Exponential Potential Energy Function!

Classical Mechanics Level 3

A particle free to move along the $X$ -axis has it's potential energy given by the expression, $U(x) =k[1-e^{-x^2}]$ , for $-\infty<x<+\infty$ , where ' $k$ ' is a positive constant with appropriate dimensions, then

(A) At any point away from the origin, the particle would be in the state of unstable equilibrium.

(B) For any finite, non zero value of x, there is a force acting on the particle directed away from the origin.

(D) The origin is point of "Unstable Equilibrium".

A D None of These B C

1 solution

Nishant Rai
May 27, 2015

It can be shown from graph (or by calculus) that the origin is a point of “local minima” for the Potential Energy expression for the given one-dimensional system, hence by definition it is a point of “stable equilibrium”.

Note : $\large \frac{\delta^2 U}{\delta x^2} < 0$ means Unstable equilibrium.

$\large \frac{\delta^2 U}{\delta x^2} > 0$ means Stable equilibrium.

$\large \frac{\delta^2 U}{\delta x^2} = 0$ means Neutral equilibrium.

Nice sol.n +1! Any other possible sol.n ?

Rishu Jaar - 4 years, 10 months ago

exactly what I did!!

Arunava Das - 3 years, 2 months ago

i think it is neutral equ. as double differential of u is zero at the origin. plz explain if i am wrong.

vidhan singh - 6 years ago

In this ques U=k[1- $e^{-x^{2}}$ ].So its double differentiation will not be zero it will be $4kx^{2}-2k^{2}e^{-x^{2}}$

Naman Kapoor - 6 years ago

ya i got it later, but thanks

vidhan singh - 6 years ago

Exponential Potential Energy Function!

1 solution

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