A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small displacement from `x=0`, the motion is simple harmonic.B. if its total mechanical energy is `(k)/(2)`, it has its minimum kinetic energy at the originC. for any finite non zero value of x, there is a force directed away from the originD.

A particle free to move along the (x - axis) hsd potential energy give

1.	A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. for small displacement from `x=0`, the motion is simple harmonic.B. if its total mechanical energy is `(k)/(2)`, it has its minimum kinetic energy at the originC. for any finite non zero value of x, there is a force directed away from the originD.
Answer» Correct Answer - A Since `F=-(dU)/(dx)=2kxexp(-x)^2` `F=0` (at equilibrium as `x=0`) U is minimum at `x=0` and `U_(min)=0` `U` is maximum `xrarr+-infty` and `U_(max)=k` The particle would oscillate about `x=0` for small displacement from the origin and it is in stable equalibrium at the origin.

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