:
\[ \frac{E}{4} = \frac{1}{2} m v^2 \]

### Step 4: Find the expression for velocity
In SHM, the velocity \( v \) at any position \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
At \( t = 0 \), if we assume the particle is at the mean position, \( x = 0 \), then:
\[ v = \omega A \]

### Step 5: Substitute the velocity into the kinetic energy equation
Now substituting \( v = \omega A \) into the kinetic energy equation:
\[ \frac{E}{4} = \frac{1}{2} m (\omega A)^2 \]
This simplifies to:
\[ \frac{E}{4} = \frac{1}{2} m \omega^2 A^2 \]

### Step 6: Relate the expressions for total energy and kinetic energy
From the expression for total energy, we have:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Substituting this into the kinetic energy equation gives:
\[ \frac{E}{4} = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{4} \]
This confirms that the kinetic energy at \( t = 0 \) is indeed \( \frac{E}{4} \).

### Step 7: Find the phase constant \( \phi \)
Using the relationship between kinetic energy and the phase constant, we know:
\[ KE = \frac{1}{2} m A^2 \omega^2 \cos^2(\phi) \]
Setting this equal to \( \frac{E}{4} \):
\[ \frac{E}{4} = \frac{1}{2} m A^2 \omega^2 \cos^2(\phi) \]
Substituting \( E \) into this equation:
\[ \frac{1}{4} \cdot \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\phi) \]
This simplifies to:
\[ \frac{1}{4} = \cos^2(\phi) \]
Thus, we find:
\[ \cos(\phi) = \pm \frac{1}{2} \]

### Step 8: Determine possible values for \( \phi \)
The values of \( \phi \) that satisfy \( \cos(\phi) = \frac{1}{2} \) are:
\[ \phi = \frac{\pi}{3}, \quad \frac{5\pi}{3} \quad (\text{or } 2\pi - \frac{\pi}{3}) \]
And for \( \cos(\phi) = -\frac{1}{2} \):
\[ \phi = \frac{2\pi}{3}, \quad \frac{4\pi}{3} \]

### Step 9: Write the equations of motion
The general equation of motion for SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
Substituting the possible values of \( \phi \):
1. \( x(t) = A \sin(\omega t + \frac{\pi}{3}) \)
2. \( x(t) = A \sin(\omega t + \frac{2\pi}{3}) \)
3. \( x(t) = A \sin(\omega t + \frac{4\pi}{3}) \)
4. \( x(t) = A \sin(\omega t + \frac{5\pi}{3}) \)

### Conclusion
Thus, the particle can be described by any of these equations depending on the initial phase \( \phi \).

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