Two particles are in `SHM` in a straight line about same equilibrium position. Amplitude `A` and time period `T` of both the particles are equal. At time `t=0`, one particle is at displacement `y_(1)=+A` and the other at `y_(2)=-A//2`, and they are approaching towards each other. after what time they cross each other?

Options

A. `T//3`
B. `T//4`
C. `5T//6`
D. `T//6`

Explanation

To solve the problem step by step, we will analyze the motion of both particles in Simple Harmonic Motion (SHM) and determine when they cross each other.

### Step 1: Understand the Initial Conditions
At time \( t = 0 \):
- Particle 1 is at maximum displacement \( y_1 = +A \).
- Particle 2 is at \( y_2 = -\frac{A}{2} \).

### Step 2: Determine the Initial Phase Angles
For SHM, the displacement can be expressed as:
\[ y = A \sin(\omega t + \phi) \]
Where:
- \( \omega = \frac{2\pi}{T} \) is the angular frequency.
- \( \phi \) is the phase constant.

For Particle 1:
- At \( t = 0 \), \( y_1 = +A \).
- Thus, \( A \sin(\phi_1) = A \) implies \( \phi_1 = \frac{\pi}{2} \) (90 degrees).

For Particle 2:
- At \( t = 0 \), \( y_2 = -\frac{A}{2} \).
- Thus, \( A \sin(\phi_2) = -\frac{A}{2} \) implies \( \sin(\phi_2) = -\frac{1}{2} \).
- This gives \( \phi_2 = -\frac{\pi}{6} \) (or \( 210^\circ \) in the unit circle).

### Step 3: Write the Equations of Motion
The equations of motion for both particles can be written as:
1. For Particle 1:
\[ y_1(t) = A \sin\left(\omega t + \frac{\pi}{2}\right) = A \cos(\omega t) \]

2. For Particle 2:
\[ y_2(t) = A \sin\left(\omega t - \frac{\pi}{6}\right) \]

### Step 4: Set Up the Condition for Crossing
The particles cross each other when their displacements are equal:
\[ y_1(t) = y_2(t) \]
Substituting the equations:
\[ A \cos(\omega t) = A \sin\left(\omega t - \frac{\pi}{6}\right) \]

### Step 5: Simplify the Equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \cos(\omega t) = \sin\left(\omega t - \frac{\pi}{6}\right) \]

Using the identity \( \sin(x) = \cos\left(\frac{\pi}{2} - x\right) \):
\[ \cos(\omega t) = \cos\left(\frac{\pi}{2} - \left(\omega t - \frac{\pi}{6}\right)\right) \]
This implies:
\[ \omega t = \frac{\pi}{2} - \left(\omega t - \frac{\pi}{6}\right) + 2n\pi \quad \text{(for some integer } n) \]

### Step 6: Solve for Time
Rearranging gives:
\[ 2\omega t = \frac{\pi}{2} + \frac{\pi}{6} + 2n\pi \]
\[ 2\omega t = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
Thus:
\[ \omega t = \frac{\pi}{3} \]
\[ t = \frac{\pi}{3\omega} \]

Since \( \omega = \frac{2\pi}{T} \):
\[ t = \frac{\pi}{3} \cdot \frac{T}{2\pi} = \frac{T}{6} \]

### Final Answer
The two particles will cross each other after a time \( t = \frac{T}{6} \).

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