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  • The potential energy of a particle moving along the x-axis varies as <img alt="U(x)=-\frac{x^2}{4}+ 2 x^3 -\frac{11 x^2}{2}+6 x+ 23" src="https://entrancecorner.oncode

The potential energy of a particle moving along the x-axis varies as

U(x)=-\frac{x^2}{4}+ 2 x^3 -\frac{11 x^2}{2}+6 x+ 23

Where U is in joules and x  is in meters. The correct statement about the stable, unstable and neutral equilibrium of a particle is

Option: 1

It is in unstable equilibrium at x=1  and x=3 and in stable equilibrium at x=2

 




 


Option: 2

It is in unstable equilibrium at x=1 and x=3 and in neutral equilibrium at x=2

 

 


Option: 3

It is in stable equilibrium at x=1  and x=3  and in unstable equilibrium at x=2

 

 


Option: 4

It is in neutral equilibrium at x=2, unstable at x=1 and stable at x=3


Answers (1)

best_answer

Potential energy function of a particle:

U(x)=-\frac{x^2}{4}+ 2 x^3 -\frac{11 x^2}{2}+6 x+ 23

The force acting on the particle:

F=-\frac{dU}{dr}

F =-\frac{d}{d x}\left(-\frac{x^2}{4}+2 x^3-11 \frac{x^2}{2}+6 x+23\right)

     \\ =-\left(\frac{2 x}{4}+6 x^2-\frac{22 x}{2}+6\right) \\ =(x-1)(x-2)(x-3)
F=0 \: \: at\: \: x=1,2,3.

Now, \frac{d^2 U}{d r^2}=-3 x^2+12 x-11

We can see that,

\frac{d^2 U}{d r^2}<0\: \: at\, \, x=1 \, \, and\, \, x=3 hence unstable equilibrium


\frac{d^2 U}{d r^2}>0\, \, at \, \, x=2 hence stable equilibrium.


 

Posted by

HARSH KANKARIA

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