Two SHM particles `P_(1)` and `p_(2)` start from `+ (A)/(2)` and `-sqrt(3A)/(2)`, both in negative directions. Find the time (in terms of T) when they collide. Both particles have same omega, `A` and `T` and the execute SHM along the same line.

Explanation

To solve the problem of when the two SHM particles \( P_1 \) and \( P_2 \) collide, we can follow these steps:

### Step 1: Understand the Initial Positions
- Particle \( P_1 \) starts from position \( +\frac{A}{2} \).
- Particle \( P_2 \) starts from position \( -\frac{\sqrt{3}A}{2} \).
- Both particles are moving in the negative direction.

### Step 2: Write the Equations of Motion
The equations of motion for both particles can be expressed as:
- For \( P_1 \):
\[ x_1(t) = \frac{A}{2} \cos(\omega t) \]
- For \( P_2 \):
\[ x_2(t) = -\frac{\sqrt{3}A}{2} \cos(\omega t) \]

### Step 3: Set Up the Collision Condition
For the particles to collide, their positions must be equal:
\[ x_1(t) = x_2(t) \]
Substituting the equations:
\[ \frac{A}{2} \cos(\omega t) = -\frac{\sqrt{3}A}{2} \cos(\omega t) \]

### Step 4: Simplify the Equation
We can divide both sides by \( \frac{A}{2} \cos(\omega t) \) (assuming \( \cos(\omega t) \neq 0 \)):
\[ 1 = -\sqrt{3} \]
This equation cannot be true, indicating that we need to find a different approach to determine the time of collision.

### Step 5: Analyze the Phasor Representation
From the phasor analysis:
- The initial angle for \( P_1 \) is \( \alpha = \frac{\pi}{3} \) (60 degrees).
- The initial angle for \( P_2 \) is \( \beta = \frac{\pi}{6} \) (30 degrees).

### Step 6: Find the Total Angle
The total angle between the two phasors at the moment of collision is:
\[ \theta = \pi - \left( \alpha + \beta \right) = \pi - \left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 7: Determine the Time of Collision
Since both particles are moving with the same angular frequency \( \omega \), the angle they cover in time \( t \) is:
\[ \omega t = \theta' \]
Where \( \theta' \) is the angle covered by \( P_1 \) from its initial position to the collision point. We can find \( \theta' \) as:
\[ \theta' = \pi - \left( \frac{\pi}{4} + \frac{\pi}{3} \right) = \frac{5\pi}{12} \]

### Step 8: Relate Time to the Period
Using the relationship \( \omega = \frac{2\pi}{T} \):
\[ t = \frac{\theta'}{\omega} = \frac{5\pi/12}{2\pi/T} = \frac{5T}{24} \]

### Final Answer
The time when the two particles collide is:
\[ t = \frac{5T}{24} \]