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(B) For a particle to be in equilibrium,the net force $F$ must be zero. From the graph,$F = 0$ at both $x = x_1$ and $x = x_2$.For stable equilibrium,

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$x = x_2$

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(B) For a particle to be in equilibrium,the net force $F$ must be zero. From the graph,$F = 0$ at both $x = x_1$ and $x = x_2$.For stable equilibrium,if the particle is slightly displaced from the equilibrium position,the force should act in a direction that brings the particle back to the equilibrium position (restoring force).At $x = x_1$,if the particle is displaced slightly to the right $(x > x_1)$,the force $F$ becomes positive,pushing it further away from $x_1$. Thus,$x_1$ is an unstable equilibrium point.At $x = x_2$,if the particle is displaced slightly to the right $(x > x_2)$,the force $F$ becomes negative (restoring force),pushing it back towards $x_2$. If displaced slightly to the left $(x < x_2)$,the force $F$ is positive (restoring force),pushing it back towards $x_2$. Therefore,$x = x_2$ is a stable equilibrium point.

$A$ force acts on a particle moving along the $x$-axis,and its variation with position is shown in the figure. At which point will the particle be in a stable equilibrium position?

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