Question

Potential energy function of a certain force field is given as$U\left(x\right)=\frac{x^4}{4}-2x^3+\frac{11}{2}{x}^2-6x$. For this field.

• A. Points of equilibrium are $x=1,x=2$ and $x=3$
• B. $x=2$ is point of unstable equilibrium
• C. $x=1$ is point of stable equilibrium
• D. $x=3$ is point of stable equilibrium

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Points of equilibrium are $x=1,x=2$ and $x=3$
B $x=2$ is point of unstable equilibrium
C $x=1$ is point of stable equilibrium
D $x=3$ is point of stable equilibrium

$F=-\frac{dU}{dx}=-x^3+6x^2-11x+6$

For equilibria, F = 0 $\Rightarrow x^3-6x^2+11x-6=0$

$x^3-3x^2-3x^2+9x+2x-6=0$

$x^2(x-3)-3x(x-3)+2(x-3)=0$

$(x^3-3x+2)(x-3)=0\Rightarrow (x-1)(x-2)(x-3)=0$


Now $dF=-(3x^2-12x+11)dx$

at $x=2$, $dF=+dx\to$ unstable

at $x=1$, $dF=-2dx\to$ stable

at $x=3$, $dF=-2dx\to$ stable