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

Options

A. `3/sqrt(2)`
B. `3sqrt(2)`
C. `9/2`
D. `2`

Answer

A

Explanation

`V(x) = (x^(4)/4-x^(2)/2)`
So when pe is minimum then ke is maximum
` or d/dx(v(x))=0Rightarrow x^(3)-x=0`
`Rightarrow x(x^(2)-1)=0`
` Rightarrowx=0,pm1`
`V(0)=0(maximum)`
` V(pm1)=-1/4(minimum)`
`RightarrowKE_(max)=TE-PE_(min)`
`1/2 mV_(max)^(2)=2-(-1/4)=9/4`
`V_(max)=sqrt(2(9//4))/(1kg)=3/sqrt(2)m//s` Ans