Phy 2511 Classical Mechanics Module Edited 18-04-2017.pdf

MODULE FOR CLASSICAL MECHANICS

THE UNIVERSITY OF ZAMBIA

CLASSICAL MECHANICS PHY 2511

UNIVERSITY OF ZAMBIA SCHOOL OF NATURAL SCIENCES

©Copyright Reccab Ochieng Manyala (2016) This module has specifically been written for distance education students of the University of Zambia. The program may use the module in all reasonable ways. However, the copyright remains with the author. No part of this module may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from the author.

RECCAB OCHIENG MANYALA SCHOOL OF NATURAL SCIENCES, DEPARTMENT OF PHYSICS.

Acknowledgement I would like to sincerely thank Dr. Mweene for his acceptance to allow me use material from his module for Fast Track Teacher Education Programme. Without his good gesture, this module could not have been completed in time as some of the materials presented here were directly taken from his module which already has a structure for a course in Classical Mechanics, Analytical Mechanics and Special Theory of Relativity for a programme such as the one for distance education.

CLASSICAL MECHANICS

Contents Contents

i

About this Module

1

How this Module is structured .......................................................................................... 1 Module overview

3

Welcome to Classical Mechanics Module (PHY 2511) ................................................... 3 CLASSICAL MECHANICS Module. Is this module for you? ........................................ 3 Timeframe ......................................................................................................................... 4 Study skills ........................................................................................................................ 4 Need help? ........................................................................................................................ 6 Assignments ...................................................................................................................... 6 Assessments ...................................................................................................................... 7 Getting around this Module

8

Margin icons ..................................................................................................................... 8 Chapter 1 Unit 1

10 Vectors and their Applications 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Introduction .................................................................................................. 11 The Cartesian Coordinate System ................................................................ 11 The Plane Polar Coordinate System ............................................................ 14 The Cylindrical Polar Coordinate System ................................................... 20 The Spherical Polar Coordinate System ...................................................... 23 Summary ...................................................................................................... 28 Exercises ...................................................................................................... 28

Chapter 2 Unit 2

10

30 Particle Dynamics in One Dimension

30

2.0 Introduction .................................................................................................. 31 2.1 Motion of a Particle in One Dimension ....................................................... 33 2.1.1 The Free Particle .......................................................................................... 33 2.1.2 Particle Moving Under the Effect of a Constant Force ................................ 34 2.1.3 The Simple Harmonic Oscillator .................................................................. 36 2.2 General Treatment of Motion in One Dimension ........................................ 38 2.2.1 Time and Velocity Dependent Forces .......................................................... 44

ii

Contents

2.2.2 Time-Dependent Forces ............................................................................... 44 2.2.3 Velocity-Dependent Forces .......................................................................... 46 2.3 Summary ...................................................................................................... 47 2.4 Exercises ...................................................................................................... 47 Chapter 3 Unit 3

50 Particle Dynamics in Two and Three Dimensions 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Introduction .................................................................................................. 51 General Theory ............................................................................................ 51 Particle in the Field of Gravity..................................................................... 53 Force and Work............................................................................................ 56 Force and Work............................................................................................ 56 Force and Work (The Del Operator) ............................................................ 58 Angular Momentum ..................................................................................... 62 Central Force Motion ................................................................................... 64 Summary ...................................................................................................... 69 Exercises ...................................................................................................... 69

Chapter 4 Unit 4

50

72 Simple Harmonic Motion 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.8

72

Introduction .................................................................................................. 73 Displacement in Simple Harmonic Motion ................................................. 74 Velocity and Acceleration in Simple Harmonic Motion ............................. 78 The Energy of a Harmonic Oscillator .......................................................... 78 Simple Harmonic Motion in Nature ............................................................ 84 Rotary Motion and Simple Harmonic Motion ............................................. 88 Damped Oscillations .................................................................................... 91 Energy Dissipation in Damped Motion ....................................................... 98 Summary .................................................................................................... 100 Exercises .................................................................................................... 101

CLASSICAL MECHANICS

About this Module This module on Classical Mechanics has been structured as outlined below.

How this Module is structured The course overview The course overview gives you a general introduction to the course. Information contained in the course overview will help you determine:  If the course is suitable for you.  What you will already need to know.  What you can expect from the course.  How much time you will need to invest to complete the course.

The overview also provides guidance on:  Study skills.  Where to get help.  Course assignments and assessments.  Activity icons.  Units.

We strongly recommend that you read the overview carefully before starting your study. The course content The course is broken down into units. Each unit comprises:  An introduction to the unit content.

Module overview

 Unit outcomes.  New terminology.  Core content of the unit with a variety of learning activities.  A unit summary.  Assignments and/or assessments, as applicable.

Resources For those interested in learning more on this subject, we provide you with a list of additional resources at the end of this Module; these may be books, articles or web sites. Your comments After completing this module on Classical Mechanics we would appreciate it if you would take a few moments to give us your feedback on any aspect of this course. Your feedback might include comments on:  Course content and structure. (Have we left out something important?

Or is the material too bulky?)  Course reading materials and resources.(Did you find these adequate

and useful?)  Course assignments. (Did you find the concepts in the assignments

consistent with what is given in the module?

Did you feel

comfortable solving the problems?  Course assessments. (Is the weighting in the assessment reasonable in

terms of marks distribution)  Course duration. (Is the duration sufficient?)  Course support (assigned tutors, technical help, etc.)

Your constructive feedback will help us to improve and enhance this course.

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Module overview

7.

Welcome to Classical Mechanics Module (PHY 2511) This Module is one of the modules in second year Physics for student taking the Distance Education Science program in which Physics as one of their subjects of study at the University of Zambia.

CLASSICAL MECHANICS Module. Is this module for you? This course is intended for people who are pursuing a degree in Science Education and studying Physics as one of the subjects. The prerequisite for this course is PHY 1010 (Introduction to Physics). It is, however, advisable that apart from PHY 1010, the learner should have a good understanding of Mathematics covering topics such as Differential Equations ( first and second order), Partial Differential Equations, Complex numbers, Integral Equations, Matrices and Trigonometry.

3

Module overview

Timeframe This is the module for Classical Mechanics that you will study during your programme.

How long?

The other module in the field of Mechanics is PHY 2522-called Analytical Mechanics and the Special Theory of Relativity. Both modules should be completed in the second year of study. The modules cover material (work) for one academic year. For both in each unit, the time frame required to study the material adequately is indicated. The time indicated is not the time for lectures but time for self study. Lectures may take less or more time than indicated. It is advisable to study PHY 2511 first before proceeding to PHY 2522, however, there are units in PHY 2522 which can be studied independent of those units in the PHY 2511 module

Study skills As an adult learner your approach to learning will be different to that from your school days: you will choose what you want to study, you will have professional and/or personal motivation for doing so and you will most likely be fitting your study activities around other professional or domestic responsibilities. Essentially you will be taking control of your learning environment. As a consequence, you will need to consider performance issues related to time management, goal setting, stress management, etc. Perhaps you will also need to reacquaint yourself in areas such as essay planning, coping with exams and using the web as a learning resource. Your most significant considerations will be time and space i.e. the time you dedicate to your learning and the environment in which you engage in that learning. We recommend that you take time now—before starting your selfstudy—to familiarize yourself with these issues. There are a number of excellent resources on the web. A few suggested links are: 

http://www.how-to-study.com/ The “How to study” web site is dedicated to study skills resources. You will find links to study preparation (a list of nine

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CLASSICAL MECHANICS

essentials for a good study place), taking notes, strategies for reading text books, using reference sources, test anxiety. 

http://www.ucc.vt.edu/stdysk/stdyhlp.html This is the web site of the Virginia Tech, Division of Student Affairs. You will find links to time scheduling (including a “where does time go?” link), a study skill checklist, basic concentration techniques, control of the study environment, note taking, how to read essays for analysis, memory skills (“remembering”).



http://www.howtostudy.org/resources.php Another “How to study” web site with useful links to time management, efficient reading, questioning/listening/observing skills, getting the most out of doing (“hands-on” learning), memory building, tips for staying motivated, developing a learning plan.

The above links are our suggestions to start you on your way. At the time of writing these web links were active. If you want to look for more go to www.google.com and type “self-study basics”, “self-study tips”, “selfstudy skills” or similar. A part from the given web site addresses, there is a free book at bookboon.com entitled-“Strategies to Fight Exam Stress and Achieve Success” ISBN 978-87-7681-917-0 by Will Stringer. This is an excellent book which gives you guidance on many aspects of preparing for an exam.

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Module overview

Need help? You may find some resources on other website or contact Prof. Reccab Ochieng Manyala at the Department of Physics, School of Natural Sciences, University of Zambia. E-mail address: [email protected].

Help

for matters related to the course. You are also free to consult with any academic member of staff in the Department of Physics; they are always willing to help students. In case I am not the one taking you in this course, you will consult with the lecturer concerned. Though we have endeavored to make this unit “mistake free” as much as possible, there could be some mistakes. We encourage you to point out these mistakes to your course mates so that they do not get the wrong ideas. Please also point out the mistakes to the lecturer concerned during the course.

Assignments You will be required to write and submit assignments based on this module. Details of these assignments will be given to you during the residential school. Be sure to write these assignments because they have Assignments

a bearing on your continuous assessment grades. During the residential school, you will be directed where to submit all written assignments. All queries or correspondence concerning your studies to be directed to the following address: The Director, Directorate of Distance Education University of Zambia

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CLASSICAL MECHANICS

P.O BOX 32379, LUSAKA

Assessments There are a number of assessment items in this module. You will be required to carry out experiments in the laboratory and write laboratory Assessments

reports that shall be graded. At the same time you will be assigned questions (tutorials) that you will be told to attempt and hand in at appropriate scheduled times. The questions will also be graded. During the course a minimum of two written tests will be administered and graded. Finally, you will be required to write a final examination in the course. The weighting of these assessment items will be as follows: Assessment item

Weight

Total

1. Two written tests

20%

20%

2. Laboratory reports

15%

15%

5%

5%

60%

60%

100%

100%

3. Tutorials 4. Final examination Total

Note. In the case where only one test is given, the test will account for 20%. Please ensure that you avail yourself for all these assessment items because they will account for your final grade.

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8

Getting around this Module Margin icons While working through this module you will notice the frequent use of margin icons. These icons serve to “signpost” a particular piece of text, a new task or change in activity; they have been included to help you to find your way around this Module. A complete icon set is shown below. We suggest that you familiarize yourself with the icons and their meaning before starting your study.

This module has been written with you as a learner in mind. On the left margins of each page, reasonable spaces have been left. This is not for the beauty of the module but the spaces are for you to interact with the module. You can carry out calculations in the spaces provided, you can make short note in these spaces or even make comments and put reminders to help you in your study of this course.

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Activity

Assessment

Assignment

Case study

Discussion

Group activity

Help

Note it!

Outcomes

Reading

Reflection

Study skills

Summary

Terminology

Time

Tip

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Chapter 1 Unit 1 Vectors and their Applications

You are supposed to spend a minimum of 10 hrs to complete this unit. The 10 hrs does not include time for the exercises which you are advised to do at your own pace in a reasonable manner OBJECTIVES (1) The first objective of this unit is to introduce the learner to the coordinate systems that are most common in the solution of problems in Physics. (2) Teach the learner how to transform coordinate systems.

After studying this unit, the learner shall be able to: (1) Differentiate between the coordinate systems that are commonly used in solution of physics problems. (2) Transform from one coordinate system to another. (3) Recognise and identify the correct and appropriate coordinate system to use for a particular problem. (4) Solve problems in different coordinate systems.

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CLASSICAL MECHANICS

1.0

Introduction To solve many problems in Mechanics or Physics for that matter, it is often necessary to be able to specify the position of its constituents. This is a very important stage in the search for a solution because the wrong choice of coordinates may render the problem difficult or not solvable at all. The position of the constituents of a system requires the use of specific coordinate system. We shall make a formal study of the different options available to be able to solve problems in this module. The four most commonly used coordinate systems are:

the Cartesian or

rectangular coordinate system, the Plane polar coordinate system, the Cylindrical coordinate system and the Spherical polar coordinate system.

1.1 The Cartesian Coordinate System The Cartesian coordinate system is the most widely used of all the coordinate systems. This is a system we encounter in our everyday lives. For example when we are walking, we normally walk in a straight line and we may label that direction x . During the walk, we may want to change direction by making a ninety degree turn. When this is done we may label that new direction y . On the other hand we may want to jump up. Jumping up is in a direction perpendicular to both x and y . These movements complete the description of the Cartesian coordinate system. Many Physics problems are framed in this coordinate system. complete Cartesian coordinate system is illustrated in Fig. 1.

A It is

constructed as follows. We first choose an origin O to which we attach three coordinate axes labeled x , y and z which are mutually orthogonal (perpendicular). The directed line from the origin to the point

P where a particle is located is called the position vector r . The position of a point such as P is specified by 3 numbers or coordinates which are the lengths of the components of the vector r , that is x , y and z .

