A particle of mass `'m'` moves on a horizont smooth line `AB` of length `'a'` such that when particle is at any general point `P` on the line two force act on it. A force `(mg(AP))/(a)` towards `A` and another force `2(mg(BP))/(a)` towards `B`.
Find its time period when released from rest from mid-point of line `AB`.

Options

A. `A. T = 2pisqrt((3a)/(g))`
B. `B. T = 2pisqrt((a)/(2g))`
C. `C. T = 2pisqrt((a)/(g))`
D. `D. T = 2pisqrt((a)/(3g))`

Explanation

To solve the problem, we will follow a systematic approach to find the time period of the particle performing simple harmonic motion (SHM) when released from the midpoint of line AB.

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Particle**:
- The particle experiences two forces when at a point P:
- A force towards A: \( F_A = \frac{mg \cdot AP}{a} \)
- A force towards B: \( F_B = \frac{2mg \cdot BP}{a} \)
- Here, \( AP \) is the distance from point A to point P and \( BP \) is the distance from point B to point P.

2. **Defining Distances**:
- Let the total length of line AB be \( a \).
- If the particle is at point P, let the distance from B be \( x \). Therefore:
- Distance from A to P: \( AP = a - x \)
- Distance from B to P: \( BP = x \)

3. **Calculating the Forces**:
- Substitute the distances into the force equations:
- \( F_A = \frac{mg(a - x)}{a} \)
- \( F_B = \frac{2mgx}{a} \)

4. **Net Force on the Particle**:
- The net force \( F_{net} \) acting on the particle is the difference between the forces towards A and B:
\[ F_{net} = F_A - F_B = \frac{mg(a - x)}{a} - \frac{2mgx}{a} \]
- Simplifying this:
\[ F_{net} = \frac{mg(a - x - 2x)}{a} = \frac{mg(a - 3x)}{a} \]

5. **Finding the Acceleration**:
- According to Newton's second law, \( F = ma \):
\[ ma = \frac{mg(a - 3x)}{a} \]
- Dividing both sides by \( m \):
\[ a = \frac{g(a - 3x)}{a} \]

6. **Relating to SHM**:
- The acceleration can be expressed in the form of SHM:
\[ a = -\omega^2 x \]
- From our equation, we can rearrange it:
\[ a = -\frac{3g}{a} x \]
- Thus, we can identify:
\[ \omega^2 = \frac{3g}{a} \]

7. **Finding the Time Period**:
- The time period \( T \) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{1}{\omega^2}} = 2\pi \sqrt{\frac{a}{3g}} \]

8. **Final Answer**:
- Therefore, the time period when the particle is released from rest from the midpoint of line AB is:
\[ T = 2\pi \sqrt{\frac{a}{3g}} \]