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91Ó°ÊÓ

A system has potential energy

Ux=x+sin2rad/mx

as a particle moves over the range0m≤x≤πm.

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 0m≤x≤πmx=π3,2π3.

(b) The equilibrium position x=Ï€3is unstable and x=2Ï€3is 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 0m≤x≤πm.

02

Explanation (part a)

To find the equilibrium positions over the range 0m≤x≤πm,

U(x)=x+sin(2x)-dU(x)dx=0-d[x+sin(2x)]dx=01+2cos(2x)=0cos(2x)=-122x=2Ï€3,4Ï€3x=Ï€3,2Ï€3

03

Given information (part b)

A system has potential energy U(x)=x+sin(2x(rad)), wherexis inm, as the particle moves over the range0m≤x≤πm.

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=Ï€3=4sin2Ï€3-d2Udx2x=Ï€3=23>0

And,

-d2Udx2=4sin(2x)-d2Udx2x=Ï€3=4sin4Ï€3-d2Udx2x=Ï€2=-23<0

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