Basic Engineering Physics PDF Third Edition WBUT-2013

Basic Engineering Physics Third Edition WBUT–2013  About the Authors Sujay Kumar Bhattacharya is Former Head of Department of Sciences and Humanities, West Bengal Survey Institute (Govt. of West Bengal), Bandel, Hoogh...

Basic Engineering Physics Third Edition WBUT–2013  About the Authors Sujay Kumar Bhattacharya is Former Head of Department of Sciences and Humanities, West Bengal Survey Institute (Govt. of West Bengal), Bandel, Hooghly. He did his MSc from Calcutta University and PhD from Jadavpur University. In the past, he has taught physics at Meghnad Saha Institute of Technology, Kolkata, for seven years. Here he served as Assistant Professor of Physics and Head of Department of Physics. He has also taught physics at the BSc (Honours) level at R K Mission Vidyamandir, Belur Math, Howrah. Prof. Bhattacharys’s research area of interest is plasma physics. He has contributed more than 32 research papers in reputed national and international journals and symposia. He is also a life member of Plasma Science Society of India (PSSI) and Indian Association for Cultivation of Science (IACS). Saumen Pal, a doctorate in physics, is currently Assistant Professor of Physics in the Department of Basic Sciences and Humanities, Meghnad Saha Institute of Technology, Kolkata. He obtained his BSc (Honours) and MSc from Dhaka University and MPhil and PhD from Pune University. He has published and presented eight research papers in various journals and conferences. Prof. Pal is also a poet and a writer in Bangla. His scientific mindedness is reflected in his poems and articles. Some of his Bangla poems and articles have been published in local magazines in Kolkata. He has also published a book of poetry titled ‘Bilambita Prayas’. He has nearly one year experience as scientific officer under Bangladesh Atomic Energy Commission, Dhaka and three and half years’ experience of teaching physics at BSc (Hons) level in Basanti Devi College, Kolkata. Additionally, he has taught physics at the BTech level for eleven years and has been involved in research in the subject for seven years. His areas of research interest include nuclear physics and biophysics. Sujay Kumar Bhattacharya and Saumen Pal have collectively authored two other books, Advanced Engineering Physics and Physics-II (PH-EE-401), both published by McGraw Hill Education (India).  Basic Engineering Physics Third Edition WBUT–2013 Sujay Kumar Bhattacharya Former Head Department of Science and Humanities West Bengal Survey Institute Bandel, Hooghly Saumen Pal Assistant Professor (Physics) Department of Basic Sciences and Humanities Meghnad Saha Institute of Technology Kolkata McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto  McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Basic Engineering Physics, 3e Copyright © 2013, 2012, 2011 by McGraw Hill Education (India) Private Limited. 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This work is published with the understand- ing that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Bharati Composers, D-6/159, Sector-VI, Rohini, Delhi 110 085, and Printed at Cover Printer: Dedicated to the science and technology students, wishing for the development of their scientific mindedness.  Contents Preface xiii Roadmap to the Syllabus xvii 1. Simple Harmonic Motion 1.1-1.26 1.1 Introduction 1.1 1.2 Relation of Simple Harmonic Motion with Circular Motion 1.1 1.3 Differential Equation of Simple Harmonic Motion 1.4 1.4 Various Characteristics of SHM 1.4 1.5 Solution of the Differential Equation of SHM 1.5 1.6 Velocity and Acceleration of the Particle Executing SHO 1.6 1.7 Energy of a Particle Executing SHM and Law of Conservation of Energy 1.7 1.8 Superposition of Waves 1.8 1.9 Derivation of the Equation of Motion from the Law of Conservation of Energy 1.12 1.10 Two Vibrations in a Plane (of Commensurate Frequencies) Acting at Right Angles to Each Other 1.12 Worked-out Examples 1.15 Review Exercises 1.21 2. Free and Damped Vibrations 2.1 2.1 Free Vibrations 2.1 2.2 Differential Equation of Free or Undamped Vibrations 2.1 2.3 Damped Vibrations 2.2 2.4 Solution of the Equation of a Damped Oscillator and its Analysis 2.3 2.5 Electrical Analogy of SHM and DV 2.8 2.6 Energy of the Damped Oscillator 2.9 Worked-out Examples 2.10 Review Exercises 2.17 viii Contents 3. Forced Vibrations 3.1 3.1 Introduction 3.1 3.2 Sympathetic Vibration or Resonance 3.1 3.3 Analysis of Forced Vibration 3.2 3.4 Energy of a Forced Vibrator 3.4 3.5 Sharpness of