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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