11

12

Z

P x, y, z 

z Y

y x X

Fig.1: Cartesian coordinates Each of the coordinate axes has a unit vector which gives its direction. These unit vectors are respectively iˆ , ˆj and kˆ for the x , y and z axes. As the names imply, each of these unit vectors has a magnitude of unity (one). The dot product of any two arbitrary vectors A and B is defined as

A  B  A B sin  AB sin

(1.1)

where  is the angle between the vectors. Since the unit vectors are mutually orthogonal (perpendicular, that is at 900 ) to each other, they obey the following relationships:

iˆ  ˆj  iˆ  kˆ  ˆj  kˆ  0

(1.2)

Again for any two vectors A and B their cross product defined as:

A  B  A B cos  AB cos  C

(1.3)

results in a third vector C which lies perpendicular to the plane containing A and B . It is therefore possible to obtain any one of the

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CLASSICAL MECHANICS

unit vectors as a cross product of the other two. Using a coordinate system, the formulas are:

iˆ  ˆj  kˆ,

ˆj  kˆ  iˆ and kˆ  iˆ  ˆj

(1.4)

In terms of the unit vectors, the position vector is

r  iˆx  ˆjy  kˆz  0

(1.5)

An alternative of writing this is

r  x, y, z 

(1.6)

The length of a vector is called its magnitude. For the position vector the magnitude is given by



r  r  r  r  x2  y 2  z 2



1

2

(1.7)

For a moving point P , the coordinates change with time and are therefore functions of time. The velocity of the point P is the time derivative of the position vector and is therefore given by

v

dr ˆ dx ˆ dy ˆ dz i  j k dt dt dt dt

(1.8)

It is usual to indicate the time derivative of a quantity by a dot above quantity in question. This notation allows us to write Eq.(1.8) as

v  r  iˆx  ˆjy  kˆz

(1.9)

The magnitude of the velocity, that is, the speed is



v  v  v  v  x 2  y 2  z 2



1

2

(1.10)

Differentiating the velocity with respect to time gives us the acceleration; this is

a  v  r  iˆx  ˆjy  kˆz

(1.11)

13

14

Example 1.1: The position of a particle at a point P in Cartesian coordinate system is given in metres by the vector r  6iˆ  5 ˆj  2kˆ . What is the distance of the particle from the origin? Solution: The required quantity is the magnitude of r and is given by

r  r  62  (5) 2  22  8.06 m Example 1.2: The position vector of a particle whose mass is 4 kg moves in space described by a vector that depends on time



 

 



as r  4t  6t 2 iˆ  t 3  2 ˆj  3t  5t 2 kˆ .

Calculate for t  2 s the

position, the velocity and the magnitude of the force acting on the particle. Solution: To obtain the position vector at t  2 s , we substitute 2 for t in the expression for r . This yields r  16iˆ  6 ˆj  26kˆ . The velocity is given by

v  r  iˆx  ˆjy  kˆz

 

 iˆ4 12t   ˆj 3t 2  kˆ3  10t  so at t  2 s , v  20iˆ  12 ˆj  23kˆ According to Newton’s second law, the force is given by ma . Since

a  12iˆ  6tˆj  10kˆ . We have at t  2 s , a  12iˆ  12 ˆj  10kˆ so that the force is F  ma  48iˆ  48 ˆj  10kˆ . The magnitude of this force is



F  F   48  482  102 2



1

2

 68.6 N

1.2 The Plane Polar Coordinate System This is a very useful coordinate system for solving problems in curved or circular motion in two dimensions. To construct this system; we start be defining an origin O , and from that origin draw a horizontal reference 14

CLASSICAL MECHANICS

line labeled X which extends to the negative side. From the origin again draw a vertical reference line labeled Y and extending in both positive and negative directions (see Fig. 2).

Y

P

y

 O

X

x

Fig. 2: Plane polar coordinates The position of the point P is measured from the origin defined by the parameters r (the length from O to P) and the angle  which the line OP makes with the horizontal reference line in the counter-clockwise direction. The coordinates of P are then given as r ,   . It is possible to obtain transformation equations which connect the plane polar coordinates and the Cartesian coordinates using Fig. 3. The reference line coincides with the positive x  axis.

By simple trigonometry, we

establish that

x  r cos

(1.12)

y  r sin

(1.13)

The inverse transformations are



r  x2  y 2



1

2

(1.14)

and

 y x



   y   sin 1    x2  y2   x2  y2     

  arctan   cos1 

x

(1.15) 15

16

with the range of coordinates being 0  r   and 0    2 .

Fig.3: Unit vectors rˆ and ˆ in plane polar coordinates In terms of the Cartesian unit vectors iˆ and ˆj , the radial unit vector is

rˆ  iˆ cos  ˆj sin

(1.16)

It is also clear that

ˆ  iˆ sin  ˆj cos

(1.17)

We can also relate these unit vectors using Fig. 4.

 

 

Fig.4: Relation between unit vectors rˆ, ˆ and iˆ, ˆj .

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Using the two equations (1.16) and (1.17) it is possible to verify that rˆ  ˆ  0 . From these equations, we can solve for the Cartesian unit vectors to obtain

iˆ  rˆ cos  ˆ sin

(1.18)

ˆj  rˆ sin  ˆ cos

(1.19)

and

The position vector of the point P can be written as

r  rrˆ

(1.20)

and its velocity if therefore

d r  (rrˆ)  rrˆ  rrˆ dt

(1.21)

As P moves in time,  changes and the direction which ˆ points also changes. Therefore, unlike the Cartesian unit vectors which are fixed and whose time derivatives are consequently zero, rˆ and its counterpart ˆ have non-vanishing time derivatives. The time derivative of the radial unit vector is

d rˆ  (iˆ cos  ˆj sin ) dt  iˆ sin  ˆj cos  (iˆ sin  ˆj cos )  ˆ

(1.22)

Hence

r  rrˆ  rˆ

(1.23)

which shows that velocity in plane polar coordinates has two components:

vr  r

(1.24)

and 17

18

v  r

(1.25)

These are respectively called the radial and angular components of the velocity and are mutually orthogonal. The radial component vr is the component along rˆ whereas v is the velocity component along ˆ . The square of the velocity is

v2  v  v  (rrˆ  rˆ)  (rrˆ  rˆ)  r2  r 22

(1.26)

so that the magnitude of the velocity is



v  r 2  r 2 2



1

2

(1.27)

The acceleration of the system is given by





dv dr d   rrˆ  rˆ dt dt dt dr drˆ d dr  ˆ d ˆ dˆ d  rˆ  r    r   r dt d dt dt dt dt dt a

(1.28)



 rrˆ  r ˆ   rˆ  rˆ  r rˆ That is,



 



a  r  r rˆ  r  2r ˆ Thus the two components of the acceleration a

(1.29) are the radial

acceleration ar and the angular acceleration a given by

ar  r  r 2

(1.30)

a  r  2r

(1.31)

Using equation (1.25) we can write the term 2

v2 v  r 2  r      r r 

(1.32)

which is called the centripetal acceleration arising from the motion in the

 direction. Furthermore, if r is held constant in time then, r  r  0 18

CLASSICAL MECHANICS

2 2 and the path is a circle with centripetal acceleration ar   r  v / r .

The term 2 r is the so called Coriolis acceleration.

Example 1.3: What are the polar coordinates of the point (2, 7) ?



Solution: Using equation (14), we find r  (2)2  72



1

2

=7.28. From

 7  0   105.9 . Therefore 2

Eq. (15) we find   arctan

(r ,  )  (7.28, 105.9). Note: When the tangent is negative, the required angle could be in the second or fourth quadrant. The correct quadrant is determined by which component of the position vector is giving rise to the negative sign. If it is the x component as in this case, the angle is in the second quadrant. If it is the y coordinate, the angle is in the fourth quadrant.

Example 1.4:

The position vector of a particle is given by

r  iˆb sin t  ˆjb cost , where b and  are constants. What are its plane polar coordinates?

Determine its acceleration in plane polar

coordinates and describe the motion of the particle. Solution: The polar coordinates are

  b

r  x2  y 2 2



1

2

sin 2 t  b 2 cos2 t



1

2

b and

  t Description of the motion: the particle is moving on the circumference of a circle of radius b centred on the origin. The angular velocity of the particle is constant value  . The acceleration is 19

20



 



a  r  r rˆ  r  2r ˆ  b 2 rˆ  b 2 rˆ where we have used r  b  0 and     0 . The particle is clearly moving under the centripetal acceleration b 2 .

1.3 The Cylindrical Polar Coordinate System This is a three-dimensional coordinate system constructed by adding a z axis at right angles to the plane polar coordinate system that we studied in the last section. The system is shown in Fig. 5.

Fig. 5: Cylindrical coordinates  ,  , z  and the corresponding unit





vectors ˆ , ˆ, zˆ .

The unit vectors are in the directions of the increasing vectors associated with them. It is important to note that zˆ  kˆ is constant, while the unit vectors ˆ and ˆ are functions of  as in the case of plane polar

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CLASSICAL MECHANICS

coordinates. The relation between the coordinates ( x, y, z ) and the cylindrical coordinates are: (see Fig. 5)

x   cos

(1.33)

y   sin 

(1.34)

zz

(1.35)

while the inverse relations are

 x 2  y 2  2 1



   x   cos1    x2  y 2   x2  y2     

 y  x

  tan 1    sin 1 

y

(1.36)

(1.37)

The unit vectors in this system are ˆ , ˆ and kˆ . Here ˆ is the same as

rˆ from the previous section, while ˆ is the same as ˆ . These unit vectors are mutually orthogonal, and so obey

ˆ  ˆ  ˆ  kˆ  ˆ  kˆ  0

(1.38)

Replacing ( r ,  ) by  ,   and with an additional Z component, we may write the new unit vectors in terms of the Cartesian unit vectors as

ˆ  iˆ cos  ˆj sin

(1.39)

ˆ  iˆ sin   ˆj cos 

(1.40)

kˆ  kˆ

(1.41)

or

zˆ  zˆ

The coordinate system is right-handed, so that

ˆ  ˆ  kˆ, ˆ  kˆ  ˆ , kˆ  ˆ  ˆ

(1.42)

From Eqs. (1.39) and (1.40), we have

dˆ dˆ ˆ   ˆ   and d d

(1.43)

21

22

In cylindrical coordinate system, the unit vectors ˆ and ˆ move with the position vector and are therefore implicitly functions of time. However, the unit vector kˆ  zˆ is constant. The position vector of the point P can therefore be represented by

r  ˆ  kˆz

(1.44)

where  is the distance of P from the Z-axis and  gives its angular rotation from the X-axis, while z gives its elevation above the XY plane. Thus we may write the velocity vector, keeping in mind that ˆ  ˆ ( ) , as



d v  r  ˆ  kˆz dt



d dˆ d dz ˆ dkˆ ˆ    kz dt d dt dt dt  ˆ   (ˆ)  zkˆ



where

dkˆ  0. dt

Hence,

v  ˆ  ˆ  zkˆ

(1.45)

Using Eq. (1.23) from the previous section with r replaced by  and  replaced by  , we see that





v2  v  v  ˆ  ˆ  zkˆ  ˆ  ˆ  zkˆ   2   22  z 2

 (1.46)

and



v   2   22  z 2



1

2

(1.47)

Similarly, the acceleration

a 22



dv d  ˆ  ˆ  zkˆ dt dt



CLASSICAL MECHANICS

can be shown using Eq. (43) to be



 



a    2 ˆ    2 ˆ  zkˆ

(1.48)

Example 1.5: A bead slides on a wire bent into the form of a helix. The motion

of

the

bead

is

given

  b,   t , z  ct , where

in

cylindrical

b,  , and c

coordinates are

by

constants.

Determine the velocity and the acceleration of the bead. Solution: In this problem   b  0,   ,   0,

z  c, z  0

Using Eqs. (1.45) and (1.48), we find that

v  bˆ  ckˆ  bˆ  czˆ and

a  b 2 ˆ

1.4 The Spherical Polar Coordinate System We now use our knowledge of the previous sections to build up the spherical polar coordinate system. Things will now be easy because we have developed all the tools we require for this work. Spherical polar coordinates or spherical coordinates are the most commonly used coordinates in situations where spherical symmetry-for example, in the case coulomb forces in atoms and gravitational forces. The coordinate system is illustrated in Fig. 6. It is formed by adding a third axis to the plane polar coordinate system, but with the component of the position vector along this axis measured by means of an angle  which gives the inclination vector to this axis.

r,  ,  .

The coordinates of a point P are

The position vector r has a length r and its z component is

given by

z  r cos

(1.49) 23

24

As shown in Fig.6, r is called the radial distance from the origin O,  is the azimuthal angle locating a plane whose angle of rotation is measured from the X-axis, while  is the polar angle measured down from the Zaxis. The polar angle  can have any value between 0 and  / 2 , while the azimuthal angle  can have any value between 0 and  . These limits restrict the description or motion of point P to the first half of the sphere on the upper plane. It is not uncommon to extend the description to include the whole sphere.