Resonance 3.5 3.6 Amplitude Resonance 3.6 3.7 Quality Factor 3.7 3.8 Forced Vibration in an LCR Circuit 3.10 Worked-out Examples 3.11 Review Exercises 3.16 4. Interference of Light 4.1 4.1 Introduction 4.1 4.2 Wave Propagation and the Wave Equation 4.2 4.3 Principle of Superposition of Waves 4.3 4.4 Huygens’ Wave Theory 4.3 4.5 Interference of Mechanical Waves 4.5 4.6 Young’s Experiment on Interference of Light Waves (Double-slit Experiment) 4.5 4.7 Theory of Interference (Analytical Treatment) 4.6 4.8 Intensity Distribution Curve and Energy Conservation 4.8 4.9 Conditions for Permanent Interference: Coherent Sources 4.9 4.10 Classification of Interference Phenomena 4.9 4.11 Calculation of Fringe Width 4.9 4.12 Shape of Interference Fringes 4.11 4.13 Coherence 4.11 4.14 Change of Phase Due to Reflection: Stokes’ Law 4.14 4.15 Interference in a Thin Film (Reflected Light) 4.14 4.16 Interference in a Wedge-Shaped Film 4.16 4.17 Newton’s Ring 4.19 Worked-out Examples 4.23 Review Exercises 4.30 5. Diffraction of Light 5.1 5.1 Introduction 5.1 5.2 Different Types of Diffraction Phenomena 5.2 5.3 Difference Between Interference and Diffraction 5.2 5.4 Fraunhofer Diffraction Due to a Single Slit 5.2 5.5 Fraunhofer Diffraction Due to a Double Slit 5.6 5.6 Difference Between Single-slit and a Double-slit Diffraction Pattern 5.11 5.7 Diffraction Due to Plane Diffraction Grating 5.12  Contents ix 5.8 Prism and Grating Spectrum 5.19 5.9 Rayleigh’s Criterion of Resolution 5.19 5.10 Resolving Power of a Grating 5.20 5.11 Resolving Power of Microscope 5.21 Worked-out Examples 5.22 Review Exercises 5.28 6. Polarization of Light 6.1 6.1 Introduction 6.1 6.2 Mechanical Demonstration of Polarization 6.1 6.3 Symbolic Representation of Unpolarized and Polarized Light 6.2 6.4 Experiment of Polarization with Tourmaline Crystal 6.3 6.5 Planes of Polarization and Vibration 6.4 6.6 Various Methods of Producing a Plane Polarized Beam of Light 6.4 6.7 Brewster’s Law 6.8 6.8 Malus’ Law 6.8 6.9 Classification of Polarized Light 6.9 6.10 Production of Plane, Circularly and Elliptically Polarized Light by Analytical Method 6.10 6.11 Optic Axis and Principal Plane of a Crystal 6.12 6.12 Nicol Prism 6.13 6.13 Retardation Plates (Quarter or Half-wave Plates) 6.15 6.14 Detection of Polarized, Partially Polarized and Unpolarized Light 6.17 6.15 Dichroic Crystals and Polaroids 6.17 Worked-out Examples 6.19 Review Exercises 6.22 7. Laser Optics 7.1 7.1 Introduction 7.1 7.2 Characteristics of Laser 7.1 7.3 Absorption and Emission of Radiations by Matter 7.4 7.4 Einstein’s A and B Coefficients: Einstein’s Theory and Relations 7.6 7.5 Working Principle of Laser 7.7 7.6 Population Inversion in Laser 7.8 7.7 Basic Components of Laser System 7.9 7.8 Optical Resonator and Q -value 7.10 7.9 Threshold Condition for Sustaining of Laser Action 7.13 7.10 Typical Lasers 7.13 7.11 Applications of Laser 7.16 Worked-out Examples 7.17 Review Exercises 7.21 x Contents 8. Holography 8.1 8.1 Introduction 8.1 8.2 Basic Principle of Holography 8.1 8.3 Recording of the Hologram 8.1 8.4 Reconstruction of the Image 8.2 8.5 Characteristics of Hologram 8.3 8.6 Requirement for an Ideal Display of Hologram 8.3 8.7 Theory of Holography 8.3 8.8 Features of Holography 8.5 8.9 Applications of Holography 8.5 Review Exercises 8.6 9. Quantum Physics 9.1 9.1 Introduction 9.1 9.2 Blackbody Radiation 9.2 9.3 Relativistic Properties of Particles with High Speed Comparable to that of Light 9.12 9.4 Compton Effect 9.13 9.5 Wave Particle Duality 9.20 9.6 Uncertainty Principle of Heisenberg 9.33 9.7 Old Quantum Theory 9.39 Worked-out Examples 9.41 Review Exercises 9.50 10. Crystallography 10.1 10.1 Introduction 10.1 10.2 Crystal Lattice and Lattice Structure 10.1 10.3 Symmetry Elements of a Crystalline Solid 10.3 10.4 Atomic radius, Coordination Number and Atomic Packing Factor (APF) of Cubic Crystal System 10.5 10.5 Cubic Crystal System 10.6 10.6 Relation between the Density (r ) of Crystalline Material and Lattice Constant (a) of a Cubic Lattice 10.9 10.7 Nomenclature of Crystal Directions 10.10 10.8 Miller Indices 10.10 10.9 Perpendicular Distance between Successive Planes in a Crystal Lattice 10.12 10.10 Important Features of Miller Indices 10.12 10.11 X-rays 10.13 10.12 Origin of X-rays 10.13 10.13 X-ray Spectrum 10.14 10.14 Moseley’s Law 10.15  Contents xi 10.15 Absorption of X-ray by a Material 10.16 10.16 Bragg’s law 10.16 Worked-out Examples 10.18 Review Exercises 10.24 Appendix-A Angular Concept A1.1-A1.2 Appendix-B Coherence of Light B1.1-B1.4 Appendix-C