Two particles are in SHM along same line with same amplitude A and same time period T. At time t=0, particle 1 is at + `A/2` and moving towards positive x-axis. At the same time particle 2 is at `-A/2` and moving towards negative x-axis. Find the timewhen they will collide.

Options

A. `(2T)/3`
B. `(5T)/12`
C. `(4T)/3`
D. `(2T)/5`

Answer

To solve the problem of when the two particles in simple harmonic motion (SHM) will collide, we can follow these steps:

### Step 1: Define the motion of the particles
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency, and
- \( \phi \) is the phase constant.

Given that both particles have the same amplitude \( A \) and time period \( T \), we can relate the angular frequency \( \omega \) to the time period:
\[ \omega = \frac{2\pi}{T} \]

### Step 2: Determine the initial conditions
At \( t = 0 \):
- Particle 1 is at \( x_1(0) = \frac{A}{2} \) and moving towards the positive x-axis.
- Particle 2 is at \( x_2(0) = -\frac{A}{2} \) and moving towards the negative x-axis.

### Step 3: Write the equations of motion for both particles
For Particle 1 (moving towards positive x):
\[ x_1(t) = A \sin\left(\omega t + \frac{\pi}{6}\right) \]
Here, we use \( \frac{\pi}{6} \) because at \( t = 0 \), \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

For Particle 2 (moving towards negative x):
\[ x_2(t) = A \sin\left(\omega t - \frac{\pi}{6}\right) \]
Here, we use \( -\frac{\pi}{6} \) because at \( t = 0 \), \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \).

### Step 4: Set the equations equal to find the collision time
To find when the particles collide, we set their positions equal:
\[ A \sin\left(\omega t + \frac{\pi}{6}\right) = A \sin\left(\omega t - \frac{\pi}{6}\right) \]

Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sin\left(\omega t + \frac{\pi}{6}\right) = \sin\left(\omega t - \frac{\pi}{6}\right) \]

### Step 5: Use the sine identity
Using the sine identity \( \sin A = \sin B \) implies:
\[ A = B + n\pi \quad \text{or} \quad A = \pi - B + n\pi \]
for some integer \( n \).

Thus, we have:
1. \( \omega t + \frac{\pi}{6} = \omega t - \frac{\pi}{6} + n\pi \)
2. \( \omega t + \frac{\pi}{6} = \pi - (\omega t - \frac{\pi}{6}) + n\pi \)

From the first equation, we can simplify:
\[ \frac{\pi}{6} = n\pi \]
This does not yield a valid solution for \( n \).

From the second equation:
\[ \omega t + \frac{\pi}{6} = \pi - \omega t + \frac{\pi}{6} + n\pi \]
This simplifies to:
\[ 2\omega t = \pi + n\pi \]
\[ \omega t = \frac{(1+n)\pi}{2} \]

### Step 6: Solve for time \( t \)
Using \( \omega = \frac{2\pi}{T} \):
\[ t = \frac{(1+n)T}{4} \]

### Step 7: Find the first collision time
For the first collision, we take \( n = 0 \):
\[ t = \frac{T}{4} \]

### Final Answer
The two particles will collide for the first time at:
\[ t = \frac{T}{4} \]