CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


A
at points away from the origin, the particle is in unstable equilibrium.
loader
B
for any finite non-zero value of x, there is a force directed away from the origin.
loader
C
if its total mechanical energy is k/2 it has its minimum kinetic energy at the origin.
loader
D
for small displacements from x = 0, the motion is SHM.
loader

Solution

The correct option is D for small displacements from x = 0, the motion is SHM.
$$exp \left ( -x^{2} \right ) = 1 - x^{2}  + \cfrac{x^{4}}{2!} + ---$$
For  small  $$x$$ :  $$exp \left ( -x^{2} \right ) = (1 - x^{2})$$
Thus  $$U(x) = K [1-(1-x^{2})] = K x^{2}$$
     $$ F = - \dfrac{dU}{dx} = - \dfrac{d(kx^{2})}{dx}  =  -2 K x$$
Thus, the motion  is  an  SHM.

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image