Question

The force acting on a particle constrained to move along x-axis varies as shown.

Identify the correct option(s)

• A. $E$ is a neutral equilibrium position
• B. $F$ is a stable equilibrium position
• C. $B, D$ are unstable equilibrium position
• D. Both (1) and (3)

Answer 1

Answer: (B)

F is a stable equilibrium position

Answer 2

To determine the type of equilibrium for a particle from a Force (F) versus position (x) graph, we use the following criteria:

1. Equilibrium Position: A particle is in equilibrium when the net force acting on it is zero (F = 0). On the given F-x graph, these are the points where the curve intersects the x-axis. From the graph, these points are A, B, C, D, E, and F.

2. Stable Equilibrium: If a particle is slightly displaced from its equilibrium position, a restoring force acts on it, tending to bring it back to the equilibrium position. On an F-x graph, this occurs when F = 0 and the slope of the F-x curve (dF/dx) is negative.

   • If x increases (moves right), F becomes negative (force to the left).
   • If x decreases (moves left), F becomes positive (force to the right).

3. Unstable Equilibrium: If a particle is slightly displaced from its equilibrium position, the force acting on it tends to move it further away from the equilibrium position. On an F-x graph, this occurs when F = 0 and the slope of the F-x curve (dF/dx) is positive.

   • If x increases (moves right), F becomes positive (force to the right).
   • If x decreases (moves left), F becomes negative (force to the left).

4. Neutral Equilibrium: If a particle is displaced from its equilibrium position, it remains in equilibrium at the new position. On an F-x graph, this occurs when F = 0 over a range of positions (a flat segment on the x-axis).

Let's analyze each equilibrium point on the graph:

• Point A: F = 0. The slope dF/dx at A is positive. If displaced to the right, F becomes positive (pushes further right). If displaced to the left, F becomes negative (pushes further left). Thus, A is an unstable equilibrium position.

• Point B: F = 0. The slope dF/dx at B is negative. If displaced to the right, F becomes negative (pushes left, back to B). If displaced to the left, F becomes positive (pushes right, back to B). Thus, B is a stable equilibrium position.

• Point C: F = 0. The slope dF/dx at C is positive. If displaced to the right, F becomes positive (pushes further right). If displaced to the left, F becomes negative (pushes further left). Thus, C is an unstable equilibrium position.

• Point D: F = 0. The slope dF/dx at D is negative. If displaced to the right, F becomes negative (pushes left, back to D). If displaced to the left, F becomes positive (pushes right, back to D). Thus, D is a stable equilibrium position.

• Point E: F = 0. The slope dF/dx at E is positive. If displaced to the right, F becomes positive (pushes further right). If displaced to the left, F becomes negative (pushes further left). Thus, E is an unstable equilibrium position.

• Point F: F = 0. The slope dF/dx at F is negative. If displaced to the right, F becomes negative (pushes left, back to F). If displaced to the left, F becomes positive (pushes right, back to F). Thus, F is a stable equilibrium position.

Now let's evaluate the given options:

(1) E is a neutral equilibrium position Based on our analysis, E is an unstable equilibrium position. So, option (1) is incorrect.

(2) F is a stable equilibrium position Based on our analysis, F is a stable equilibrium position. So, option (2) is correct.

(3) B, D are unstable equilibrium position Based on our analysis, B and D are stable equilibrium positions. So, option (3) is incorrect.

(4) Both (1) and (3) Since options (1) and (3) are incorrect, option (4) is also incorrect.

Therefore, only option (2) is correct based on the standard definitions of equilibrium.