Fig. 6: (a) Spherical polar coordinates r ,  ,   and the corresponding









unit vectors rˆ, ˆ, ˆ . (b) Orientation of unit vectors rˆ, ˆ, ˆ relative to the coordinate system XYZ and polar angle  .

The x and y components are visualized as follows. We imagine a light shining straight down the Z-axis so that it casts a shadow of the position vector in the XY-plane. This shadow is clearly the same as the quantity

 of the cylindrical coordinate system.

In terms of this shadow

therefore, the x and y coordinates are

x   cos

24

(1.50)

CLASSICAL MECHANICS

y   sin 

(1.51)

Clearly from the cylindrical coordinate system or Fig. 6, we note that

  r sin

(1.52)

Substituting this into Eq. (1.50) and (1.51) yield

x  r sin cos

(1.53)

y  r sin sin 

(1.54)

The inverse relations are



r  x2  y 2  z 2



1

2

(1.55)

Now, according to Fig.6,

 2  x2  y 2

(1.56)

Using Eqs. (1.49) and (1.52), we observe that

tan  



(1.57)

z

Therefore,

 x2  y 2     z  

  tan 1 

(1.58)

Using Eqs. (1.53) and (1.54) we obtain

  tan 1

y x

(1.59)

The three mutually perpendicular unit vectors used in spherical coordinates are rˆ, ˆ, and ˆ as shown in Fig. 6 (a) and (b). Also shown are the unit vectors iˆ, ˆj , and zˆ (  kˆ ) , and ˆ .

The unit

vector ˆ lies in the XY plane, while rˆ, ˆ, ˆ , and zˆ all lie in one vertical plane. In terms of the Cartesian unit vectors, the unit vectors of

25

26

the spherical polar coordinate system are as follows. The radial unit vector is

rˆ  sin cosiˆ  sin sinˆj  coskˆ

(1.60)

For the unit vector in the  direction, the z component is easily deduced to be  sin . To obtain the x and y coordinates of this unit vector, we shine a light on it from the top parallel to the Z axis. The shadow of the unit vector ˆ in the XY plane has a length cos so that its x and y components are cos cos and cos sin respectively. As a result

ˆ  cos cosiˆ  cos sinˆj  sinkˆ

(1.61)

From the previous two sections and Fig.6, we see that

ˆ   siniˆ  cosˆj

(1.62)

The unit vectors are mutually orthogonal and therefore obey the following rules

rˆ  ˆ  rˆ  ˆ  ˆ  ˆ  0

(1.63)

and as the coordinate system is right-handed,

ˆ  rˆ  ˆ, rˆ  ˆ  ˆ,

ˆ  ˆ  rˆ

(1.64)

Differentiating Eqs. (1.60)-(1.62), we obtain the following relations:

rˆ ˆ   ˆ   rˆ  ˆ 0 

rˆ ˆ   sin  ˆ ˆ   cos  ˆ   ˆ   rˆ sin  ˆ cos 

(1.65)

These relations can also be derived from geometrical considerations by drawing figures similar to the ones in the case of plane polar coordinates. In spherical coordinates, the position vector of a point P in space is given by the position vector r :

r  rrˆ  rrˆ( ,  ) 26

(1.66)

CLASSICAL MECHANICS

We can now find expressions for velocity and acceleration by making use of the preceding relations. Thus

d d dr drˆ drˆ v  r  (rrˆ)  rrˆ( ,  )  rˆ  r  rrˆ  r dt dt dt dt dt

(1.67)

Using Eq. (1.65),

drˆ drˆ d drˆ d ˆ  ˆ ( ,  )       sin dt d dt d dt

(1.68)

Hence we obtain

v  rrˆ  rˆ  (r sin )ˆ

(1.69)

Similarly,



dv d a  v  r   rrˆ  rˆ  (r sin )ˆ dt dt



(1.70)

which on simplification yields

a  (r  r 2  r sin 2 2 )rˆ  (r  2r  r sin cos2 )ˆ

(1.71)

 (r sin  2r sin  2r cos )ˆ Example 1.6. The position vector of a particle is

r  brˆ,   B sin t , where b,

B,

  ct

 , and c are constants. Determine the velocity and

acceleration of the particle. Give a complete description of the trajectory of the particle. Solution: Since in general r  rrˆ , it follows that r  b . Therefore

r  b  0,   B cost,    2 B sint,   c,   0 Inserting these values into Eqs. (1.69) and (1.71), we obtain

v  bB costˆ  bc sin(B sint )ˆ and 27

28

a  (b 2 B 2 cos2 t  bc2 sin2 ( B sin t ))rˆ  (b 2 B sin t  bc2 sin( B sin t ) cos(B sin t ))ˆ  2bcB cost cos(B cost ) ˆ Description of trajectory: The particle moves on the surface of the sphere of radius b thus keeping a fixed distance from the origin. From

  ct , we see that its motion in the XY plane is a rotation about the Zaxis with constant angular velocity   c . At the same time the polar angle changes as if the particle is performing simple harmonic motion in the plane instantaneously defined the position vector and the Z-axis. Viewed from the top, the particle seems as if it is a pendulum of amplitude B and angular frequency  whose plane of swing is rotating with angular velocity   c .

1.5 Summary In this unit we have familiarized ourselves with some of the most common coordinate systems used to solve problems in Physics. There are a number of other coordinate systems, but they are generally used at higher level. A good mastery of the particular coordinate systems treated in this unit will be sufficient for any problems you as the learner will encounter in this course.

1.6 Exercises 1.0

A honey bee homes in its hive in a spiral path in such a way that

the radial distance decreases at a constant rate, so that r  b  ct , while the angular speed increases at the constant rate   kt , where b , c and

k are constants. Find the speed of the bee as a function of time.

28

CLASSICAL MECHANICS

1.1

On a horizontal turntable rotating with constant angular velocity,

an insect crawls outward on a radial line in such a way that its radial distance increases quadratically with time, so that

r  bt 2 and   t , where b and  are constants. Determine the acceleration of the insect. 1.2

A racing car moves in a circle of radius b . If the speed of the

car varies with time t according to the equation v  ct , where c is a positive constant, show that the angle between the velocity vector and the acceleration vector is 450 at time t  b / c . 1.3

The position vector of a certain particle as a function of time is

r (t )  iˆ(1  ekt )  ˆjekt where k is a positive constant. Find the velocity and the acceleration of the particle and sketch its trajectory. 1.4

An ant crawls on the surface of a ball of radius b in such a way

that the position of the ant is given by the spherical polar coordinate

r  b,

  t ,





1  1  cos 4t  2 4 

Determine the speed of the ant as a function of time and describe the sort of path represented by this equation. 1.5

The

position

vector

of

a

particle

is

given

by

r  iˆa sin t  ˆjb cost , where a , b and  are constants.

By

eliminating t , show that the path of the particle is an ellipse. Obtain the speed of the particle. 1.6

Find the force acting on a particle of mass 3 kg at t  3 s if it

has a velocity in m/s given by v  iˆ2t 3  ˆj (1  t 2 ) .

29

30

Chapter 2 Unit 2 Particle Dynamics in One Dimension

You should be able to cover this unit in 20 hrs. This time period does not include the time for solving problems in the exercises. You may solve the exercises at your own convenient pace. OBJECTIVES (1) To introduce the learner to the fundamentals of dynamics and kinematics (2) To teach the learner how to solve one-dimensional problems involving position and velocity dependent forces. (3) To teach the learner how to solve the dynamical problem in two and three dimensions. (4) To acquaint the learner with the properties of motion under a conservative force. (5) To introduce the learner to the properties of central force motion.

After studying this unit, the learner should be able to: (1) Differentiate between a kinematic and dynamic problem. (2) Solve one-dimensional problems involving position dependent forces 30

CLASSICAL MECHANICS

(3) Solve dynamical problems in two and three dimensions. (4) Identify central force problems. (5) Differential between conservative and non-conservative forces. (6) Solve problems involving conservative forces (7) Identify and enumerate the properties of central force motion.

2.0

Introduction The fundamental problem of mechanics is to determine how a particle moves under the action of a particular force. For example, the motion of a particle of mass m along a straight line which we may consider to be the x  axis under the direction of a force directed along the x  axis , constitutes a one dimensional problem. Newton’s second law is the fundamental equation for treating problems in dynamics. According to Newton’s second law, if a force F acts on a particle of mass m , the effect is to impart to the particle an acceleration a given by

a

F m

(2.1)

Newton’s second law (Eq. 2.1), leads in two or three dimensions to the vector equation. In Cartesian coordinates, two dimension equations is equivalent to two component equations and in three dimensions to three component equations. These are

m

d 2x  Fx , dt 2

m

d2y  Fy , dt 2

m

d 2z  Fz dt 2

(2.2)

These are differential equations of second order in time, and the dynamical problems reduce to solving these equations. It is usual in dynamics to indicate the time derivative by placing a dot over the quantity being differentiated and so we can write the vector equation as 31

32

m

d 2r  F  mr dt 2

(2.3)

When the force F is specified, this is called the equation of motion. It is clear that if the force is zero, the equation of motion is

mr  0

(2.4)

In other words

m

dv 0 dt

(2.5)

Integrating Eq. (2.5) once with respect to time t , we deduce that

mv  C

(2.6)

where C is a constant. Since mv is the momentum, the constant has the dimensions of momentum and we denote it by p0 . This allows us to write

mv  p0

(2.7)

Eq. (2.7) tells us that the linear momentum p  mv is constant when the force is zero.

The equation is therefore a statement of the law of

conservation of linear momentum and states that: The linear momentum of a particle which is isolated so that no force acts on it is conserved. Let us divide Eq. (2.6) by m to obtain

v

dr  p0 / m  v dt

(2.8)

From Eq. (2.8) we can write

dr  v0 dt

(2.9)

Eq. (2.9) can be integrated one more to obtain

r  v0t  D where D

(2.10)

is a constant of integration with the dimensions of

displacement. Since r  D when t  0 , this constant is just the initial 32

CLASSICAL MECHANICS

displacement of the particle. It may be zero or not depending on the reference point. Example 2.1: The velocity of a particle moving with constant velocity

v  (3, 1,  7)

has a value

when the particle is at position

r  (1,  5,  2) . What is the distance the particle travels in 5 s ? Solution: We take the instance the particle is at r  (1,  5,  2) as

t  0 . This implies that D  (1,  5,  2) . At this point, v  v0  (3, 1,  7) . Substituting t  5 s and D  (1,  5,  2) into Eq. (2.10), we obtain the position of the particle 5 s later as

r  (16, 0,  37) . The distance travelled is

s

r  D   r  D  

(16  1) 2  (0  5) 2  (37  2) 2

 152  52  372  225  25  1,369  1,619  40.24

2.1 Motion of a Particle in One Dimension 2.1.1 The Free Particle We start our study of dynamics by considering the case of a particle moving in one dimension (one direction only). We assume that the motion is along the x axis, so that the equation of motion is

mx  F ( x, t )

(2.11)

If there is no force acting on the particle, the equation of motion of the particle becomes

mx  0

(2.12)

Dividing Eq. (2.12) by m results in the differential equation

x  0

(2.13)

The solution of this equation is the x component of Eq. (2.10): 33

34

x  v0t  x0

(2.14)

where v0 is the velocity at time t  0 with corresponding position x0 .

2.1.2 Particle Moving Under the Effect of a Constant Force Next we consider the case of a constant force k acting on a particle. We assume that the particle is moving in one dimension taken as the

x  direction. The equation of motion is then

mx  k

(2.15)

dv dt

(2.16)

Using

x 

and following the procedure of Eq. (2.9), we are able to write this as

k dt m

(2.17)

v  at  A

(2.18)

dv  whose solution is

where we have used the fact that k / m  a , the acceleration. (Remember that we assumed a constant for F  k ) The result then follows from Newton’s second law (Eq. 2.1). If the velocity is v0 at t  0 , we find that

A  v0

(2.19)

and Eq. (2.18) becomes

v  at  v0

(2.20)

dx dt

(2.21)

Using

v

34

CLASSICAL MECHANICS

we can recast Eq. (2.20) into the form

dx  at  v0 dt

(2.22)

Integrating this gives us the solution which is

x

1 2 at  v0t  B 2

(2.23)

with B as a constant of integration which has the dimensions of displacement. If we suppose that the initial conditions of the particle is

x  x0 when t  0

(2.24)

This fixes B , and gives

B  x0

(2.25)

The position as a function of time is then

x

1 2 at  v0t  x0 2

Using the initial conditions t  0,

x

(2.26)

x0  0 , we obtain

1 2 at  v0t 2

(2.27)

which we recognize as one of the five equations of rectilinear motion. When a  g as in the case of a body moving under gravity, then

y

1 2 gt  v0t 2

(2.28)

In general, Eqs. (2.26) and (2.27) are sometimes written as

1 s  ut  at 2 2

(2.29)

where u  v0 is the initial velocity.

35

36

Example 2.2:

A particle of mass 2 kg moves under the force

F  25 N . When passing the origin, the particle has a velocity of  5 m/s . Where was the particle 10 s earlier? Solution: The instance the particle passes the origin is at t  0 s . We require the position at t  10 s .

