The potential energy "(U)" of a particle in a conservative field varies with position "(x)" as `U=(4x^(3))/(3)-x^(2)`, where "U" and "x" are in SI unit.Then

The position of unstable equilibrium is x=0

The position of unstable equilibrium is x= - `(1)/(2)`

The position of stable equilibrium is x=0

The position of stable equilibrium is x=`(1)/(2)`

The potential energy of a particle under a conservative force is given by U ( x) = ( x^(2) -3x) J. The equilibrium position of the particle is at

The potential energy of a particle varies with position X according to the relation U(x) = [(X^3/3)-(3X^2/2)+2X] then

Assertion :Mean positon of SHM is the stable equlibrium position. Reason: In stable equilibrium , position potential energy is minimum.

The potential energy of a particle in a conservative field is U =a/r^3-b/r^2, where a and b are positive constants and r is the distance of particle from the centre of field. For equilibrium, the value of r is

A particle located in a one-dimensional potential field has its potential energy function as U(x)(a)/(x^4)-(b)/(x^2) , where a and b are positive constants. The position of equilibrium x corresponds to

potential enrgy of a particle along x-axis varies as, U=-20 + (x-2)^(2) , where U is in joule and x in meter. Find the equilibrium position and state whether it is stable or unstable equilibrium.

The potential energy of a conservative force field is given by U=ax^(2)-bx where, a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral.

The potential energy of a particle in a force field is: U = (A)/(r^(2)) - (B)/(r ) ,. Where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium the distance of the particle is

A particle located in one dimensional potential field has potential energy function U(x)=(a)/(x^(2))-(b)/(x^(3)) , where a and b are positive constants. The position of equilibrium corresponds to x equal to

Potential energy function along x- axis in a certain force field is given as U(x)=(x^(4))/(4)-2x^(2)+(11)/(2)x^(2)-6x For the given force field :- (i)the points of equilibrium are x=1 , x=2 and x=3 (ii) the point x=2 is a point of unstable equilibrium. (iii) the points x=1 and x=3 are points of stable equilibrium. (iv) there exists no point of neutral equilibrium. The correct option is :-