Two small particles P and Q each of mass m are fixed along x-axis at points (a,0) and (-a,0). A third particle R is kept at origin. Then

Options

A. if particle R is displaced along x-axis it will start oscillating.
B. oscillations of R along x-axis are simple harmonic in nature
C. if R is displaced of R along y-axis, it starts oscillating
D. oscillations along y-axis may be simple harmonic in nature.

Answer

To solve the problem step by step, we will analyze the forces acting on the third particle R when it is displaced from its equilibrium position at the origin.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two particles P and Q, each of mass \( m \), fixed at points \( (a, 0) \) and \( (-a, 0) \) respectively.
- A third particle R, also of mass \( m \), is placed at the origin \( (0, 0) \).

2. **Displacing Particle R**:
- We will first consider the case when particle R is displaced along the x-axis by a small distance \( x \).

3. **Calculating Forces Along the X-axis**:
- When R is displaced to \( (x, 0) \), the gravitational force exerted by particle P at \( (a, 0) \) on R is given by:
\[ F_1 = \frac{Gm^2}{(a - x)^2} \]
- The gravitational force exerted by particle Q at \( (-a, 0) \) on R is:
\[ F_2 = \frac{Gm^2}{(-a - x)^2} \]
- Since \( |a - x| < | -a - x| \) when \( x \) is small, \( F_1 \) will be greater than \( F_2 \).
- The net force \( F_{net} \) acting on R will be directed towards the right (positive x-direction):
\[ F_{net} = F_1 - F_2 \]
- This indicates that the net force acts in the direction of displacement, which means the equilibrium is unstable.

4. **Displacing Particle R Along the Y-axis**:
- Now, consider when particle R is displaced along the y-axis by a small distance \( y \).

5. **Calculating Forces Along the Y-axis**:
- When R is displaced to \( (0, y) \), the gravitational forces from P and Q will act at angles towards R.
- The forces can be resolved into components. The vertical components of the forces from P and Q will be equal due to symmetry, while the horizontal components will cancel out.
- The net force acting on R will be directed downwards (negative y-direction):
\[ F_{net} = -2F \cos(\theta) \]
- Here, \( F \) is the gravitational force magnitude from either P or Q, and \( \theta \) is the angle between the line connecting R to P or Q and the vertical.

6. **Determining the Nature of Motion**:
- For small displacements, we can approximate the forces and find that:
\[ F_{net} \propto -y \]
- This indicates that the motion is oscillatory and can be classified as Simple Harmonic Motion (SHM).

7. **Conclusion**:
- The particle R, when displaced along the y-axis, will oscillate about the equilibrium position, confirming that options 3 and 4 are correct.

### Final Answer:
The correct options are 3 and 4.

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