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. `sqrt2`
B. `(1)//(sqrt2)`
C. `2`
D. `(3)//(sqrt2)`

Answer

D

Explanation

`V(x)=(x^4)/(4)-(x^2)/(2)`
`F=-(dV_(x))/(dx)=-[x^3-x]=0impliesx(x^2-1)=0`
`x=0`,`x=+-1`
`(d^2V_(x))/(dx)=3x^2-1`
At `x=+-1`,`(d^2V_(x))/(dx)=+ve`, i.e., at `x=+-1`, P.E. is minimum
`V_(min)=-(1)/(4)`
`E=K_(max)+V_(min)`
`implies2=(1)/(2)xx1xxv_(max)^2-(1)/(4)impliesv_(max)=(3)/(sqrt2)(m)//(s)`