/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 57 A system has potential energy聽U... [FREE SOLUTION] | 91影视

91影视

A system has potential energy

Ux=x+sin2rad/mx

as a particle moves over the range0mxm.

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 0mxmx=3,23.

(b) The equilibrium position x=3is unstable and x=23is 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 0mxm.

02

Explanation (part a)

To find the equilibrium positions over the range 0mxm,

U(x)=x+sin(2x)-dU(x)dx=0-d[x+sin(2x)]dx=01+2cos(2x)=0cos(2x)=-122x=23,43x=3,23

03

Given information (part b)

A system has potential energy U(x)=x+sin(2x(rad)), wherexis inm, as the particle moves over the range0mxm.

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=3=4sin23-d2Udx2x=3=23>0

And,

-d2Udx2=4sin(2x)-d2Udx2x=3=4sin43-d2Udx2x=2=-23<0

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.