Relativistic Concept C1.1-C1.16 Appendix-D Lab Experiments D1.1-D1.33 Appendix-E Useful Physical Constants E1.1 Model Question Papers M1.1-M1.13 Solved WBUT Question Papers (2008–2012) S1.1-S1.46  Preface Science and technology are very closely related subjects. Science (more specifically physics) can be thought of as the mother of engineering and technology. The progress of engineering and technology will stop fully without the help of science, in general, and that of physics, in particular. So, the progress of engineering and technology is directly related to that of science, and the progress of science depends on research in science as well as new inventions in the field. Engineering and technology are the applied aspects of science. Science provides concepts, theories as well as formulae to engineering and technology and thus aids in their progress. Any technological development depends on the scientific understanding of the working of nature. Though physical science and technology are very closely interlinked, some modern technologies like biotechnology, medical technology, food technology, etc., are also very much dependent on biological science. It is thus clear that the students of engineering and technology must have appropriate knowledge of advanced topics of science, in general, and physics in particular, to face the challenges of the future world as engineers and technologists. Aim The West Bengal University of Technology recently reorganized the physics syllabus of first-year BTech by selecting those topics which are commonly studied by students of all streams of engineering and technology. The university has introduced some specific advanced topics of physics in the syllabus of second-year BTech selectively for a few streams only where more knowledge of physics is essential. The first-year syllabus of physics consists of topics like waves and acoustics, classical optics (i.e., interference, diffraction and polar- ization), modern optics (i.e., laser and fiber optics), foundation of quantum theory, x-ray, and crystallography. So, we have put all the aforesaid topics of physics in the present book Basic Engineering Physics. Though many books covering the said topics are available in the market, none of them cover entirely the new syllabus and a major part of them discuss topics either insufficiently or in an exaggarated manner. For this reason, these books are not completely suitable for BTech first-year students of WBUT to understand their physics course. Keeping these points in mind, we decided to present this textbook to the first-year BTech students of WBUT for comprehensive yet lucid study. This book has been written on the basis of the class lecture notes we have prepared and used to teach this subject during the last nine and half years. The text has been written in the most coherent and exhaustive manner so that students can grasp the subject with minimum labor as well as time. Much effort has been given to clarify the fundamental concepts of the subject in an easy way. Apart from appropriate examples and illustrations, we have enriched the book with a large number of solved problems at the end of each chapter so that students can develop a thorough understanding of the topics. xiv Preface Salient Features Coverage and chapter organization as per the latest WBUT syllabus Excellent coverage of forced vibrations, interference of light, quantum physics, and crystallography Exhaustive and varied pedagogical features to provide practice for solving all types of questions for WBUT examination Useful lists of Greek alphabets, derived units, prefixes as multiplying factors, and physical constants for easy reference 10 lab experiments with detailed step-wise procedures, clear illustrations and aids to viva voce Model Question Papers based on WBUT pattern for self-practice Latest Solved WBUT Question Papers Pedagogy: 140 Illustrations 120 Worked-out Examples 340 Multiple Choice Questions 145 Short Questions with Answers 280 Long Questions with Answers 140 Numerical Problems Chapter Organization This book comprises 10 chapters. The chapters have been judiciously arranged in such a way that the stu- dents can develop the concepts of the subject step by step without skipping any required topic. Chapter 1 of the book gives an idea of simple harmonic motion. Chapter 2 discusses free and damped vibration. Forced vibration has been presented in Chapter 3. As (classical) optics is a broad topic, it has been divided into three chapters. Interference of light is covered in Chapter 