Question

Force (in $N$) varies with position as $F=2x^2-3x-2$, Choose correct option :-

• A. $x=-1/2$ is position of stable equilibrium
• B. $x=2$ is position of stable equilibrium
• C. $x=-1/2$ is position of unstable equilibrium
• D. $x=2$ is position of neutral equilibrium

Answer 1

The correct option is A $x=-1/2$ is position of stable equilibrium

$F=2x^2-3x-2=-\frac{dU}{dx}$
$\frac{dU}{dx}=-(2x^2-3x-2)$
For equilibrium
$\frac{dU}{dx}=0\Rightarrow 2x^2-3x-2=0$
$2x^2-4x+x-2=0$
$2x(x-2)+1(x-2)=0$
$x=-\frac{1}{2}$ or $x=2$
Now,
$\frac{d^{2}U}{d{x}^2}=-4x+3$

At $x=-\frac{1}{2};\frac{d^{2}U}{d{x}^2}=-4\left(\frac{-1}{2}\right)+3=5=+ve$
$\therefore$ stable equilibrium.

At $x=2,\frac{d^{2}U}{d{x}^2}=-4\left(2\right)+3=-5=-ve$
$\therefore$ unstable equilibrium.