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Q.

A particle free to move along the x-axis has potential energy given by U(x) = K[1-exp(-x2)]  for -∞ < x< +∞ where t is a positive constant of appropriate dimensions. Then

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a

for small displacement from x= 0, the motion is simple harmonic

b

if its total mechanical energy is k2, it has its minimum kinetic energy at the origin.

c

for any finite nor-zero value of x, there is a force directed away from the origin.

d

at points away from the origin, the particles is in unstable equilibrium.

answer is A.

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Detailed Solution

Since F = -dUdx = 2kx exp(-x2)F = 0 (at equilibrium as x = 0)U is minimum at x = 0 and Umin = 0U is maximum at x →±∞ and Umax = kThe particle would oscillate about x = 0 for small displacement from the origin and it is in stable equilibrium at the origin.
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