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

Total mechanical energy of particle,
`E_(T) = 2J`
When kinetic energy is maximum, the potential energy should be minimum.
The potential of the particle is given by
`V (x) = (x^(4))/(4) -(x^(2))/(2)`
or `(dV)/(dx) =(4x^(3))/(4)-(2x)/(2) = x^(3) - x = x (x^(2) -1)`
For `V` to be minimum, `(dV)/(dx) = 0`
`:. x(x^(2) -1) = 0`, or `x = 0 pm 1`
At `x = 0, V (x) = 0`
At `x = "pm" 1, V (x) = - (1)/(4) J`
`:. ("Kinetic energy")_(max) = E_(T) - V_(min)`.
or `("Kinetic energy")_(max) = 2 2-(-(1)/(4)) = (9)/(4) J`
or `(1)/(2) mv_(m)^(2) = (9)/(4)`
or `v_(m)^(2) = (9xx2)/(mxx4)`
or `v_(m)^(2) = (9xx2)/(1xx4) = (9)/(2)`
`rArr v_(m) = (3)/(sqrt(2)) m//s`.