In a certain field, the potential energy is U= axbx,where a and b constants. The particle is in stableequilibrium at x equal to(1) Zero(2)3b2a(4)2a(E)3b

Q. The potential energy of a particle in a certain field is given by $U=\frac{a}{r^2}-\frac{b}{r},$where a and b are positive constants and r is the distance from the centre of the field. The distance of particle in the stable equilibrium position is

Q. The potential energy of a particle in a certain field has the form $U=\frac{a}{r^2}-\frac{b}{r},$ where $a$ and $b$ are postive constants, $r$ is the distance from the centre of the field.

Q.

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = $\frac{a}{x^4}-\frac{b}{x^2}$, where a and b are positive constants. The position of equilibrium x corresponds to

Q. A particle of mass m is located in a one dimensional potential field where potential energy is given by U(x) = A(1 - cos px), where A and p are constants. The period of small oscillations of the particle