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A particle free to move along the (x - a...

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

at points away from the origin, the particle is in unstable equilibrium

B

for any finite non-zero value of `x`, there is a force directed away from the origin.

C

if its total mechanical energy is `k/2`, It has its minimum kinetic energy away from the origin

D

from small displacements from `x = 0`, the motion is simple harmonic.

Text Solution

Verified by Experts

The correct Answer is:
D
`U(x) = k(1-e^(-x^(2)))`
It is an exponential incresing graph of potential energy `(U)` with `x^(2)`. Therefore `U` versus x graph will be as shown.

From the graph it is clear that at origin. Potential energy U is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because
`F = (-dU)/(dx) = - ("slope of U-x graph") = 0`
Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about `x = 0` for small displacements. Therefore, correct option is (d). (a), (b) and (c) options are wrong due to following reasons :
(a) At equilibrium position `F = (-dU)/(dx) = 0` i.e. shape of `U-x` graph should be zero from the graph we can see that slope is zero at `x = 0` and `x = +-oo`.
Now among these equilibrium stable equilibrium position is that where U is minimum (Here `x = 0`).
Unstable equilibrium position is that where U is maximum (Here none).
Netural equilibrium position is that where U is constant (Here `x = +-oo`). Therefore, option (a) is wrong.
(b) For any finite non-zero value of x, force is directed towards the origin because origin is in stable equilibrium position. Therefore, option, (b) is incorrect.
(c) At origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (c) is also wrong.
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