The potential energy- position graph of a particle is shown in figure. Then particle :-

Question

The potential energy- position graph of a particle is shown in figure. Then particle :- (A) is in stable equilibrium at x 1 (B) is in stable equilibrium at x 2 (C) is in unstable equilibrium at x 2 (D) is in stable equilibrium at x 3

Tardigrade Admin · Accepted Answer

∵ F =-( d U / dx )=0 for equilibrium and ( d 2 U / dx 2)>0 for minimum potential energy and stable equilibrium. ( d 2 U / dx 2)<0 for maximum potential energy and unstable equilibrium.

Q. The potential energy- position graph of a particle is shown in figure. Then particle :-

is in stable equilibrium at $x_{1}$

is in stable equilibrium at $x_{2}$

is in unstable equilibrium at $x_{2}$

is in stable equilibrium at $x_{3}$

$∵ F = - \frac{d U}{d x} = 0$ for equilibrium
and $\frac{d ^{2} U}{d x ^{2}} > 0$ for minimum potential energy and stable equilibrium.
$\frac{d ^{2} U}{d x ^{2}} < 0$ for maximum potential energy and unstable equilibrium.

Q. The potential energy- position graph of a particle is shown in figure. Then particle :-

is in stable equilibrium at x1​

is in stable equilibrium at x2​

is in unstable equilibrium at x2​

is in stable equilibrium at x3​

∵F=−dxdU​=0 for equilibrium and dx2d2U​>0 for minimum potential energy and stable equilibrium. dx2d2U​<0 for maximum potential energy and unstable equilibrium.

is in stable equilibrium at $x_{1}$

is in stable equilibrium at $x_{2}$

is in unstable equilibrium at $x_{2}$

is in stable equilibrium at $x_{3}$

$∵ F = - \frac{d U}{d x} = 0$ for equilibrium
and $\frac{d ^{2} U}{d x ^{2}} > 0$ for minimum potential energy and stable equilibrium.
$\frac{d ^{2} U}{d x ^{2}} < 0$ for maximum potential energy and unstable equilibrium.