Chapter 10: Q. 58 (page 259)

A particle that can move along the x-axis is part of a system with potential energy
$U (x) = \frac{A}{x^{2}} - \frac{B}{x}$
where A and B are positive constants.
a. Where are the particle’s equilibrium positions?
b. For each, is it a point of stable or unstable equilibrium?

Short Answer

Expert verified

(a) The equilibrium position is $x = \frac{2 A}{B}$

(b) Since $- {\frac{d^{2} U}{d x^{2}}|}_{x = \frac{2 A}{B}} < 0$ , the equilibrium position $x = \frac{2 A}{B}$ is stable.

Step by step solution

Given information (part a)

A system has potential energy $U (x) = \frac{A}{x^{2}} - \frac{B}{x}$ , whereA and B are positive constants.

Explanation (part a)

To find the equilibrium positions,

$U (x) = \frac{A}{x^{2}} - \frac{B}{x} - \frac{d U (x)}{d x} = 0 - \frac{d}{d x} (\frac{A}{x^{2}} - \frac{B}{x}) = 0 \frac{2 A}{x^{3}} - \frac{B}{x^{2}} = 0 2 A - B x = 0 x = \frac{2 A}{B}$

Given information (part b)

A system has potential energy $U (x) = \frac{A}{x^{2}} - \frac{B}{x}$ , whereA and B are positive constants.

Explanation (part b)

To determine the stability,

$- \frac{d^{2} U}{d x^{2}} = - \frac{d^{2}}{d x^{2}} (\frac{A}{x^{2}} - \frac{B}{x}) - \frac{d^{2} U}{d x^{2}} = - \frac{d}{d x} (\frac{B}{x^{2}} - \frac{2 A}{x^{3}}) - \frac{d^{2} U}{d x^{2}} = \frac{2 B}{x^{3}} - \frac{6 A}{x^{4}} - {\frac{d^{2} U}{d x^{2}}|}_{x = \frac{2 A}{B}} = \frac{2 B}{{(\frac{2 A}{B})}^{3}} - \frac{6 A}{{(\frac{2 A}{B})}^{4}} - {\frac{d^{2} U}{d x^{2}}|}_{x = \frac{2 A}{B}} = \frac{B^{4}}{4 A^{3}} - \frac{3 B^{4}}{8 A^{3}} - {\frac{d^{2} U}{d x^{2}}|}_{x = \frac{2 A}{B}} = - \frac{B^{4}}{8 A^{3}} < 0, since A > 0, B > 0$

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A particle that can move along the x-axis is part of a system with potential energy
$U (x) = \frac{A}{x^{2}} - \frac{B}{x}$
where A and B are positive constants.
a. Where are the particle’s equilibrium positions?
b. For each, is it a point of stable or unstable equilibrium?

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

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A particle that can move along the x-axis is part of a system with potential energyUx=Ax2-Bxwhere A and B are positive constants.a. Where are the particle’s equilibrium positions?b. For each, is it a point of stable or unstable equilibrium?

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Study anywhere. Anytime. Across all devices.

A particle that can move along the x-axis is part of a system with potential energy
$U (x) = \frac{A}{x^{2}} - \frac{B}{x}$
where A and B are positive constants.
a. Where are the particle’s equilibrium positions?
b. For each, is it a point of stable or unstable equilibrium?