A particle free to move along the x-axis has potential energy given by $$U(x)=k[1-\exp {{(-x)}^{2}}]$$ for $$-\infty \le x\le +\infty $$, where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999; UPSEAT 2003]

Question

A particle free to move along the x-axis has potential energy given by $$U(x)=k[1-\exp {{(-x)}^{2}}]$$ for $$-\infty \le x\le +\infty $$, where k is a positive constant of appropriate dimensions. Then                   [IIT-JEE 1999; UPSEAT 2003]

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	A particle free to move along the x-axis has potential energy given by \[U(x)=k[1-\exp {{(-x)}^{2}}]\] for \[-\infty \le x\le +\infty \], where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999; UPSEAT 2003]
A.	At point 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 at the origin
D.	For small displacements from x = 0, the motion is simple harmonic
Answer» E.

A particle free to move along the x-axis has potential energy given by \[U(x)=k[1-\exp {{(-x)}^{2}}]\] for \[-\infty \le x\le +\infty \], where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999; UPSEAT 2003]

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