A system has potential energy

Ux=x+sin2rad/mx

as a particle moves over the range0mxπm.

a. Where are the equilibrium positions in this range?

b. For each, is it a point of stable or unstable equilibrium?

Short Answer

Expert verified

(a) The equilibrium position in the range 0mxπmx=π3,2π3.

(b) The equilibrium position x=π3is unstable and x=2π3is a stable equilibrium.

Step by step solution

01

Given information (part a)

A system has potential energy U(x)=x+sin(2x(rad)), where x is in m,as the particle moves over the range 0mxπm.

02

Explanation (part a)

To find the equilibrium positions over the range 0mxπm,

U(x)=x+sin(2x)-dU(x)dx=0-d[x+sin(2x)]dx=01+2cos(2x)=0cos(2x)=-122x=2π3,4π3x=π3,2π3

03

Given information (part b)

A system has potential energy U(x)=x+sin(2x(rad)), wherexis inm, as the particle moves over the range0mxπm.

04

Explanation (part b)

To determine the stability,

-d2Udx2=-d2[x+sin(2x)]dx2-d2Udx2=-d[1+2cos(2x)]dx-d2Udx2=-[-4sin(2x)]-d2Udx2=4sin(2x)-d2Udx2x=π3=4sin2π3-d2Udx2x=π3=23>0

And,

-d2Udx2=4sin(2x)-d2Udx2x=π3=4sin4π3-d2Udx2x=π2=-23<0

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