From the given information, the

acceleration is a  15 / 2  7.5 m/s2 . Thus using Eq.(2.27), we find that the required position x  325 m.

2.1.3 The Simple Harmonic Oscillator Simple harmonic motion is the case involving a particle acted upon by a restoring force that is proportional to its displacement from the equilibrium position. Denoting the displacement by x , the resulting equation of motion is

mx  kx

(2.30)

if the oscillating particle has mass m and k is the spring constant. By making the substitution

2 

k m

(2.31)

where  is called the angular frequency, the equation of motion becomes

x   2 x

(2.32)

By means of an example, we shall show that the solution to this equation is

x  x0 sint   

(2.33)

where  is called the phase angle. We shall analyze this further when we carry out detailed study on simple harmonic oscillators Eq. (2.33) can be expressed as 36

CLASSICAL MECHANICS

x  x0 sin t cos  x0 cost sin 

(2.34)

Example 2.3: A harmonic oscillator is at the positive extreme of its motion at time t  0 . Determine the expression, x (t ) , that describes the motion of the particle at any subsequent later time t . Solution: Let the extreme be x  A , where A is the amplitude. The initial conditions are therefore x  A and x  0 when t  0 . Now

x  x0 cost   

(2.35)

By expanding Eq. (2.35), we have

x  x0 cost cos  x0 sin t sin 

(2.36)

Substituting the initial conditions in Eqs. (2.35) and (2.36), we obtain the following system of equations

x0 sin   A

(2.37)

x0 cos  0

(2.38)

If we divide Eq. (2.38) by  the system of equations become

x0 sin   A

(2.39)

x0 cos  0

(2.40)

Squaring the new system of equations and adding them gives





x02 sin2   x02 cos2   x02 sin2   cos2   A2

(2.41)

so that

x02  A2  x0  A

(2.42)

In order to deduce the phase angle  , we proceed as follows: Divide Eq. (2.39) by Eq. (2.40) to get

tan  

A  0

(2.43) 37

38

The smallest angle for which this is true is    / 2, which means that

  x  A sin t    A cost 2 

(2.44)

2.2 General Treatment of Motion in One Dimension

When the force is a function only, there is a systematic way of treating motion in one-dimension. Such a force which does not depend on time is called static or conservative force. This means that Eq. (2.11) is modified to read

mx  F (x)

(2.45)

In order to solve this equation we apply some mathematical procedures. First we multiply both sides of this equation by x to obtain

mxx  F ( x) x

(2.46)

This can be written as

mx

dx dx  F ( x) dt dt

(2.47)

or

mxdx  F ( x)dx

(2.48)

Integrating once we obtain

1 2 mx   F ( x)dx  E 2

(2.49)

where E is a constant of integration. Since the left-hand side is the kinetic energy, the term on the right-hand side must also be an energy, as is the constant of integration. Rewriting this as

1 2 mx   F ( x)dx  E 2 38

(2.50)

CLASSICAL MECHANICS

The second term on the left-hand side is called the potential (energy and is denoted by V (x) and defined as

V ( x)   F ( x)dx

(2.51)

and is interpreted as the work done by the force F (x ) in moving the particle through a distance x. Eq. (2.50) the reads

1 2 mx  V ( x)  E 2

(2.52)

The equation is telling us that the sum of the two energies on the left hand side is always constant. It is a statement of the conservation of energy. We can solve for the velocity from Eq. (2.52) and obtain

x  

2 E  V ( x )  m

(2.53)

By taking the positive root,

dx 2 E  V ( x )   dt m

(2.54)

or

dt 

dx

(2.55)

2 E  V ( x )  m

By integrating this equation we find that

t

dx 2 E  V ( x )  m



(2.56)

where  is a constant of integration which has the dimensions of time. Once we find t  t (x) , we can invert the expression to find x  x(t ) . We now give some examples to illustrate these concepts.

39

40

Example 2.4: Starting from the fact that a particle moving under a constant force F ( x)  k obey the stated principles above, show that if it starts from the origin with velocity v0 at time t  0 , then its motion can be described using Eq. (2.27). Solution: For this case F ( x)  k and the potential energy is

V ( x)   kdx  kx

(2.57)

In consequence, Eq. (2.56) can be written as

dx

t

2 E  kx m



(2.58)

Since

   bx

1

2

dx 

1 2   bx 2 b

(2.59)

it follows that

m 2  t   ( E  kx)  k m 

1

2



(2.60)

 t 

(2.61)

so that

m 2   ( E  kx)  k m 

1

Using the initial conditions x  0,

2

v  v0 at t  0 , then from Eq. (2.5),

we see that

E

1 2 mv0 2

(2.62)

Employing this in Eq. (2.61) and remembering that this condition arises when t  0 , we obtain

  Hence 40

mv0 k

(2.63)

CLASSICAL MECHANICS

m 2   ( E  kx)  k m 

1

2

t

mv0 k

(2.64)

By squaring both sides we obtain

2mE 2mx 2 2mv0t m 2v 2  t   2 k2 k k k

(2.65)

We can rewrite this as

2mx 2 2mv0t m 2v 2 2mE t   2  2 k k k k

(2.66)

Dividing each term in Eq. (2.66) by 2m and multiplying by k , we obtain

x

k 2 mv 2 E t  v0t   2m 2k k

Using Eq. (2.62) while solving for x , we find

x

k 2 mv 2 mv02 t  v0t   2m 2k 2k

Now bearing in mind that at t  0,

(2.67)

x  0 , we see that Eq. (2.67) reveal

that

mv 2 mv02  2k 2k

(2.68)

Hence

x

1 2 at  v0t 2

(2.69)

where we have used k / m  a . This is the same as Eq.(2.27) is the required solution for the problem.

41

42

Example 2.5: Using the same approach as in example 2.4 above. (a) show that for a simple harmonic oscillator in which the force is

F ( x)   kx , the position at any time is given by x 

2E sin  (t  t0 ) . k

If the amplitude is b , (b) show that the total energy is given by

E

1 2 kb . 2

Solution: (a) We similarly use Eq. (2.58) to find x (t ) . The potential energy is

V ( x)     kxdx 

1 2 kx 2

(2.70)

so that

dx

t

2 1 2  E  kx  m 2 

 t0

(2.71)

Therefore

m t  t0    k

1

2

 b

dx 2

 x2



(2.72)

where

2E k

b

(2.73)

This integral can be solved by the substitution

x  b sin

(2.74)

and b 2  x 2  b 2 (1  sin2  )  b 2 cos2 

(2.75)

With

dx  b cos we have

m t  t0    k 42

1

2

1

b cosd  m  2  b cos   k  

(2.78)

CLASSICAL MECHANICS

Solving for  , we obtain 1

k 2     (t  t0 ) m

(2.79)

 k  12  x  b sin   (t  t0 )  m  

(2.80)

and

Recalling that

k   m

1

2



(2.81)

and using Eq. (2.73), we can write Eq. (2.80) as

x (b)

2E sin (t  t0 ) k

(2.82)

We know that the total energy must be the sum of the kinetic

energy and the potential energy, that is

1 Etotal  KE  PE  KE  kx2 2

(2.83)

We need to work out the kinetic energy and also obtain the maximum potential energy.

KE 

1 2 mx 2

(2.84)

Using Eq. (2.82), we have

x  b cos (t  t0 )

(2.85)

For simplicity let (t  t0 )   , then Eq. (2.83) becomes

Etotal 

1 2 2 1 mb  cos2   kb2 sin 2  2 2

(2.86)

43

44

We now use Eq. (2.81) to eliminate m and  2 from the first term of this expression. The result is

Etotal 

1 2 1 kb cos2   kb2 sin 2  2 2

(2.87)

Factoring out the common terms in Eq. (2.87) and using the trigonometric identity cos2   sin2   1 , where    in this case, we finally get

Etotal 

1 2 kb 2

(2.88)

2.2.1 Time and Velocity Dependent Forces Two categories of forces that we have not discussed in our treatment of one dimensional dynamics are the time and velocity dependent forces. There are the purely time-dependent and purely velocity dependent forces. Sometimes we say that they are explicitly dependent. We shall only discuss these forces very briefly because for the time-dependent forces, the solutions can be obtained in two successive integrations with respect to time. However, the velocity-dependent forces constitute a very large topic which we have no time for in such a compressed module.

2.2.2 Time-Dependent Forces If the force is explicitly dependent on time t , the differential equation of motion for constant mass is

F (t )  m

dv dt

(2.89)

which can directly integrated to give t

 F (t )dt  mv(t )  mv

0

0

44

(2.90)

CLASSICAL MECHANICS

The integral on the left-hand side of Eq. (2.90) is called the IMPULSE and it is equal to the change of momentum imparted to a body by a force

F (t ) acting over a certain interval of time. The position of the body as a function of time is obtained by a second integration, that is, we can write from Eq.(2.90) t

dx F (t )  v(t )  v0   dt dt m 0

(2.91)

Integrating both sides of this equation, we get t t t  F (t )  x  x0   v(t )dt  v0t     dt dt m 0 0 0 

(2.92)

Provided the form of F (t ) and the initial conditions are given, the position as a function of time can easily be obtained. We shall consider as an example the case where the force is constant. For this case we have t

F Ft v(t )  v0   dt  v0  m0 m

(2.93)

and t

x(t )  x0  v0t 

F dt m 0

(2.94)

x(t )  x0  v0t 

Ft 2 2m

(2.95)

or

which is equivalent to the equation of uniformly accelerated motion with

a

F , giving m

x(t )  x0  v0t 

at 2 2

(2.96)

45

46

For x0  0

at 2 x(t )  v0t  2

(2.97)

The other time-dependent forces that we are not going to discuss here are the step forces. However, just to understand what these step forces are, we provide a brief discussion. Suppose that a body of mass m is subjected to a constant force F1 acting for a time interval t1 and the force suddenly changes to a different constant value of say F2 , then F2 is a step force. Another type of force that is time-dependent is called “ the jerk”. It is a uniformly increasing force.. Such a force can be represented by F (t )  ct , where c is a constant. It is obvious that this force increases with time.

2.2.3 Velocity-Dependent Forces Many situations in everyday occurrences exist in addition to constant applied forces where the forces are functions of velocity. For example, when a body is in a gravitational field, in addition to the gravitational force, there exists a force of air resistance on the falling or rising body. This resisting force is some complicated function of velocity. The same is true for objects moving through fluids (gases and liquids).

Such

opposing forces to the motion of objects through fluids are called VISCOUS FORCES or VISCOUS RESISTANCE. In cases where they exist, Newton’s second law may be applied in the following form:

F (v )  m

dv dt

(2.98)

F (v )  m

dv dx dv  mv dx dt dx

(2.99)

or

46

CLASSICAL MECHANICS

Knowing the form of the force F (v ) , either of the two equations Eq. (2.98) or (2.99) may be solved to analyze the motion, that is, to calculate

x as a function of time t .

2.3 Summary In this unit we have learned how to treat dynamical problems in one dimension. The methods that have been presented are instructive and therefore need to be generalized to the case of two and three dimensions. This is the task in the next unit. From a formal point of view, the equations that have been given can be used to obtain exact solutions of the problem; however, there are many practical difficulties in implementing them. For example, the integrals that need to be performed may be intractable, but there are so-called numerical methods for treating such cases. As far as we are concerned in this unit we have been able to solve the one dimensional problem under position dependent-force completely.

2.4 Exercises 2.0

A pendulum of length l is pulled to the side and by an angle  0

to the vertical and allowed to oscillate. Ignoring friction, (i)

Obtain the equation of motion of the pendulum.

(ii)

Show that if the angle of swing is small, the pendulum performs

simple harmonic motion. (iii)

Obtain the equation if the displacement is maximum at t  0 .

2.1

The position of a particle moving along the x axis depends on

the time t according to the equation

47

48

x

v0 (1  e  kt ) k

where v0 and k are constants. (i) Obtain the velocity and the force acting on the particle (ii) Calculate the maximum distance covered by the particle. 2.3

Calculate the potentials of the following forces

(i)

F ( x)  ax  b (ii)

(iv)

F ( x)  F0 sin kx

2.4

The force acting on a particle of mass 1.5 kg is

F ( x)  F0e x / a (iii)

F ( x)  x / a

F ( x)  7t 5  3t 2  2 in Newtons. Calculate (i) The momentum of the particle. (ii) The acceleration of the particle. (iii) The position of the particle as a function of time. 2.5

A particle moves in a straight line under the action of a force

whose potential energy is given by

V ( x)  ax2 (b  x) (i) Find the force acting on the particle. (ii) Obtain the position of the particle as a function of time. 2.6

A body falls from a height of 1,000 km towards the surface of

the earth. Neglecting air resistance, write down the equation of motion of the body and hence find the velocity with which it hits the earth. 2.7

A particle of mass m is released from rest a distance b from a

fixed origin of force that attracts the particle according to the inversesquare law

F ( x)  kx2 48

CLASSICAL MECHANICS

Show that the time required to reach the origin is

 mb3   T     8k  2.8

1

2

Find the velocity as a function of displacement x for a particle of

mass m which starts from rest at x  0 , subject to the following forces:  cx (i) F ( x)  F0  cx (ii) F ( x )  F0 e

2.9

(iii) F ( x)  F0 coscx

A particle of mass m moves along a frictionless horizontal plane

with speed v ( x )   / x , where x is its distance from the origin and  is a positive constant. Find the force F (x ) which acts on the particle. 2.10

A particle moving in a straight line along the positive x axis is

subject to a force F ( x)  b / x 3 , where b is a positive constant. (i)

Determine the dimensions of b .

