[Solved] A particle free to move along the x-axis has potential energy given by U(x)=k[1−exp(−x)2] for −∞≤x≤+∞ , where k is a positive constant of appropriate dimensions. Then - Principles of Anatomy and Physiology 14th Edition(Edition)

Potential energy of the particle U=k(1−e−x2) Force on particle F=dx−dU=−k[−e−x2×(−2x)] F=−2kxe−x2=−2kx[1−x2+2!x4−……] For small displacement F=−2kx ⇒F∝−x i.e. motion is simple harmonic motion.

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Principles of Anatomy and Physiology

A particle free to move along the x-axis has potential energy

Principles of Anatomy and Physiology 14th Edition by Gerard J. Tortora

Principles of Anatomy and Physiology 14th Edition Edition

Chapter 1: Chapter 1

16 questions

Chapter 2: Chapter 2

32 questions

Chapter 3: Chapter 3

41 questions

Chapter 4: Chapter 4

13 questions

Chapter 5: Chapter 5

15 questions

Chapter 6: Chapter 6

22 questions

Chapter 7: Chapter 7

32 questions

Chapter 8: Chapter 8

18 questions

Chapter 9: Chapter 9

20 questions

Chapter 10: Chapter 10

29 questions

Chapter 11: Chapter 11

33 questions

Chapter 12: Chapter 12

41 questions

Chapter 13: Chapter 13

30 questions

Chapter 14: Chapter 14

60 questions

Chapter 15: Chapter 15

24 questions

Chapter 16: Chapter 16

24 questions

Chapter 17: Chapter 17

33 questions

Chapter 18: Chapter 18

49 questions

Chapter 19: Chapter 19

32 questions

Chapter 20: Chapter 20

38 questions

Chapter 21: Chapter 21

70 questions

Chapter 22: Chapter 22

39 questions

Chapter 23: Chapter 23

48 questions

Chapter 24: Chapter 24

56 questions

Chapter 25: Chapter 25

44 questions

Chapter 26: Chapter 26

39 questions

Chapter 27: Chapter 27

18 questions

Chapter 28: Chapter 28

49 questions

Chapter 29: Chapter 29

46 questions

Question

Hard

Solving time: 4 mins

A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - exp (- x)^{2}]$ for $- \infty \leq x \leq + \infty$ , where k is a positive constant of appropriate dimensions. Then

At point away from the origin, the particle is in unstable equilibrium

For any finite non-zero value of x, there is a force directed away from the origin

If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

For small displacements from x = 0, the motion is simple harmonic

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Updated on: Oct 23, 2023

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Principles of Anatomy and Physiology

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Harper Discussed

A particle free to move along the x-axis has potential energy given by

U (x) = k [1 - exp (- x)^{2}]

for

- \infty \leq x \leq + \infty

, where k is a positive constant of appropriate dimensions. Then

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

Potential energy of the particle U = k (1 - e^{- x^{2}})

Force on particle

F = \frac{- d U}{d x} = - k [- e^{- x^{2}} \times (- 2 x)]

F = - 2 k x e^{- x^{2}} = - 2 k x [1 - x^{2} + \frac{x ^{4}}{2 !} - \dots\dots]

For small displacement

F = - 2 k x

\Rightarrow F \propto - x

i.e. motion is simple harmonic motion.

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Question Text	A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - exp (- x)^{2}]$ for $- \infty \leq x \leq + \infty$ , where k is a positive constant of appropriate dimensions. Then
Updated On	Oct 23, 2023
Topic	Work, Energy and Power
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	352
Avg. Video Duration	11 min

A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - exp (- x)^{2}]$ for $- \infty \leq x \leq + \infty$ , where k is a positive constant of appropriate dimensions. Then

Text solutionVerified

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A particle free to move along the x-axis has potential energy given by U(x)=k[1−exp(−x)2] for −∞≤x≤+∞ , where k is a positive constant of appropriate dimensions. Then

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A particle free to move along the x-axis has potential energy given by $U (x) = k [1 - exp (- x)^{2}]$ for $- \infty \leq x \leq + \infty$ , where k is a positive constant of appropriate dimensions. Then