4. Diffraction of light is discussed in Chapter 5. Polarization of light is dealt with in Chapter 6. Laser optics is discussed in Chapter 7. Holography is dis- cussed in Chapter 8. The foundation of quantum theory is discussed in Chapter 9 on quantum physics. X–ray and crystallography are explained in Chapter 10. Besides these, angular concept, coherence of light, and relativistic concept are presented respectively in Appendices A, B and C. A set of 10 experiments of physics practicals with detailed step-wise procedures, clear illustrations, and aids to viva voce is presented in Appendix D. Review Exercises given at the end of each chapter are divided into three parts. Part 1 contains multiple choice questions and short questions followed by their answers. Part 2 consists of descriptive questions and Part 3 consists of unsolved numerical problems. For solving such problems, students will have to apply critically the text which they have already learnt. This exercise will help them develop an analytical view of the topics. This book also has five model question papers which have been set as per the pattern of WBUT question papers. Further, the questions from WBUT exam papers of years 2008, 2009, 2010, 2011 and 2012 have been solved year-wise and provided toward the end of this book. We were very much surprised to observe that in this country, the study of science fails to have an impact in the development of scientific mindedness among most of the science as well as engineering and technol- ogy students. It may be due to the faulty method of teaching science in India. We feel that practical use of the knowledge of science in one’s daily life will help remove superstitions. Having kept this view in mind, the use  Preface xv of scientific theories learnt in various chapters of this book has been discussed at the end of each chapter. We hope that it will help develop a scientific attitude among the students. Acknowledgements We are indebted to the following reviewers who took out time to go through various chapters of this book and share their valuable comments. Ipsita Chakraborty Adamas Institute of Technology, Kolkata, West Bengal Kanchan Kumar Patra RCC Institute of Information Technology, Kolkata, West Bengal Mahadeb Das Abacus Institute of Engineering and Management, Hooghly, West Bengal Sabyasachi Sen JIS College of Engineering, Kolkata, West Bengal Srinjoy Bid Academy of Technology, Hooghly, Kolkata, West Bengal We would also like to express our sincere gratitude to our colleagues for their encouragement and constructive suggestions during the period of preparation of the manuscript. Special thanks to our family mem- bers who have cooperated with us during the writing of this book under extremely precarious situations. We would also like to thank the entire team at McGraw Hill Education (India) for bringing out this book in a very short span of time. Constructive criticism and suggestions are always welcome, which can be sent to [email protected] and [email protected]. Sujay Kumar Bhattacharya Saumen Pal Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments, and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and author’s name in the subject line. In case you spot piracy of this book, please do let us know.  Roadmap to the Syllabus This text will be useful for subject codes: PH101 and PH201 – Physics-I Module 1: Oscillation 1.1 Simple harmonic motion, Preliminary concepts, Superposition of SHMs in two mutually perpendicular direction, Lissajous figure 1.2 Damped vibration, Differential equation and its solution, Critical damping, Logarithmic decrement, Quality factor 1.3 Forced vibration, Differential equation and its solution, Amplitude resonance, Velocity resonance, Sharpness of resonance, Application LCR Circuit GO TO: CHAPTER 1. Simple Harmonic Motion CHAPTER 2. Free and Damped Vibrations CHAPTER 3. Forced Vibrations Module 2: Optics 1 2.1 Interference of electromagnetic waves, Conditions for sustained interference, Double-slit as an example, Qualitative idea of spatial and temporal coherence, Conservation of energy and intensity distribution, Newton’s ring (no deduction necessary) 2.2 Diffraction of light – Fresnel and Fraunhofer class, Fraunhofer diffraction for single-slit and double- slits. Intensity distribution of N-slits and plane transmission grating (no deduction of the intensity dis- tributions is necessary), Missing orders, Rayleigh criterion, Resolving power of grating and microscope (definition and formulae) GO TO Chapter 4. Interference of Light Chapter 