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 `(2mg(BP))/(a)` towards `B`. particle released from rest from mid point of AB.
Find the minimum distance of the particle from `B` during the motion.

Options

A. `A. (a)/(6)`
B. `B. (a)/(4)`
C. `C. (a)/(3)`
D. `D. (a)/(8)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the particle and derive the equations of motion.

### Step 1: Understanding the Forces
The particle of mass `m` is subjected to two forces when it is at a point `P` on the line `AB`:
1. A force towards `A`: \( F_A = \frac{mg \cdot AP}{a} \)
2. A force towards `B`: \( F_B = \frac{2mg \cdot BP}{a} \)

Where:
- \( AP \) is the distance from point `A` to point `P`.
- \( BP \) is the distance from point `B` to point `P`.

### Step 2: Defining Distances
Let:
- The total length of the line `AB` be `a`.
- The distance from `A` to `P` be \( AP = \frac{a}{2} - x \) (where `x` is the distance from `B` to `P`).
- The distance from `B` to `P` be \( BP = x \).

### Step 3: Expressing the Forces
Substituting the distances into the force equations:
- Force towards `A`:
\[ F_A = \frac{mg \left(\frac{a}{2} - x\right)}{a} = \frac{mg}{a} \left(\frac{a}{2} - x\right) \]
- Force towards `B`:
\[ F_B = \frac{2mg \cdot x}{a} \]

### Step 4: Finding the Net Force
The net force \( F_{net} \) acting on the particle is the difference between the forces:
\[ F_{net} = F_B - F_A = \frac{2mg \cdot x}{a} - \frac{mg}{a} \left(\frac{a}{2} - x\right) \]
Simplifying this:
\[ F_{net} = \frac{mg}{a} \left(2x - \left(\frac{a}{2} - x\right)\right) = \frac{mg}{a} \left(3x - \frac{a}{2}\right) \]

### Step 5: Setting Up the Equation of Motion
Using Newton's second law, \( F = ma \):
\[ ma = \frac{mg}{a} \left(3x - \frac{a}{2}\right) \]
Dividing both sides by `m`:
\[ a = \frac{g}{a} \left(3x - \frac{a}{2}\right) \]

### Step 6: Identifying Simple Harmonic Motion
This equation can be rearranged to resemble the standard form of simple harmonic motion:
\[ \frac{d^2x}{dt^2} = -\frac{3g}{a} x + \frac{g}{2a} \]
This indicates that the particle undergoes simple harmonic motion with an equilibrium position at \( x = \frac{a}{6} \).

### Step 7: Finding the Minimum Distance from B
The amplitude of the motion can be calculated as:
\[ \text{Amplitude} = \frac{a}{2} - \frac{a}{6} = \frac{3a}{6} - \frac{a}{6} = \frac{2a}{6} = \frac{a}{3} \]
Thus, the minimum distance from `B` during the motion is:
\[ \text{Minimum distance} = \frac{a}{3} - \frac{a}{6} = \frac{2a}{6} = \frac{a}{6} \]

### Final Answer
The minimum distance of the particle from `B` during the motion is:
\[ \boxed{\frac{a}{6}} \]