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

Correct option is 1for small displacement from x= 0, the motion is simple harmonic

Questions

A particle free to move along the x-axis has potential energy given by $U_{(x)} = K [1 - \exp (- x^{2})] for - \infty < x < + \infty$ where t is a positive constant of appropriate dimensions. Then

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

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

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

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

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detailed solution

Correct option is A

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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