The force acting on a body moving along the x | vedclass

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(B) For a particle to be in stable equilibrium,the force $F$ must be zero at the equilibrium position,and if the particle is slightly displaced,the fo

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$x = x_2$

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(B) For a particle to be in stable equilibrium,the force $F$ must be zero at the equilibrium position,and if the particle is slightly displaced,the force should act in a direction that brings it back to the equilibrium position.At $x = x_1$,the force is zero. For $x > x_1$ (slightly to the right),the force $F$ is positive (repulsive),which pushes the particle further away from $x_1$. Thus,$x_1$ is a point of unstable equilibrium.At $x = x_2$,the force is zero. For $x > x_2$ (slightly to the right),the force $F$ is negative (attractive),which pulls the particle back towards $x_2$. For $x < x_2$ (slightly to the left),the force $F$ is positive (repulsive),which pushes the particle back towards $x_2$. Since the force acts to restore the particle to its original position,$x_2$ is a point of stable equilibrium.

The force acting on a body moving along the $x-$axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at

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