5. Diffraction of Light xviii Roadmap to the Syllabus Module 3: Optics 2 3.1 Polarization — General concept of polarization, Plane of vibration and plane of polarization, Qualitative discussion on plane, Circulary and elliptically polarized light, Polarization through reflection and Brewster’s law, Double refraction (birefringence), Ordinary and extraordinary rays, Nicol’s prism, Polaroid. Half-wave plate and quarter-wave plate 3.2 Lasar — Spontaneous and stimulated emission of radiation, Population inversion, Einstein’s A and B coefficients (derivation of the mutual relation), Optical resonator and condition necessary for active laser action, Ruby laser, He–Ne laser — Application of laser 3.3 Holography — Theory of holography, Viewing the hologram, Applications GO TO Chapter 6. Polarization of Light Chapter 7. Laser Optics Chapter 8. Holography Module 4: Quantum Physics 4.1 Concept of dependence of mass with velocity, Mass energy equivalence, Energy momentum relation (no deduction required). Black-body radiation— Rayleigh Jeans’ law (Derivation without the calculation of number of states)’ Wien’s law, Ultraviolet catastrophy, Planck’s radiation law (Calculation of the aver- age energy of the oscillator), Derivation of Wien’s displacement law and Stephan’s law from Planck’s radiation law. Raylegh Jean’s law and Wien’s law as limiting cases of Planck’s law, Compton effect (Calculation of Compton wavelength is required) 4.2 Wave-particle duality and de Broglie’s hypothesis, Concept of matter waves, Davisson-Germer experiment, Concept of wave packets and Heisenberg’s uncertainty principle GO TO Chapter 9. Quantum Physics Module 5: Crystallography 5.1 Elementary ideas of crystal structure — Lattice, basis, unit cell, Fundamental types of lattices – Bravais lattice, Simple cubic, f.c.c. and b.c.c. lattices (Use of models in the class during teaching is desirable) Miller indices and miller planes, Coordination number and Atomic packing factor 5.2 X-rays — Origin of characteristic and continuous X-ray, Bragg’s law (No derivation), Determination of lattice constant GO TO Chapter 10. Crystallography  The Greek Alphabet Small Capital Name Roman equivalent a A alpha a b B beta b g G gamma g d D delta d e E epsilon e V Z zeta z h H eta e q Q theta th i I iota i k K kappa k l L lambda l m M mu m n N nu n x X xi x o O omicron o p P pi p r r rho r s S sigma s t T tau t u g upsilon u j F phi ph c c chi kh y Y psi ps w W omega o Some Useful Derived Units Physical Quantity Derived unit Symbol Area square meter m2 Volume cubic meter or litre m3 or 1 Force newton N Pressure and Stress pascal Pa (or N/m2) Work, Energy, Heat joule J Power watt W Moment of inertia kgm2 I Torque Nm T Specific heat capacity J/(kg k) c Thermal conductivity W/(mk) k Luminance lux lx  Luminous flux lumen lm Electric charge coulomb C Electric potential volt V Electric resistance ohm W Frequency hertz Hz Prefixes as Multiplying Factors Prefix Symbol (Multiplying) factor value atto – a 10–18 femto – f 10–15 pico – p 10–12 nano – n 10–9 micro – m 10–6 milli – m 10–3 centi – c 10–2 deci – d 10–1 deca – da 101 hecto – h 102 kilo – k 103 mega – M 106 giga – G 109 tera – T 1012 peta – P 1015 exa – E 1018 Examples: 1 fm = 1 × 10–15 m, 1 mg = 1 × 10–3 g, 1 M watt = 1 × 106 watt etc.  CHAPTER 1 Simple Harmonic Motion 1.1 INTRODUCTION The simple harmonic motion is a special case of the generalized periodic motion. In our daily life, we come across numerous things. We can classify them into two categories: The bodies which remain at rest with respect to us and the bodies which move with respect to us. The motions of all moving bodies can be catego- rized into two categories, namely, (i) the motion in which the body moves about a mean position (i.e., fixed point), and (ii) the motion in which the body moves from one place to another with respect to time. The first category of motion is called periodic motion while the second one is called translatory motion. A moving bus, a flying aeroplane, a moving football, etc., are examples of translational motion while the motion of a simple pendulum, a spring-mass system, vibrations of a stretched string, etc., are examples of periodic motion. Sometimes both the categories of motion can be observed in the same phenomenon depending on the observer’s point of view. The waves in the sea appear to us to move towards the beach but the water moves up and down about a mean position. When someone displaces a stretched string, the displacement pulse trav- els from one end to the other end of the string but the material of the string vibrates about a mean position without getting itself translated forward. The periodic motion in which the concerned body moves in a straight line is known as simple harmonic motion. 