(ii)

Find an expression for the total energy of the particle.

(iii)

Find the position of the particle as a function of time.

49

50

Chapter 3 Unit 3 Particle Dynamics in Two and Three Dimensions

You should be able to cover this unit in 20 hrs. This time period does not include the time for solving problems in the exercises. You may solve the exercises at your own convenient pace. OBJECTIVES (1) To use the knowledge acquired in unit 2 to tackle problems in two and three dimensions (2) To solve specific problems concerning central force.

After studying this unit, the learner should be able to: (1) Solve two and three dimensional problems involving position dependent forces. (2) Identify the salient properties of central forces in different systems. (3) Solve problems involving central force motion. (4) Solve specific standard conservative force problems in two and three dimensions.

50

CLASSICAL MECHANICS

(5) Identify and enumerate the properties of central force motion. (6) Define work. (7) Solve problems involving conservation of energy. (8) Use the Del. Operator to solve conservative force problems. (9) Use the Del. Operator to generate the components of angular momentum in three dimensions.

3.0

Introduction In the previous unit, we studied the dynamics of a particle in one dimension. Many important dynamical systems fall in this category. But nature is three-dimensional and we naturally have to extend our discussion to the case of particles moving in three dimensions. There are also important situations where two-dimensional motion exists. In this unit we generalize the theory we developed earlier to the case of two and three-dimensional motion.

3.1

General Theory In three-dimensional motion, the force is in general given by F ( r , t ) . The equation of motion is then

mr  F (r , t )

(3.1)

We shall confine ourselves to static or time-independent forces. Thus

F  F (r ) , this makes the treatment of the problem somewhat simpler. The vector equation (3.1) can be written in Cartesian-component form as

mx  Fx my  Fy

(3.2)

mz  Fz In some cases the form of the force may warrant that the motion be treated using a different coordinate system. If spherical polar coordinates are used, then the component form of the equation of motion is 51

52

m(r  r 2  r sin 2 2 )  Fr m(r  2r  r sin cos2 )  F



(3.3)

m(r sin  2r sin  2r cos )  F If the force is directed along the position vector, it is called a central force. For this type of force, the equation of motion takes a simple form because the force does not depend on the radial coordinate  and  and so can be expressed as

F  f (r )rˆ

(3.4)

In that case, although the problem is normally three-dimensional, the motion takes place on a plane (may be xy,

xz, or yz ). This will

depend on the plane of choice. The components of the equation of motion are

m(r  r 2 )  f (r ) m(r  2r)  0

(3.5)

For the general case however, the procedure to be followed in tackling Eq. (3.1) is analogous to the one that was adopted for the onedimensional motion case. We take the dot product of Eq. (3.1) with r to obtain

mr  r  F  v

(3.6)

This can be written as

mv  v  F  v

(3.7)

v2  v  v

(3.8)

We know that

and

d (v 2 ) d dv dv dv  (v  v )  v    v  2v  dt dt dt dt dt

(3.9)

It follows that by dividing the last expression and first expression by 2 we obtain 52

CLASSICAL MECHANICS

1 d (v 2 ) dv v 2 dt dt

(3.10)

We can then substitute these expressions in Eq. (3.7) to obtain

1 d (v 2 ) dr m F 2 dt dt

(3.11)

1 m d ( v 2 )  F  dr 2

(3.12)

which yields

When integrated, this gives us

1 m v 2   F  dr  E 2

(3.13)

where E is a constant of integration and is identified as the total energy. This equation is evidently the counterpart of Eq. (2.50) and is a statement of the conservation of energy and takes the form

K  V (r )  E

(3.14)

for the three-dimensional motion. Here K represents the kinetic energy while the potential energy

V (r )    F  r

(3.15)

Many important systems can be treated by means of these results.

3.2

Particle in the Field of Gravity Problems of particle motion in the field of gravity are quite common as occurring near the surface of the earth so that the force acting on the particle is taken to be constant. We shall now take into account the variation of this force with distance from the centre of the earth. According to Newton’s law of universal gravitation, the force with which the earth acts on a particle of mass m in its vicinity is given by 53

54

Fr  

GMm r2

(3.16)

where M is the mass of the earth, G is the universal gravitational constant and r is the distance of the particle measured from the centre of the earth to the centre of the particle. We have taken in this case the earth to be a uniform sphere so that its gravitational attraction is equivalent to that of a point mass M located at its centre. Since we are dealing with a spherical system (3-Dimensions), the equation of motion has three components. For the present problem, we shall ignore the components corresponding to the directions ˆ and ˆ . We shall focus on the radial component and write

mr  

GMm r2

(3.17)

If the particle is at a distance x above the earth’s surface then the distance of separation is

r  Rx

(3.18)

It is clear from the above equation that r  x , since the radius of the earth R is constant. Eq. (3.17) may then be written as

mx  

GMm  Fx ( R  x) 2

(3.19)

The potential energy is therefore

GMm dx ( R  x) 2 GMm  Rx

V ( x)    

(3.20)

And the energy conservation equation is

E

1 2 GMm mv  2 Rx

(3.21)

Now, if the body is projected upwards with a velocity v0 at the surface of the earth, it follows from Eq. (3.21) that 54

CLASSICAL MECHANICS

1 2 GMm 1 2 GMm mv0   mv  2 R 2 Rx

(3.22)

From this we have

v02  v 2 

2GM 2GM  Rx R

(3.23)

The maximum height H reached by the body will correspond to v  0 and is therefore such that

v02 

2GM 2GM  RH R

(3.24)

If we want to project a particle fast enough to make sure that it escapes the gravitational pull of the earth, we must ensure that H   . This gives the speed of projection as

2GM R

(3.25)

GMm  mg R2

(3.26)

v0  From

the equation which holds at the surface of the earth, the acceleration due to gravity g can be written as

g

GM R2

(3.27)

so that Eq. (3.25) can be expressed as

v0  2 gR

(3.28)

This quantity v0 is called the escape velocity and is the minimum velocity which in the absence of air resistance or air friction, we must project a particle from the surface of the earth to ensure that it escapes the earth’s

gravity.

Since

g  9.8 m/s2

and

R  6.4  106 m , 55

56

v0  11 km/s . Similar calculations can be carried out for other planets when their gravitational constants are known.

3.3

Force and Work We can now carry out a more detailed study on the potential energy Eq. (3.15). We start by considering a force F which acts on a particle and moves it through the displacement s .

The work done cannot be

calculated by taking the scalar product of the force and the displacement because the force is position dependent and therefore not constant. Additionally, to reach the end of the displacement there are many paths that can be followed. However, if the displacement dr is small enough, the force is essentially constant over the infinitesimal distance dr and the work done on the particle is

dW  F  dr

(3.29)

The work when the particle is moved from ri to r f is rf

W   F  dr

(3.30)

ri

The work done in moving the particle from r to r f is by definition the potential energy or potential or potential energy function. This is given by rf

r

V (r )   F  dr    F  dr    F  dr  C r

(3.31)

rf

where C is a constant of integration.

3.4

Force and Work Statement of the work principle: The work done on a particle by a force increases the kinetic energy of the particle by the same amount. To demonstrate this, we start by looking again at Eq. (3.11) 56

CLASSICAL MECHANICS

1 d (v 2 ) dr m F 2 dt dt

(3.32)

This can be rewritten as

d 1 2 dr  mv   F  dt  2 dt 

(3.33)

By defining

K

1 2 mv 2

(3.34)

this shows that

dK  F  dr

(3.35)

 dK   F  dr

(3.36)

Therefore

Since the right-hand side of this equation is the work done on the particle by the force, we see that this equation states that the work done on a particle by a force increases the kinetic energy of the particle by the same amount.

Example 3.1: A body of mass 3 kg and velocity 2.3 m/s is acted upon by the force F ( x)  5 x  2 x 2 from point x  0 to x  2 . What is the final speed of the body?

Solution: The change in the kinetic energy of the body is 2

K   0

2

 5 x 2 2 x3  5 x  2 x dx     4.667J 3  0  2



2



(3.37)

57

58

The original kinetic energy was K i 

1 2 mv i  7.935J . This shows that 2

if we add the difference to this value we should obtain the final energy. This works out to be K f  12.602 J . Using Eq. (3.34) the corresponding final velocity is

vf 

3.5

2K f m

 2.90 m/s

(3.38)

Force and Work (The Del Operator) We have already studied the different types of forces and seen how they arise in different situations.

We shall go into more details on

conservative forces and develop a more general approach in dealing with them. The integral

 F  dr

is called a line or path integral because it is

evaluated along the path of the particle. The exact value of the integral depends on the path taken by the particle between the initial point ri and the final point r f . However, for certain forces, this integral is path independent. This is true for all position-dependent forces F (r ) . For such forces, it must be that the integral is given by the difference in the values of some function at the end points. The function is denoted by

U (r ) , so that rf

 F  dr  U (r )  U (r ) f

(3.39)

i

ri

This equation means that the integrand is an exact differential meaning that

 F  dr   dU  U (r )

(3.40)

This of course can be written in Cartesian components as

 F  dr   F dx  F dy  F dz x

58

y

z

(3.41)

CLASSICAL MECHANICS

In order to proceed from here, we digress a bit and give a brief mathematical discussion on functions and their dependence on independent variables. If an arbitrary function G depends on the independent variables

x, y and z , then the change in G when the variables change from x to x  dx, y to y  dy and z to z  dz is

dG 

G G G dx  dy  dz x y z

(3.42)

and dG is an exact differential such that

 dG  G

(3.43)

This shows us that if F  dr is an exact differential, there is a scalar function such that the components of F are just the partial derivatives of this function with respect to the coordinates. If we let this function be

U (r ) , then we have

Fx 

U U U , Fy  , Fz  x y z

(3.44)

so that

F  dr 

U U U dx  dy  dz x y z

(3.45)

This ensures that F  dr is an exact differential and the integral of this quantity between two points is given by the difference in the values of U at those points. It is customary to introduce a negative sign so that the scalar function is

V (r )  U (r )

(3.46)

dV (r )  dU (r )   F  dr

(3.47)

Then

and 59

60

V (r )   F  dr

(3.48)

V (r ) is called the potential energy. By employing Eq. (3.45), F  dr  dV (r ) can be expressed in rectangular coordinates as

Fx dx  Fy dy  Fz dz  

V V V dx  dy  dz x y z

(3.49)

This implies that

Fx  

V , x

Fy  

V , and y

Fz  

V z

(3.50)

We can then express the force F as

F 

V ˆ V ˆ V ˆ i j k x y z

(3.51)

By defining



 ˆ  ˆ  ˆ i j k x y z

(3.52)

we see that Eq. (3.51) can be written in a compact form as

F  V  

V ˆ V ˆ V ˆ i j k x y z

(3.53)

This can be compared with the one dimensional form

Fx  

dV ( x) dx

(3.54)

with similar expressions for the y and z components. The differential operator defined in Eq. (3.51) is a vector operator called the del operator. The expression  V is called the gradient of V and is sometimes written

grad V . The negative sign in Eq. (3.53) implies that the particle is urged to move in the direction of increasing potential energy rather than in the opposite direction 60

CLASSICAL MECHANICS

Suppose the potential energy function exists so that Eq. (3.50) holds, then we can take the derivative of Fx with respect to y and the derivative of

Fy with respect to x to obtain

Fy

Fx  2V  , y yx

x



 2V xy

(3.55)

Since the order of differentiation can be reversed on the right-hand side of each expression in Eq. (3.55), we see that the expressions are equal and we can write

Fx Fy  y x

(3.56)

Performing similar operations on Fz we get

Eqs.

(3.56)

Fx , Fy

Fx Fz  z z

and

and

are

(3.57)

Fy z the



Fy

(3.57)

y

necessary

conditions

on

and Fz for the potential energy function to exist and they

express that condition that

F  dr  Fx dx  Fy dy  Fz dz

(3.58)

We can now introduce the cross product of the del operator on F given as

 F F   F F   F F    F   z  y iˆ   z  x  ˆj   y  x kˆ z   x z   x y   y

(3.59)

This is sometimes written with a change of sign in the second term and the terms in the bracket switched giving

 F F   F F   F F    F   z  y iˆ   x  z  ˆj   y  x kˆ z   z x   x y   y

(3.60)

Eq. (3.59) or (3.60) is what we call the CURL of F . If we substitute Eq. (3.56) and (3.57) into Eq. (3.60) we see that 61

62

F 0

(3.61)

since each term vanishes separately. This is the condition for a force to be conservative. We see from Eq. (3.60) that we can write

iˆ  F  x Fx

3.6

ˆj  y Fy

kˆ  z Fz

(3.62)

Angular Momentum There are other dynamical quantities of interest that arise for a particle moving in three dimensions.

