Two particles are in SHM with same amplitude A and same regualr frequency `omega`. At time t=0, one is at x =`+A/2` and the other is at `x=-A/2`. Both are moving in the same direction.

Options

A. phase difference between the two particles is `pi/3`
B. phase difference between the two particles is `(2pi)/3`
C. they will collide after time t= `pi/(2omega)`
D. they will collide after time t=`(3pi)/(4omega)`

Answer

To solve the problem step by step, we will analyze the motion of the two particles in Simple Harmonic Motion (SHM) and determine their phase difference and the time at which they collide.

### Step 1: Understand the Initial Positions
At time \( t = 0 \):
- Particle P is at \( x = +\frac{A}{2} \)
- Particle Q is at \( x = -\frac{A}{2} \)

Both particles have the same amplitude \( A \) and angular frequency \( \omega \).

### Step 2: Determine the Direction of Motion
Since both particles are moving in the same direction, we can assume they are both moving towards the equilibrium position (i.e., \( x = 0 \)).

### Step 3: Use the SHM Equation
The general equation for SHM is given by:
\[ x(t) = A \sin(\omega t + \phi) \]
Where \( \phi \) is the phase constant.

### Step 4: Find the Phase Constants
For Particle P at \( x = +\frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega \cdot 0 + \phi_P) \implies \sin(\phi_P) = \frac{1}{2} \]
This gives:
\[ \phi_P = \frac{\pi}{6} \quad \text{(since it is in the first quadrant)} \]

For Particle Q at \( x = -\frac{A}{2} \):
\[ -\frac{A}{2} = A \sin(\omega \cdot 0 + \phi_Q) \implies \sin(\phi_Q) = -\frac{1}{2} \]
This gives:
\[ \phi_Q = -\frac{\pi}{6} \quad \text{(since it is in the fourth quadrant)} \]

### Step 5: Calculate the Phase Difference
The phase difference \( \Delta \phi \) between the two particles is given by:
\[ \Delta \phi = \phi_P - \phi_Q = \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} \]

### Step 6: Determine the Time of Collision
Since both particles are moving in the same direction, we can set their positions equal to find the time \( t \) when they collide:
\[ A \sin(\omega t + \phi_P) = A \sin(\omega t + \phi_Q) \]
This simplifies to:
\[ \sin(\omega t + \frac{\pi}{6}) = \sin(\omega t - \frac{\pi}{6}) \]
Using the sine identity, this leads to:
\[ \omega t + \frac{\pi}{6} = \pi - (\omega t - \frac{\pi}{6}) \quad \text{or} \quad \omega t + \frac{\pi}{6} = \pi + (\omega t - \frac{\pi}{6}) \]
Solving the first equation:
\[ \omega t + \frac{\pi}{6} = \pi - \omega t + \frac{\pi}{6} \]
This simplifies to:
\[ 2\omega t = \pi \implies t = \frac{\pi}{2\omega} \]

### Final Answers
- The phase difference \( \Delta \phi = \frac{\pi}{3} \)
- The time of collision \( t = \frac{\pi}{2\omega} \)