Two particle of same time period `(T)` and amplitude undergo `SHM` along the same line with initial and phase of `pi//6`. If they start at the same point along the opposite directions. Find the time other which they will meet again for the first time

Options

A. `T//8`
B. `T//4`
C. `T//2`
D. `T`

Explanation

To solve the problem, we need to analyze the motion of the two particles undergoing Simple Harmonic Motion (SHM) with the same amplitude and time period but with a phase difference of \(\frac{\pi}{6}\).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- Both particles have the same amplitude \(A\) and time period \(T\).
- The first particle starts at the maximum position \(A\) and moves in the positive direction.
- The second particle starts at the same point \(A\) but moves in the negative direction.

2. **Equations of Motion**:
- The displacement of the first particle can be expressed as:
\[ x_1(t) = A \sin\left(\omega t\right) \]
- The displacement of the second particle, which has a phase difference of \(\frac{\pi}{6}\), can be expressed as:
\[ x_2(t) = A \sin\left(\omega t + \frac{\pi}{6}\right) \]
- Here, \(\omega = \frac{2\pi}{T}\) is the angular frequency.

3. **Finding the Condition for Meeting**:
- The particles will meet when their displacements are equal:
\[ x_1(t) = x_2(t) \]
- This gives us:
\[ A \sin\left(\omega t\right) = A \sin\left(\omega t + \frac{\pi}{6}\right) \]
- Dividing both sides by \(A\) (assuming \(A \neq 0\)):
\[ \sin\left(\omega t\right) = \sin\left(\omega t + \frac{\pi}{6}\right) \]

4. **Using the Sine Function Property**:
- The sine function has the property that \(\sin(x) = \sin(y)\) implies:
\[ x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]
- Applying this to our equation:
- For the first case:
\[ \omega t = \omega t + \frac{\pi}{6} + 2n\pi \quad \Rightarrow \quad \text{No solution} \]
- For the second case:
\[ \omega t = \pi - \left(\omega t + \frac{\pi}{6}\right) + 2n\pi \]
Simplifying this gives:
\[ 2\omega t = \pi - \frac{\pi}{6} + 2n\pi \]
\[ 2\omega t = \frac{5\pi}{6} + 2n\pi \]
\[ \omega t = \frac{5\pi}{12} + n\pi \]

5. **Finding the First Meeting Time**:
- For \(n = 0\):
\[ \omega t = \frac{5\pi}{12} \]
\[ t = \frac{5\pi}{12\omega} \]
- Since \(\omega = \frac{2\pi}{T}\):
\[ t = \frac{5\pi}{12 \cdot \frac{2\pi}{T}} = \frac{5T}{24} \]

### Final Answer:
The two particles will meet again for the first time after a time of \(\frac{5T}{24}\).