One of these quantities is the angular

momentum L . The quantity is conserved whenever the moment of the force about the origin is zero. To show this we consider the equation of motion given by Eq. (2.3)

mr  F

(3.63)

Taking the cross product of this equation with the position vector gives

r  mr  r  F

(3.64)

The right-hand side is the moment of the force F about the origin. We want to analyze the left-hand side. We can rewrite Eq. (3.64) as

r F  r m

dv dp r dt dt

(3.64)

The right-hand side can be simplified further by noting that

d dp dr dp (r  p )  r   pr v p dt dt dt dt

(3.65)

and since v  p  v  mv  m(v  v )  0 because (v  v )  0 , Eq (3.64) can be written as 62

CLASSICAL MECHANICS

rF 

d (r  p ) dt

(3.66)

If we define L  (r  p )  r  mr , then Eq. (3.66) becomes

dL d dr  dr   (r  mr )     mr  r  m dt dt dt  dt 

(3.67)

This is similar to Eq. (3.65) and the first term vanishes and we conclude that

dL d dr  (r  mr )  r  m  r  F  M dt dt dt

(3.68)

Thus the moment of the force F , about the origin is equal to the time derivative of the angular momentum. If the moment is zero, then

dL 0 dt

(3.69)

L  constant

(3.70)

This implies that

We thus arrive at the law of conservation of angular momentum for a particle. The law states that: If the moment of the net force acting on a particle is zero, the angular momentum of the particle about the origin is constant. We have seen that the angular momentum is just the cross product of the position vector and the linear momentum. This means that

L rp

(3.71)

and its direction is given by the right-hand rule. We move our right hand as if we are manipulating a screw driver. If the rotary motion is from the vector r to the vector p then L points in the direction in which the screw would go In view of Eq. (3.59) and (3.62) we can express the components of L in a compact form as

63

64

iˆ

ˆj

kˆ

L rp x

y

z

px

py

pz

(3.72)

Therefore

Lx  ypz  zpy Ly  zpx  xpz

(3.73)

Lz  xpy  ypx

3.7

Central Force Motion Now that we understand what angular momentum is, we shall treat an important category of three-dimensional motion governed by a central force. A central force is a static (conservative) force whose magnitude depends only on the radial coordinates and whose direction is along the position vector, so that it points either from the origin of the particle or from the particle to the origin. For a central force therefore, we have Eq. (3.4).

F  f (r )rˆ

(3.74)

We have seen that the angular momentum L of a particle is related to the moment M on the force acting on the particle by Eq. (3.68)

dL rF  M dt

(3.75)

and for a central force, it is

M  rrˆ  f (r )rˆ  rf (r )rˆ  rˆ  0

(3.76)

So a central force has no torque about the origin and therefore

dL 0 dt so that

64

(3.77)

CLASSICAL MECHANICS

L  constant

(3.78)

We see that the angular momentum of a particle moving under the effect of a central force is a constant. This result is important and implies that far from being three-dimensional, central force motion takes place in a plane. The proof of this is as follows. Since the angular momentum is defined as

L rp

(3.79)

it is a vector at right angles to the plane containing the position vector and the linear momentum. But since L is constant, it is in particular constant in both magnitude and direction. Hence the plane containing position vector r and the velocity vector v is fixed in orientation because if this orientation is changing, then L would also change. At all times the particle has a position vector lying in this fixed plane. We conclude that motion under a central force always takes place in a plane of fixed orientation. Central force motion owes a great deal of its importance to the fact that the gravitational force is central. Thus the motion of the earth about the sun is characterized by a fixed angular momentum and is planer. The motion about the moon is the same. In fact any celestial body orbiting another moves in a plane.

Another important central force is the

electrostatic force between charged particles.

These two forces, the

gravitational and electrostatic, are among the most important in nature. For a central force, the potential energy is

V (r )   f (r )rˆ  dr   f (r )dr

(3.80)

where we have used the definition and property of the dot product of two vectors as given by Eq. (1.1). Using the expressions for the components of the acceleration in spherical polar coordinates, we find from Newton’s second law

mr  F

(3.81) 65

66

that components of the equation of motion for motion in a central force field are

m(r  r 2  r sin2 2 )  Fr  f (r ) m(r  2r  r sin cos2 )  F  0 

(3.82)

m(r sin  2r sin  2r cos )  F  0 which follow from Eq. (1.71) and (3.3). If we chose polar coordinates

r ,  in the plane of motion, only the first and second equations in Eq. (3.82) hold and can be rewritten as

m(r  r 2 )  f (r ) m(r  2r)  0

(3.83)

since the motion does not depend on  which is therefore suppressed. Now from

r  rrˆ( )

(3.84)

we have

d rrˆ( )  r  rrˆ  rˆ dt

v

(3.85)

as given by Eq. (1.23). The acceleration follows from Eq. (3.85) and is



 



a  r  r rˆ  r  2r ˆ

(3.86)

as given by Eq. (1.29). Thus the two components of the acceleration a are the radial acceleration ar and the angular acceleration a given by Eq. (1.30) and (1.31)

ar  r  r 2

(3.87)

a  r  2r

(3.88)

Since the force is central, the F  0 and F  0 . Hence, Eq. (3.83). Let us rewrite the second equation in Eq. (3.83) as follows

66

CLASSICAL MECHANICS

d d dL (mr 2)  (mrr)  0 dt dt dt

(3.89)

Integrating this we obtain

mr 2  L  0

(3.90)

The constant L is to be evaluated from the initial conditions. Since the force is conservative we must have

T V  E

(3.91)

Using Eq. (3.85) we have

T









1 2 1 1 mv  m rrˆ  rˆ  rrˆ  rˆ  m r 2  r 2 2 2 2 2



(3.92)

or

T



 

1 1 m r 2  m r 2 2 2 2



(3.93)

which we can see as composed of translational and rotational energy. Eq. (3.91) can then be written as

T V 

1 2 1 2 2 mr  mr   V (r )  E 2 2

(3.94)

where V (r ) is the potential energy or potential energy function given by Eq. (3.48) rf

V (r )  V (r )    F  dr

(3.95)

ri

and E the constant energy to be evaluated from the initial conditions Now from

mr 2 2  L

(3.96)

we can rewrite Eq. (3.94) as

67

68

T V 

1 2 L2 mr   V (r )  E 2 2mr 2

(3.97)

We can solve this equation for r to get

v  r 

2 L2   E  V (r )   m 2mr 2 

(3.98)

Therefore r

 r0

dr 2 L   E  V (r )   m 2mr 2  2



2 t m

(3.99)

The integral can be evaluated and the resulting equation solved for r (t ) . For  (t ) we use mr 2 2  L to get t

L mr 2 0

  0  

(3.100)

We have thus so far obtained the solutions of Eq. (3.83) in terms of the four constants L, E , r0 and 0 which can be evaluated when the initial position and the velocity in the plane are known. Furthermore the potential energy function can have many forms such as the Yukawa potential in nuclear particles the potential function is taken to be of the form

V (r ) 

ke ar r

(3.101)

where k and a are constants and k  0 For two masses M and m separated by a distance r in a gravitational field, the potential (potential energy) or potential energy function is

V (r )    

GMm GMm dr   C 2 r r

(3.102)

with C the constant of integration and has the same units as V (r ) . Physically C represents the initial potential energy. 68

CLASSICAL MECHANICS

3.8

Summary In this unit, we have generalized to the case of two and three dimensions the methods seen in unit 2 for handling dynamical problems. From the knowledge gained, it is now possible to treat such important problems such as the revolution of the planets about the sun and motion of charged particles in an electric or magnetic field force. The theory we have developed so far can be use very conveniently to point particles and the dynamics of aggregate of particles and rigid bodies will be presented later in the module.

3.9

Exercises 3.0

Show that the velocity and acceleration of a particle moving

in a circle are perpendicular if and only if the particle is moving with a constant velocity. 3.1

A particle of mass m moves according to the equations

x  x0  at 2 , y  bt 2 , z  ct where x0 , a, b, c are constants. (a) Find the momentum L at any time t . (b) Find the force F and from it the torque N acting on the

particle (c) Verify that the angular momentum theorem

dL / dt  r  F  N is satisfied.

69

70

3.2

Consider the one-dimensional potential



 Wd 2 x 2  d 2 U ( x)  x 4  8d 4



Sketch the potential and discuss the motion at various values of x. Is the motion bounded or unbounded? Where are the equilibrium values? Are they stable or unstable? Find the turning points for E  W / 8 . The value of W is a positive constant. 3.3

A particle is under the influence of a force F  kx 

kx3 a2

where k and a are positive constants. (a) Obtain an expression for the velocity (b) Determine U (x) and discuss the motion, 3.4

A particle is projected vertically upwards with a speed v from a height h . Show that the particle hits the ground with speed v  v02  2 gh

3.5

A particle of mass m is subject to a force  x3  F  k   x  2  a  

(i)

What are the SI units for a and k ?

(ii)

Obtain an expression for the potential energy V (x) assuming that V (0)  0 .

70

CLASSICAL MECHANICS

3.6

A locomotive of mass m moving with a velocity v in a straight line is pulled with a force P against a resistance R . Show that the distance travelled by the locomotive in

accelerating from a velocity v1 to v 2 is given by v2

vdv . PR v1

d  m

When the speed of the locomotive is v 3 , its engine is turned off and it comes to rest under the action of a resistance

R  a  bv . Find an expression for the distance travelled. 3.7

If a block is projected with an initial velocity v 0 on a smooth horizontal surface and that there is no air resistance such that the block moves under a force F (v)  c1v . Write down the differential equation describing the motion if the block has a mass m . Find an expression for the position at any later time t .

71

72

Chapter 4 Unit 4 Simple Harmonic Motion

You should be able to cover this unit in 20 hrs. This time period does not include the time for solving problems in the exercises. You may solve the exercises at your own convenient pace. OBJECTIVES (1) To teach the learner how to treat both free and damped harmonic oscillators. (2) To teach the learner how to obtain the frequency of vibration of disturbed systems.

After studying this unit, the learner should be able to: (1) Identify a free and damped simple harmonic oscillating system. (2) Write down the equation of motion for the identified system in (1) above. (3) Solve the problem of the system whose equation has been written in (2) above. (4) Obtain solutions of free, underdamped (weakly damped), overdamped (strongly damped) and critically damped systems.

72

CLASSICAL MECHANICS

(5) Calculate amplitudes, frequencies, positions, velocities and accelerations of different oscillating systems.

4.0

Introduction All around us there are objects which vibrate or oscillate. Swings in playgrounds, musical instruments, water waves, atoms and molecules moving in matter are some examples of vibrations and oscillations. A force acting on a particle displacing it from its equilibrium in the direction opposed to its displacement will always result in to and fro motion.

If the magnitude of this force is f (x ) , where x is the

displacement then for vibratory motion to occur, we must have

ma   f (x)

(4.1)

where m is the mass and a is the acceleration. Since the force  f (x) always pulls the mass m back towards the position of equilibrium, it is called a restoring force. Most important vibrations correspond to the case where f (x ) is directly proportional to the displacement, so that we can write

f ( x)  kx

(4.2)

where k is a constant called the force constant. In some literature it is also called the spring constant because a vibrating body behaves as if is connected to a spring. In view of E. (4.2), the equation of motion is

mx  kx

(4.3)

A body whose motion is governed by this equation is called a simple harmonic oscillator. We see that the conditions for a body to undergo simple harmonic motion (often abbreviated SHM) are: (i) There must be a restoring force (ii) The magnitude of this restoring force must be proportional to the displacement 73

74

There are many examples of simple harmonic oscillators such as a mass connected to a spring and hung vertically so that the mass is free to move up and down when displaced, a pendulum displaced slightly from its equilibrium position and a floating body pushed further into the liquid and released. Simple harmonic motion is so widespread that any system whatsoever will perform simple harmonic motion if displacement from the equilibrium position is small enough. Such is the case when a guitar string is plucked or when a drumhead is beaten. At microscopic level, atoms and molecules perform simple harmonic motion when excited by thermal energy from the environment.

4.1

Displacement in Simple Harmonic Motion We now proceed to solve Eq.(4.3) by introducing the angular frequency through the equation

k m



(4.4)

the equation of motion can be written as

x   2 x

(4.5)

Simple harmonic motion can in fact be thought of as the projection of the position vector of a particle which rotates with constant angular velocity

 on the circumference of a circle. If the frequency of the motion is f , then the particle makes f revolutions per unit time and in that time the angle rotated is 2f . We can also say that the period of oscillation is

1 / f . In this period, the angle rotated is 2 . If the angular velocity is

 , we see that time taken for one revolution is 2 /  . The connection between  and f is therefore

2





1 T f

(4.6)

This is normally written as

f  74

 2

(4.7)

CLASSICAL MECHANICS

An inspection of Eq. (4.5) shows that the solution that we seek is a function with the property that when it is differentiated twice, it remains unchanged but acquires a multiplicative constant.

