x-t equation of a particle moving along x-axis is given as
`x=A+A(1-cosomegat)`

Options

A. particle oscillates simple harmonically between `x=2A` and `x=A`
B. velocity of particle is maximum at `x=2A`
C. time taken by particles in travelling from x=A to x=3A is `pi/omega`
D. time taken by particles in travelling from x=A to x=2A is `pi/(2omega)`

Answer

To solve the problem step by step, we start with the given equation of motion for the particle:

**Step 1: Rewrite the equation**
The equation given is:
\[ x = A + A(1 - \cos(\omega t)) \]
We can simplify this:
\[ x = A + A - A\cos(\omega t) \]
\[ x = 2A - A\cos(\omega t) \]

**Step 2: Identify the mean position and amplitude**
From the simplified equation, we can see that:
- The mean position (equilibrium position) of the particle is at \( x = 2A \).
- The amplitude of the motion is \( A \).

**Step 3: Determine the extreme positions**
The particle oscillates around the mean position. The maximum and minimum positions can be calculated as follows:
- Maximum position: \( x = 2A + A = 3A \)
- Minimum position: \( x = 2A - A = A \)

Thus, the particle oscillates between \( x = A \) and \( x = 3A \).

**Step 4: Analyze the velocity of the particle**
The velocity of the particle is maximum at the mean position. Since the mean position is at \( x = 2A \), the velocity is maximum when \( x = 2A \).

**Step 5: Calculate the time taken to travel between positions**
1. **From \( x = A \) to \( x = 3A \)**:
- This represents the full oscillation from maximum compression to maximum stretching, which is half of the time period \( T \).
- The time period \( T \) is given by \( T = \frac{2\pi}{\omega} \).
- Therefore, the time taken is:
\[ \text{Time} = \frac{T}{2} = \frac{\pi}{\omega} \]

2. **From \( x = A \) to \( x = 2A \)**:
- This represents a quarter of the oscillation period.
- The time taken is:
\[ \text{Time} = \frac{T}{4} = \frac{\pi}{2\omega} \]

**Final Summary of Results**:
- The mean position is \( x = 2A \).
- The amplitude is \( A \).
- The particle oscillates between \( x = A \) and \( x = 3A \).
- Maximum velocity occurs at \( x = 2A \).
- Time taken from \( x = A \) to \( x = 3A \) is \( \frac{\pi}{\omega} \).
- Time taken from \( x = A \) to \( x = 2A \) is \( \frac{\pi}{2\omega} \).

**Correct Options**:
- The correct options are 2, 3, and 4.

---