The potential enery of a particle varies with posiion `x` according to the relation `U(x) = 2x^(4) - 27 x` the point `x = (3)/(2)` is point of

Options

A. unstable equilibrium
B. stable equilibrium
C. neutral equilibrium
D. None of these

Answer

B

Explanation

`U (x) =2x^(4) -27 x`
`F = -(du)/(dx) =-(8x^(3)-27) = 0` for equilibrium
`x = (3)/(2) , (d^(2)u)/(dx^(2)) = 24 x^(2) gt 0 at x = (3)/(2)`
Hence `U` is min at `x = (3)/(3)` so stable.