Of all the known

functions, the only ones with this property are the sine, the cosine and the exponential functions. Since sine or cosine functions can be expressed in exponential form we choose to express the required solution in exponential form because the ease with which exponential functions can be manipulated in terms of differentiation. We let the solution be

x  e t

(4.8)

where  is to be determined. Differentiating this once we get

x  et

(4.9)

and a second differentiation gives

x  2 e t

(4.10)

Substituting these expressions in Eq. (4.5) we obtain

2et   2et

(4.11)

Cancelling the non-zero factor et we have

2   2

(4.12)

which is satisfied by   i and   i (where i 2  1 ) and so

eit and eit

(4.13)

are both solutions of the SHM differential equation. Since the equation is linear the general solution will be a linear combination of these two solutions with arbitrary constant coefficients, that is

x  Ae  it  Be  it

(4.14)

The two constants can be determined from the so-called initial conditions

x(t  t0 )  x0

(4.15) 75

76

and

x (t  t0 )  v0

(4.16)

These constants have been introduced because the solution of any second-degree differential equation demands this. The reason is that when the differential equation was obtained by differentiating the solution twice, two constants must have been lost and they have to be recovered.

The initial conditions evidently give us the value of the

solution and of its first time derivative at a certain instant of time. Since there are two unknowns and two equations from the initial conditions, these constants can be determined from the initial conditions. The information required to fix A and B can also be provided in the form of the so-called boundary conditions

x(t  t1 )  x1

(4.17)

x(t  t2 )  x2

(4.18)

x (t  t1 )  x1  v1

(4.19)

x(t  t2 )  x2  v2

(4.20)

or in the form of

By means of Euler’s formula

e  i  cos  i sin

(4.21)

we can expand Eq. (4.14) as

x  A cost  iA sin t  B cost  iB sin t  ( A  B) cost  i( A  B) sin t  x0 sin  cost  x0 cos sin t

(4.22)

 x0 sin(t   ) where

x0 sin   ( A  B) and

76

(4.23)

CLASSICAL MECHANICS

x0 cos  i( A  B)

(4.24)

We can similarly convert expression (4.12) to the cosine form

x  x0 cos(t   )

(4.25)

x0 is the amplitude while  and   are called phase angles- ( some literature use  and   ). When the phase angle is zero, the general form of x (t ) is illustrated in Fig. 4.1

x

x(t )  A sin t

0

Time Period T

Fig. 4.1: Position plotted against time in SHM

The frequency f is just the number of complete oscillations the oscillator makes in unit time. What of  ? The angular frequency is so named because of the intimate connection between the circular motion and simple harmonic motion. We suppose that a particle moves with a constant angular velocity  in the xy -plane in a circle of radius x0 centered on the origin. If we shine light straight down the y axis, the shadow of the particle will be seen on a horizontal screen placed below the x axis at any distance not greater than x0 . It can be shown that the shadow performs horizontal simple harmonic motion along the x axis with the origin as the equilibrium position. The 77

78

period of this motion corresponds to the time it takes the particle to make one complete revolution. The angular frequency is just the angle through which the particle rotates in unit time. This is clearly 2f , the expression for  (Eq. (4.6) or (4.7)).

4.2

Velocity and Acceleration in Simple Harmonic Motion Now that we know the displacement of simple harmonic motion, we can proceed to calculate the velocity and acceleration of such motion. The velocity and acceleration in simple harmonic motion given by Eq. (4.22)

x  x0 sin(t   )

(4.26)

dx  x  x0 cos(t   ) dt

(4.27)

d 2x  x  x0 2 sin(t   ) dt 2

(4.28)

are given by

and

One can proceed and use Eq. (2.25) instead. The maximum value of the velocity x0 is called the velocity amplitude 2 and the acceleration amplitude is given by x0 .

4.3

The Energy of a Harmonic Oscillator The fact that the velocity is zero at maximum displacement in simple harmonic motion and is a maximum at zero displacement illustrates the important concept of an exchange between kinetic and potential energy. In an ideal case, the total energy remains constant but this is never realized in practice because of losses. If no energy is dissipated then all the potential energy becomes kinetic energy and vice versa, so that the value of (a) the total energy at any time (b) the maximum potential energy and (c) the maximum kinetic energy will be equal; that is 78

CLASSICAL MECHANICS

Etotal  KE  PE  KEmax  PEmax

(4.29)

From this equation, it is evident that

Etotal  E 

1 2 mx  V ( x) 2

(4.30)

For a spring

F  kx

(4.31)

and so

1 2 kx 2

(4.32)

1 2 1 2 mx  kx 2 2

(4.33)

V ( x)     kxdx  Hence

Etotal  E  Since E is constant we have

dE d  1 1  1 d mxx  kxx  0   mxx  kxx  dt dt  2 2  2 dt 1  mxx  mxx  kxx  kxx   0 2 2  mxx  kxx   mxx  kxx  0 2

(4.34)

By canceling the non-zero common term x , we again obtain the equation of motion

mx  kx  0

(4.35)

The maximum potential energy occurs at x   x0 and is therefore

PEmax 

1 2 kx0 2

(4.36)

The maximum kinetic energy is

79

80





1 1  KEmax   mx 2   mx02 2 cos2 (t   ) max 2  max 2 1  mx02 2 2

(4.37)

when the cosine factor is unity. But from Eq. (4.2) m 2  k so the maximum value of the potential energy and kinetic energies are equal, showing that the energy exchange is complete. The total energy at any instant of time or value of x is

1 2 1 2 mx  kx 2 2 1 2 2  mx0  cos2 (t   )  sin 2 (t   ) 2 1  mx 02 2 2 1  kx02 2

E



 (4.38)

as we expected Fig. 4.2 shows the distribution of energy versus displacement for simple harmonic motion. Note that the potential energy curve

PE 

1 2 1 2 2 2 kx  mx0  sin (t   ) 2 2

(4.39)

is parabolic with respect to x and is symmetric about x  0 , so that the energy is stored in the oscillator both when x is positive and when it is negative, e.g. a spring stores energy whether compressed or extended, as does a gas in compression or rarefaction. The kinetic energy curve

KE 

1 2 2 mx0  cos2 (t   ) 2

(4.40)

Is parabolic with respect to both x and x . The inversion of one curve with respect to the other displays the  / 2 phase difference between the displacement (related to the potential energy) and the velocity (related to the kinetic energy). 80

CLASSICAL MECHANICS

For any value of the displacement x the sum of the ordinates of both curves equals the total constant energy E .

Fig. 4.2: Parabolic representation of potential energy and kinetic energy of simple harmonic motion versus displacement. Inversion of one curve with respect to the other shows a 900 phase difference.

At any

displacement value the sum of the ordinates of the curves equals the total constant energy E.

From the foregoing theory, we now take a look at the specific example of a simple pendulum.

81

82

Example 4.1: The simple pendulum

L cos



L

h

Fig. 4.3: The motion of a simple pendulum For a simple pendulum, the potential energy is gravitational as illustrated in Fig. 4.3. The pendulum is of length L and is at an angle  to the vertical. Its vertical height from the lowest position is

h  L  L cos

(4.41)

and its potential energy is

V  mgh  mgL(1  cos )

(4.42)

Therefore

Etotal  E 

1 2 1 2 mx  kx 2 2

(4.43)

Since  is very small

cos  1 

2 2

(4.44)

and

sin  

(4.45)

Also, since the sideways displacement of the pendulum is

x  L sin  L

(4.46)

x  L

(4.47)

then

Hence 82

CLASSICAL MECHANICS

E

1 2 2 1 mL   mgL 2 2 2

(4.48)

Example 4.2: A harmonic oscillator passes through the equilibrium position when t  0 , Use the solution

x  Ae  it  Be  it

(4.49)

to show that the position as a function of time is

x(t )  R sin t

(4.50)

where R is the amplitude

Solution: The boundary condition given is x  0 when t  0 . This gives

A B  0

or

B  A

(4.51)

Thus,

x  Ae  it  Ae  it

(4.52)

Making use of Eq. (4.21) we get



x(t )  A e  it  e  it



 e  it  e  it  2iA 2i   R sin t

  

(4.53)

where R  2iA

83

84

4.4

Simple Harmonic Motion in Nature Whenever a slight displacement from equilibrium of a physical system takes place, the resulting vibration is simple harmonic. We now proceed to prove this fact. Consider a particle in one dimensional-motion which is in stable equilibrium before it is disturbed.

Let the particle be in stable

equilibrium and held by a potential V (x) and be located at a minimum of the potential. Such an arbitrary potential is shown in Fig. 4.3 (where the potential is denoted by U (x ) rather than V (x) ). Let a minimum of such a potential be at x0 . Such a location is a point of stable equilibrium. At the minimum, the potential does not change with position and therefore satisfies

dV 0 dx

(4.54)

and since the force is given by

F ( x)  

dV dx

(4.55)

it vanishes at x  x0 . This is precisely why the particle is in equilibrium at that point.

E

U (x )

a

x0

b

Fig. 4.4: Minimum of an arbitrary potential

84

c

x

CLASSICAL MECHANICS

When the particle is slightly displaced, the new position of the particle is

x and the value of the force at that point can be expressed in terms of its value at x0 by means of the Taylor expansion. We have

1  d 2F   dF  F ( x)  F ( x0 )    ( x  x0 )   2  ( x  x0 ) 2 2!  dx  x  dx  x0 0 1  d 3F    3  ( x  x0 )3   3!  dx  x

(4.56)

0

If ( x  x0 )  1 , then only the first two terms are sufficient and we have

 dF  F ( x)  F ( x0 )    ( x  x0 )  dx  x0

(4.57)

But F ( x 0 )  0 and from Eq. (4.55)

dF d 2V  2 dx dx

(4.58)

This means that the force at x can be written as

 d 2V  F ( x)    2  ( x  x0 )  dx  x0

(4.59)

When evaluated at x0 the second derivative of the potential gives a constant which we denote by k :

 d 2V  k 2  dx  x0

(4.60)

This constant is positive because the second derivative of any function at a minimum of the function is positive. The quantity gives the rate of change of the slope at that point and we see that the slope is gradually changing from negative to positive value at the minimum, this quantity is necessarily positive. It is technically called the curvature of the function. We can then write the equation for the force as

F ( x)  k ( x  x0 )

(4.61) 85

86

and the equation of motion is

mx  F ( x)  k ( x  x0 )

(4.62)

If we denote the displacement by   ( x  x0 ) , we find that   x so that the equation of motion is

m  k

(4.63)

This is just the defining equation for simple harmonic motion as already presented by Eq. (4.3). We have established that whenever any particle or body is slightly displaced from its equilibrium position, it subsequently performs simple harmonic motion with a force constant given by the second derivative of the potential evaluated at the position of stable equilibrium. In nature, slight displacements of bodies from equilibrium positions are numerous. As a result simple harmonic motion occurs everywhere, but normally with such small amplitudes that cannot be detected by the naked eye. However, such vibrations tend to produce sound if they fall in the correct frequency range. Sound is therefore one of the proofs of the prevalence of simple harmonic motion.

Example 4.3: A particle of mass m moving along the positive x axis is acted upon by a force whose potential is

V ( x) 

c1 c2  x4 x2

(4.64)

where c1 and c2 are positive constants. Show that the period of small oscillations about the position of stable equilibrium is

T  2

c1 c2

m c2

Solution: We first find the point of stable equilibrium. From

86

(4.65)

CLASSICAL MECHANICS

dV 2c2 4c1  3  5 0 dx x x

(4.66)

we find that the critical point is at

x

2c1 c2

(4.67)

To determine which of these points correspond to stable equilibrium, we compute the second derivative of the potential at each of these points. Now

d 2V 20c1 6c2  6  4 dx2 x x

(4.68)

2c1 c2

(4.69)

When

x

is inserted into Eq. (4.68), the result is

d 2V c23  0 dx2 c12

(4.70)

since the constants are both positive. Hence the point is a minimum. The force constant for small oscillations about a point of stable equilibrium x  x0 is given by Eq. (4.60)

 d 2V  k 2  dx  x  x0

(4.71)

In this case therefore,

c23 k 2 c1

(4.72)

and from Eq. (4.4) and (4.6) we find

87

88

k  m



c23 c  2 2 mc1 c1

c2 m

(4.73)

and

T

4.5

2



 2

c2 c1

c2 m

(4.74)

Rotary Motion and Simple Harmonic Motion Simple harmonic motion occurs in many mechanical systems as the projection of rotary motion onto a line. We may understand this by referring to Fig. 4.5. The figure shows a particle rotating in the xy plane with constant angular velocity  in a circle of radius A centered on the origin. The position vector of the particle has magnitude A but changing indirection all the time.

y

A

Asint x

0

Acost

Fig. 4.5: SHM from rotary motion

The angle   t is the angle which the position vector makes with the positive x axis in the counter-clockwise direction. Hence the x and y coordinates of the particle are

x  Acos  Acost 88

(4.75)

CLASSICAL MECHANICS

y  A sin  A sin t

(4.76)

If when t  t0 the position vector lies along the positive x axis, then

  t and at any time x  A cos (t  t0 )

(4.77)

y  A sin  (t  t0 )

(4.78)

We see that the coordinates of the particle perform simple harmonic motion.