1.2 RELATION OF SIMPLE HARMONIC MOTION WITH CIRCULAR MOTION Simple harmonic motion (i.e., oscillation) is a special case of circular motion. To understand this relationship, let us discuss this with the help of a diagram. Figure 1.1 shows that PAB is a circle with its centre at C. Let us assume that P is a moving point on the circumference of the circle PAB. At time t = 0, it was at the point O and it has taken t units of time to reach the position P. So, the angle made by it with the line CO is q = w t where w is the uniform angular velocity of the particle. Let us drop a perpendicular PM from the point P to the diameter YY¢ of the circle. So, the point M also will be a moving point. While P will make an angle of 2p radian (q = 360°), the point M will make a complete oscillation. That is, when P will start moving from O towards the point P, M will start moving from C towards the point M. M will make a complete oscillation by moving from C to Y, then from Y to Y¢ and lastly from Y¢ to C during the time when P will make one complete 1.2 Basic Engineering Physics Y D P P¢ y A M q y a o q = wt F O¢ t t C O Q T/4 T/2 3T/4 T d B R Y¢ E l (a) (b) Fig. 1.1 Graph of simple harmonic oscillation along with its reference circle. revolution on the circumference of the said circle by making an angular displacement of q = 2p. If one plots a graph of y against t, then one will get the graph OP¢DEF. The circle along which the point P moves is called the reference circle. From Fig. 1.1 we can write, MC __ ____ y = = sin q CP a or, y = a sin q...(1.1) Equation (1.1) represents the oscillation (i.e., vibration) of the moving particle M on the straight path YCY¢ while the other moving particle P circulates on the circumference of the circle OPABRO. Now if we consider M as our simple harmonic oscillator, we can consider P as the reference point which circulates uniformly on the reference circle with an angular velocity w. If P remains at R instead of O, at t = 0, then the reference point P will start moving from the point R instead of the point O. In that case we can represent the motion of M by the following equation y = a sin (q – d)...(1.2) So Eq. (1.2) represents the generalized equation of a simple harmonic oscillator. If now, we put q = wt, then we can write Eq. (1.2) as y = a sin (wt – d)...(1.3) CM (= y) is called the displacement of the oscillator M, i.e., the vibrating particle. The displacement of a vibrating particle at any instant of time can be defined as its distance from the mean position of rest. In this case, C is the position of rest of the oscillating particle M. The maximum displacement of a vibrating particle is called its amplitude of vibration. The instantaneous displacement y is given by, y = a sin (q – d) or, y = a sin (w t – d)...(1.3) where d is the initial phase of the oscillator and a is the amplitude \ ymax = a The rate of change of displacement is called the velocity of the vibrating (i.e., oscillating) particle. dy \ velocity, v = ___ = a w cos (w t – d)...(1.4) dt The rate of change of velocity of an oscillating particle is called acceleration.  Simple Harmonic Motion 1.3 \ the acceleration is given by d d2 f = __ (v) = ___2 (y) dt dt = – aw 2 sin (wt – d) or, f = – aw 2 sin (wt – d)...(1.5) where f represents the acceleration of the oscillating particle. Now using Eq. (1.3) we can write the Eq. (1.5) as follows: f = – w2 y...(1.6) 2 dy d y At the extreme positions, where y is maximum, the velocity v = ___ = 0 and the acceleration f = ___2 is maxi- dt dt mum and it is directed towards the mean position. The returning force induces a negative velocity at the point of return. When the displacement y becomes zero, the velocity v becomes maximum and when the velocity v becomes zero, the acceleration becomes maximum in magnitude. The returning force again induces a veloc- ity in the opposite direction. It becomes maximum when the displacement again becomes zero. The particle overshots the mean position due to its velocity. The process repeats itself periodically. Thus, the system vibrates. And in this way, displacement y, velocity v and acceleration f continuously keep on changing with respect to time. Thus, the velocity of the vibrating particle becomes maximum (either in the direction of CY or that of CY¢) at the mean position of rest and it becomes zero at the