The potential energy of a conservative force field is given by
`U=ax^(2)-bx`
where, a and b are positive constants. Find the equilibrium position and type of equilibrium

Options

A. `x=(b)/(2a)` , Stable Equilibrium
B. `x=(b)/(a)` Unstable Equilibrium
C. `x=(b)/(2a)` Neutral Equilibrium
D. None of the Above

Answer

A

Explanation

In a conservative field,
`F=-(dU)/(dx)`
`:. F=-(d)/(dx)(ax^(2)-bx)=b-2ax`
For equilibrium " " F=0 " or " b-2ax=0
`:. X=(b)/(2a)`
From the given equation, we can see that `(d^(2)U)/(dx^(2))=2a` (positive),ltbrlt i.e., U is minimum.
Therefore, `x=(b)/(2a)` is the stable equilibrium position.