Simple harmonic motion may therefore be viewed as the

projection of rotary motion on the coordinate axis

Example 4.4: A vertical spring attached to the ceiling supports a mass

m under whose weight the mass stretches by an amount L1 .. If further pulled down by a distance L2 and released at time t  0 , the mass oscillates. Determine the velocity of the mass as it passes through the equilibrium position and the acceleration of the mass when it is at the top of the motion. Solution: Let the stiffness (spring) constant be k and let the downward direction be taken as positive. Then by Hooke’s law we have

kL1  mg

(4.79)

and find

k

mg L1

(4.80)

Hence the angular frequency is



k  m

g L1

(4.81)

The solution is therefore a linear combination of the two solutions

y  A cost  B sin t

(4.82) 89

90

and the velocity at any time is

dy  y  A sin t  B cost dt

(4.83)

Employing the initial conditions

y  L2 ,

y  0,

when

t 0

(4.84)

and applying the velocity condition first gives us

0  0  B,

B0

(4.85)

Hence

y  A cost ,

y  A sin t

(4.86)

The position condition gives

L2  A

(4.87)

y  L2 cost

(4.88)

so that

Therefore the velocity is

y  L2 sint

(4.89)

while the acceleration is

y   2 L2 cost   2 y

(4.90)

As the mass passes through the equilibrium position, the time is at

t  T / 4 , where T is the period. Hence g  2 1  y  L2 sin t  L2 sin  T   L2   L2 L1  T 4 

(4.91)

At the top of the motion, the position is y   L2 therefore,

y   2 L2 cost   2 y   2 L2 90

(4.92)

CLASSICAL MECHANICS

Substituting the expression for  from Eq. (4.81), we get

y 

4.6

gL2 L1

(4.93)

Damped Oscillations Theoretically, a linear or simple harmonic oscillator once set into motion will continue oscillating forever. oscillations.

Such oscillations are called free

In practice, however, in any physical system there are

dissipative or damping forces, and the oscillating system will lose energy with time. Thus the oscillating system is damped and eventually comes to rest.

Damping is not just a nuisance but is an effect that finds

important applications in many areas of technology. For example, the shock absorbers in cars are meant to quickly subdue any oscillations of the springs even as they perform their function of making the ride smooth.

Damping forces are usually functions of velocity. Roughly speaking, the air friction experience by a car is proportional to the square of the velocity at which it is moving.

We need to know the functional

dependence of velocity on any damping force acting on a physical system if we have to treat it theoretically.

In a system which is damped, the differential equation for a linear oscillator given in Eq. (4.3) must be modified to include the effect of damping. We consider a mass m tied to a spring as shown in Fig. 4.6, as a prototype and restrict its motion to one dimension. As the mass moves in a fluid, air or liquid, the frictional force is the viscous force that produces the damping. As long as the speed of the 91

92

mass is small so as not to cause turbulence, the frictional force or damping force Fd may be assumed to be directly proportional to the velocity. That is

Fd  bv  bx

(4.94)

where b must be a positive constant.

Fig. 4.6: Force acting on a prototype of a damped harmonic oscillator.

The net force Fnet due to forces acting on the mass m as shown in Fig. 4.6 is

Fnet  F  Fd  kx  bx

(4.95)

Using Newton’s second law and substituting Fnet  mx in Eq. (4.95), we get

mx  bx  kx  0

(4.96)

Eq. (4.96) is a second-order differential equation for the damped harmonic oscillator of the mass spring system shown in Fig. 4.6. There are a number of techniques of solving this equation. The one we shall use is as follows: We divide both sides of the equation by m and substitute

92

CLASSICAL MECHANICS



b 2m

(4.97)

k m

(4.98)

and

02  to obtain

x  2x  02 x  0

(4.99)

As with the free harmonic oscillator, let us try an exponential solution of the form

x  e t

(4.100)

where  is to be determined. Differentiating this once we get

x  et

(4.101)

and a second differentiation gives

x  2 e t

(4.102)

Substituting these expressions in Eq. (4.99) we obtain





et 2  2  02  0

(4.103)

Since e t  0 , we must have



2



 2  02  0

(4.104)

This auxiliary or sometimes called the characteristic equation has the roots

1     2  02

(4.105)

1     2  02

(4.106)

and

93

94

The solution of Eq. (4.99) is with A1 and A2 as arbitrary constants

x(t )  A1e1t  A2e2 t

(4.107)

or

x(t )  et  A1e 

  2  02 t

 A2e

  2  02 t

 

(4.108)

The following cases of this solution are of special interest and will be discussed in some detail. Case (I)

Underdamped or

02   2

weakly damped

1 and 2 are imaginary roots

motion (complex) (oscillatory) Case (II)

Critically damped

02   2

and are equal

(not oscillatory) Case (III)

1 and 2 are real

02   2

Overdamped

1 and 2 are real roots

(not oscillatory)

Case (I) Underdamped Oscillations, 02   2 : For this case it is convenient to make a substitution:

1      2 0

2

k b2  m 4m 2

Since 02   2 , this makes

1      2 0

imaginary, and we may write

94

2

k b2  m 4m 2

(4.109)

CLASSICAL MECHANICS



x(t )  et A1e i1t  A2ei1t



(4.110)

which is a solution of an under-damped oscillator. Using Eq. (4.21), we may write Eq. (4.110) as

x(t )  et i A1  A2 sin 1t   A1  A2 cos1t 

(4.111)

Substituting i A1  A2   B and  A1  A2   C , we obtain an alternative solution

x(t )  et B sin 1t  C cos1t 

(4.112)

This may be written in a slightly different form by making the following substitution in Eq. (4.112).

A  B 2  C 2 and tan   

C B

(4.113)

Thus we obtain

x(t )  Ae t cos(1t   )

(4.114)

Of the three solutions given by Eqs. (4.110), (4.112) and (4.114), we shall concentrate on Eq. (4.114). It may be pointed out that the constants A1 and A2 in Eq. (4.110) are complex quantities, while B and C in Eq. (4.112) and A and  in Eq. (4.114) are all real quantities. The solution given by Eq. (4.114) indicate that for a damped oscillator the motion is oscillatory due to the existence of the cosine function, but the amplitude of the oscillation decays exponentially as shown in Fig. 4.7. The natural angular frequency, 1 , or the frequency of the damped oscillator is always less than the free oscillation frequency 0 . The natural frequency 1 is not a frequency in the true sense of the word because the oscillator never passes through the same point twice with the same velocity; that is the motion is not periodic. But if  is very small, then 1  0 and we call 1 the “frequency”. If  is small, we can expand Eq. (4.109) (using the binomial expansion) as 95

96

1     2 0

2



1

2

 2   0 1  2   0 

1

2

  2  0 1    2  20 

(4.115)

 2    0 1  2   20  If   0 , then

1  0

Fig. 4.7: Motion of damped harmonic oscillator

Case (II) Critically damped, 02   2 : For this case, the two roots given by Eqs. (4.105) and (106) are equal, that is

1  2  

(4.116)

and the general solution given by Eq. (4.108) takes the form

x(t )   A1  A2 et  B1et

(4.117)

where ( A1  A2 )  B1 =constant. This is not a general solution because it contains only one constant. In such cases, we show that if e  t is a solution, then 96

CLASSICAL MECHANICS

x  tet

(4.118)

is also a solution. Substituting Eq. (4.118) into the differentia equation (Eq. (4.99)), we get



2 0



  2 et  0

(4.119)

Since 0    , the equation is satisfied, and te  t is also a solution. Thus, for a critically damped case, the general solution is a linear combination of e  t and te  t ; that is

x(t )  B1  B2t e t

(4.120)

where B1 and B2 are constants to be determined by the initial conditions.

Case (III) Overdamped (Strongly damped), 02   2 : If the damping increases such that  2  02 , then the roots 1 and 2 are real. If we represent



2

 02



1

2

 2

(4.121)

the general solution given by Eq. (4.108) takes the form



x(t )  et A1e 2 t  A2e 2 t



(4.122)

Note that 2 is no longer a frequency because the motion is not oscillatory. Due to the decay term e  t multiplying both terms in the parenthesis, the terms decay exponentially, one faster than the other and become zero when sufficient time has elapsed. This is called strong, over-damping or heavy damping. Once displaced or released, the body or particle merely return to the equilibrium position although it may overshoot. Fig. 4.8 shows the behavior of underdamped, overdamped and critically damped harmonic oscillators from their equilibrium positions.

97

98

Fig. 4.8: Return of harmonic oscillator to equilibrium (a) Underdamped. (b) Overdamped. (c) Critically damped.

4.7

Energy Dissipation in Damped Motion The total energy E (t ) of a damped harmonic system at any time t is given by

E(t )  E(0)  Wf

(4.123)

where E (0) is the total energy at time t  0 and W f is the work done by friction in the time interval 0 to t .

Assuming the dissipative

frictional force f  bx  bv , we can calculate W f as follows:

W f   f dx   f

1

dx dt   f v dt    b v 2 dt dt 0

(4.124)

Thus the rate of energy loss by friction may be written as

dE  dW f      bv2 dt  dt 

98

(4.125)

CLASSICAL MECHANICS

which is negative and represents the rate at which energy is dissipated into heat.

Since Wf  0 , the energy at any time Et continuously

decreases with time and may be calculated in the following manner:

E (t )  K (t )  U (t ) 

1 2 1 2 mx  kx 2 2

(4.126)

From Eq. (4.114)

x(t )  Ae t cos(1t   ) and

   x (t )  1 Ae t sin(1t   )  cos(1t   ) 1  

(4.127)

We shall now assume that the system is lightly damped so that

 / 1  1, and neglecting the second term on the right in the preceding expression for x , we can substitute for x and x in Eq. (4.126) to obtain

E (t ) 



1 2  2t A e m12 sin 2 (1t   )  k cos2 (1t   ) 2



(4.128)

Since we assumed light damping, we may write 12  02  k / m ; hence the equation takes the form

E (t ) 

1 2  2t kA e 2

(4.129)

while the initial energy of the system is obtained by substituting t  0 in Eq. (4.129); that is

E0 

1 2 kA 2

(4.130)

Thus

E ( t )  E 0 e  2 t

(4.131)

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100

We see from this equation that the energy decreases ( or decays)

  than the rate at which the

exponentially at a much faster rate e 2t

 

amplitude decreases or decays et . The time  in which E decreases to 1 / e ( ( 0.368) of its initial value is called the characteristic time or decay constant and may be evaluated by substituting E (t )  E0 / e and t   in Eq. (4.131):

E0  E0e  2 e

(4.132)

2  1

(4.133)

or

That is,



1 2m m   2 2b b

(4.134)

If  is small,    , and if  is very large,   0 . We shall end our study of damped harmonic systems here. Another very important and interesting system is forced harmonic oscillator (or driven harmonic oscillator). Because of lack of time, you are asked to go and read on this topic.

4.8

Summary The harmonic oscillator in its various forms is one of the most important problems in Physics. In this unit, the learner has been introduced to the treatment of this problem for the case of free motion and three degrees of damping. These topics should serve as a springboard for further study of the problem in harmonic motion. Further problems include the driven oscillator, the anharmonic oscillator and the oscillator damped by a force with non-linear dependence of velocity.

100

CLASSICAL MECHANICS

4.8

Exercises A block of mass m lying on a horizontal frictionless

4.0

surface is attached to two identical springs of spring constant k as shown below. k m k

The mass is displaced to a position x to the right and then released. (i) Write down the equation of motion which describes the motion of the mass at any time t . (ii) Solve the equation of motion to obtain an expression of the position of the mass at any time t . (iii) Show that your solution found in (ii) above can also be written as B sin  0 t  C sin  0 t

where B and C are constants. (iv) Obtain the relationship between the constants A and  to your solution in (ii) and the new constant B and C in (iii) above. (v) Write down the expression of the natural frequency in terms of k and the mass m . (vi) Will the new natural frequency be higher or lower if one of the springs is removed? (vii) Explain two methods which you may use to damp the motion of the mass 101

102

The springs of a car of mass 1,200 kg give the car a vertical

4.1

oscillatory period of 0.5 s when the car is empty. How far does the car sink when the driver and three passengers, each of mass 75 kg, get in the car? 4.2

A weakly damped harmonic oscillator is at the point x0 and is at

rest at t  0 . What is its position as a function of time? 4.3

A particle of mass m moves under the conservative force with

potential energy

V ( x) 

kx x  a2 2

where a and k are positive constants. Obtain the frequency of small oscillations about the position of stable equilibrium. 4.4

A particle of mass m moves in a circular orbit under the action

of an attractive inverse-square central force

F (r )  

 r2

(i) Prove that if the particle experiences a small radial disturbance, then the periodic time of the perturbations is equal to the period of revolution in the circular orbit. (ii) If  is reduced instantaneously by half, then the particle will subsequently move in a parabolic orbit.

102