Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement `x_(1)=+A` and the other `x_(2)=(-A/2)` and they are approaching towards each other. After what time they across each other? `T/4`

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Initial Positions**:
- Particle P is at position \( x_1 = +A \) (maximum amplitude).
- Particle Q is at position \( x_2 = -\frac{A}{2} \) (halfway towards the mean position).

2. **Identifying the Motion**:
- Both particles have the same amplitude \( A \) and the same time period \( T \).
- Since they are in SHM, their positions can be described by the equations:
- For Particle P: \( x_1(t) = A \sin(\omega t + \phi_1) \)
- For Particle Q: \( x_2(t) = A \sin(\omega t + \phi_2) \)
- Here, \( \omega = \frac{2\pi}{T} \).

3. **Determining the Phase Angles**:
- At \( t = 0 \), Particle P is at its maximum position, so \( \phi_1 = \frac{\pi}{2} \) (since \( \sin(\frac{\pi}{2}) = 1 \)).
- For Particle Q, at \( t = 0 \), \( x_2 = -\frac{A}{2} \). We need to find \( \phi_2 \):
\[ -\frac{A}{2} = A \sin(\phi_2) \implies \sin(\phi_2) = -\frac{1}{2} \implies \phi_2 = -\frac{\pi}{6} \text{ (in the 4th quadrant)} \]

4. **Setting Up the Equations**:
- The equations for the positions of the particles become:
- \( x_1(t) = A \sin\left(\frac{2\pi}{T} t + \frac{\pi}{2}\right) \)
- \( x_2(t) = A \sin\left(\frac{2\pi}{T} t - \frac{\pi}{6}\right) \)

5. **Finding the Time When They Cross**:
- They will cross each other when \( x_1(t) = x_2(t) \):
\[ A \sin\left(\frac{2\pi}{T} t + \frac{\pi}{2}\right) = A \sin\left(\frac{2\pi}{T} t - \frac{\pi}{6}\right) \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sin\left(\frac{2\pi}{T} t + \frac{\pi}{2}\right) = \sin\left(\frac{2\pi}{T} t - \frac{\pi}{6}\right) \]

6. **Using the Sine Identity**:
- The sine function is equal when:
\[ \frac{2\pi}{T} t + \frac{\pi}{2} = \frac{2\pi}{T} t - \frac{\pi}{6} + 2n\pi \text{ or } \frac{2\pi}{T} t + \frac{\pi}{2} = \pi - \left(\frac{2\pi}{T} t - \frac{\pi}{6}\right) + 2n\pi \]
- Solving the first equation:
\[ \frac{\pi}{2} + \frac{\pi}{6} = 2n\pi \implies \frac{3\pi + \pi}{6} = 2n\pi \implies \frac{4\pi}{6} = 2n\pi \implies n = 0 \implies t = \frac{T}{3} \]

7. **Conclusion**:
- The two particles will cross each other after \( \frac{T}{3} \).