extreme positions of vibration. The acceleration of the particle becomes zero at the mean position of rest and it becomes maximum at the extreme positions of vibration. And the acceleration is always directed towards the mean position of rest and it is also directly proportional to the displacement of the vibrating particle. So, we can define simple harmonic motion as follows: It is such a motion where the acceleration is always directed towards a fixed point (i.e., mean position of rest) and is proportional to the displacement of the oscillating particle. Again, acceleration f is given by d2y f = ___2 = – w 2y dt \ acceleration = – (angular velocity)2 × displacement acceleration or, angular velocity2 = ___________ (considering the magnitude only) displacement ___________ ÷ acceleration or, angular velocity = ___________ displacement It implies that, ___________ ÷ acceleration angular velocity = ___________ displacement Symbolically, we can write, __ ÷ f w = 2pn = __y where n is the frequency of oscillation. __ ÷ 2p ___ f or, = __y , where T is the time period of vibration T __ y or, T = 2p __ f ÷ 1.4 Basic Engineering Physics __ or, T = 2p ÷g where g is the displacement per unit acceleration. 1.3 DIFFERENTIAL EQUATION OF SIMPLE HARMONIC MOTION Let us consider an oscillating particle which is executing simple harmonic motion. The general equation of its displacement is given by y = a sin (w t + d)...(1.3) where y is the displacement and a is the amplitude and d is the phase (or epoch) of the oscillating particle. Now, differentiating Eq. (1.3) with respect to time t, we get dy ___ = aw cos (wt + d) dt dy Here ___ (= v) represents the velocity of the particle. dt dy Again, differentiating ___ with respect to time t, we get dt d2y ___ 2 = – aw 2 sin (w t + d) dt But, we know that y = a sin (w t + d) d2y ___ \ = – w 2y dt2 d2y ___ or, + w 2y = 0...(1.7) dt2 2 dy The second derivative ___2 represents the acceleration of the oscillating particle. Equation (1.7) represents dt the differential equation of simple harmonic motion. In any phenomenon, where an equation similar to Eq. (1.7) is obtained, the related body executes simple harmonic motion (SHM). The general solution of Eq. (1.7) is given by y = a sin (w t + d) One can also calculate the time period of oscillation of Eq. (1.7), ______ ÷ d2y/dt2 Numerically, w = ______ y ___________ ÷ acceleration or, w = ___________ displacement ___________ ÷ 2p displacement \ T = ___ = 2p ___________ w acceleration 1.4 VARIOUS CHARACTERISTICS OF SHM When one particle executes simple harmonic motion, one can observe the following characteristics in its motion:  Simple Harmonic Motion 1.5 (a) The motion is periodic as well as oscillatory. (b) The restoring force acting on the particle (and hence the acceleration) is directly proportional to the displacement of the particle and directs oppositely to the displacement measured from the mean position. (c) The acceleration of the oscillating particle is always directed towards the mean position of its path. (d) The motion of the simple harmonic oscillator (SHO) always takes place in a straight line. 1.5 SOLUTION OF THE DIFFERENTIAL EQUATION OF SHM Let y = aeat. Now differentiating y with respect to t, we get dy ___ = a aeat dt d2y ___ or, 2 = a2 (aeat) = a2y dt Now, substituting this value in Eq. (1.7), we get a2y + w 2y = 0 or, a2 = – w 2 ___ or, a = ± iw where i = ÷–1 So, y = ae iwt or y = ae–iwt Hence, the general solution is given by y = a1 e+iw t + a2 e–iw t...(1.8) where a1 and a2 are arbitrary constants. \ y = a1 (cos w t + i sin w t) + a2 (cos w t – i sin w t) [ eiq = cos q + i sin q and e– iq = cos q – i sin q] or, y = (a1 + a2) cos wt + i (a1 – a2) sin w t or, y = a cos wt + b sin w t …(1.9) where a = (a1 + a2) and b = i(a1 – a2) and d where a = q cos d and b = q sin d Let us now replace a and b by q______ \ q = ÷a2 + b2 and ( ) b d = tan–1 __ a Now, putting these values of a and b in Eq. (1.9), we get y = q cos d cos wt + q sin d sin w t or, y = q cos (wt – d)...(1.10) Hence, Eq. (1.10) represents the general solution of the differential equation d2y ___ 2 + w 2y = 0 dt The maximum value of the displacement y is given by ymax = q q is known as amplitude of the simple harmonic motion. 1.6 Basic Engineering Physics If T be the period of oscillation, the same displacement repeats after an interval of time T, i.e., yt = yt + T or, a cos (w t – d) = a cos {w (t + T) – d} or, a cos (w t – d) = a cos {(w t – d) + w T} or, cos (w t – d) = cos {(w t – d) + w T} Hence, w T = 2p 2p \ T = ___ w...