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

Correct option is 4For small displacements from x = 0, the motion is simple harmonic

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

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At point away from the origin, the particle is in unstable equilibrium

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

If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

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

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detailed solution

Correct option is D

Potential energy of the particle U=k(1−e−x2)Force on particle F=−dUdx=−k[−e−x2×(−2x)]F= −2kxe−x2=−2kx1−x2+x42 !−......For small displacement F=−2kx⇒F(x)∝−x i.e. motion is simple harmonic motion.

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

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