The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J`
The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) is

Options

A. `1//sqrt(2)`
B. `2`
C. `3//sqrt(2)`
D. `sqrt(2)`

Answer

C

Explanation

`U =(x^(4))/(4)-(x^(2))/(2),(dU)/(dx) =x^(3) -x = 0 rArrx =0, x = pm 1`
`(d^(2)U)/(dx^(2)) = 3x^(2) -1(d^(2)U)/(dx^(2)) = +ve` for `x =pm1`
`U(pm 1)=-(1)/(4)`
`K_(max) =Ulmin=T.E = 2J`
`K_(max) =(9)/(4)`
`K_(max) =(1)/(2) mv^(2) , v = (3)/(sqrt(2))`.