(1.10) 2p or, w = ___ [w is angular velocity] T The frequency of SHM is given by 1 w n = __ = ___...(1.11) T 2p At t = 0, y = a cos d y \ d = cos–1 __a () The variable d is known as the initial phase or epoch. 1.6 VELOCITY AND ACCELERATION OF THE PARTICLE EXECUTING SHO The displacement (y) of a particle, which executes simple harmonic motion, at any time t is given by the fol- lowing generalized equation, y = a sin (w t + d) \ the velocity of the particle, v is given by dy v = ___ = aw cos (w t + d)...(1.12) dt When the particle reaches its mean position, the phase d=0 _________ \ v = aw cos (wt) = ± aw ÷1 – sin2 w t ______ ÷ y2 or, v = ± aw 1 – __2 a ______ \ v = ± w ÷a2 – y2 where w = 2pn When the particle is at its mean position, we get y = 0, so in this position the particle velocity is maximum and it is given by vmax = ± aw And at either of the extreme positions (i.e., y = ± a), velocity becomes minimum and it is given by vmin = 0 From Eq. (1.12), we get v = aw cos (wt + d)  Simple Harmonic Motion 1.7 \ the acceleration of the particle is given by dv f = ___ = – aw 2 sin (w t + d)...(1.13) dt When the particle is at its mean position, we get d = 0, so in this position the acceleration of the particle is given by f = – aw 2 sin wt or, f = – w 2y [ t = 0 and d = 0 fi y = a sin w t] So, when the particle is at its extreme position, y = a and the magnitude of the acceleration is given by fmax = aw 2 [It is maximum] And when the particle is at its mean position, y = 0 and the magnitude of the acceleration is given by fmin = 0 [It is minimum] 1.7 ENERGY OF A PARTICLE EXECUTING SHM AND LAW OF CONSERVATION OF ENERGY When a particle executes simple harmonic motion, it possesses both kinds of mechanical energy, namely potential and kinetic energy at any instant of time. Kinetic Energy (Ek) In case of a particle of mass m and velocity v which is executing SHM, the kinetic energy is given by 1 Ek = __ mv2 2 Now, y = a sin (w t + d) ______ ÷ dy y2 \ v = ___ = aw cos (w t + d) = ± aw 1 – __2 dt a ______ or, v = ± w ÷a – y 2 2 1 \ Ek = __ mw 2 (a2 – y2)...(1.14) 2 Potential Energy (EP) In case of a particle of mass m and velocity v which is executing SHM, the poten- tial energy is given by y 1 Ep = Ú (mw 2y) dy = __ mw 2y2...(1.15) o 2 Because the potential energy of an oscillating particle at any instant of time can be calculated from the total amount of work done in overcoming the effect of the restoring force, in this case the restoring force is given by, F = – mw 2y When the particle comes to an extreme point, its kinetic and potential energies become Ek = Ek(min) = 0 1 and Ep = Ep(max) = __ mw 2 a2 [ in this case, y = a] 2 Similarly, when it is at the mean position, its kinetic and potential energies are given by 1.8 Basic Engineering Physics 1 Ek = Ek(max) = __ mw 2 a2 2 and Ep = Ep(min) = 0 So, the kinetic energy is maximum when the potential energy is minimum and vice versa. Total Energy At any instant of time, the total energy of the oscillating particle is given by E = Ek + Ep 1 1 or, E = __ mw 2 (a2 – y2) + __ mw 2y2 2 2 1 or, E = __ ma2w 2 = 2p2 ma2 n2...(1.16) E 2 E = Ek + Ep where w = 2pn \ E = constant. Ep Hence, we can conclude that in case of simple harmonic oscillation the total energy is conserved. And it is propor- tional to the square of the amplitude of Ek oscillation. The energy distribution of a simple harmonic oscillator has been –a O +a y shown in Fig. 1.2. Both kinetic and poten- Fig. 1.2 The energy distribution curve of an SHO. y is tial energies have been plotted against displacement, Ep is potential energy and Ek is the displacement. kinetic energy. 1.8 SUPERPOSITION OF WAVES It is an experimental fact that two or more waves can traverse through the same space independent of each other. For this reason, the displacement of any particle at a given instant of time is simply the algebraic sum of the individual displacements, this process of vector addition of the displacement of a particle is known as superposition. Principle of Superposition If the wave equations governing the wave motions are linear, the displacement of any particle, at a particular instant of time, is simply the algebraic sum of the individual displacements due to the different waves. 1.8.1 Superposition of Two Collinear SHMs of Frequency n Let two collinear simple harmonic motions be represented by the following two equations: y1 = a1 cos (w t – f1) and y2 = a2 cos (w t – f2) where y1 and y2 are instantaneous displacements, a1 and a2 are the amplitudes, the w (= 2pn) is, the angular velocity and f1 and f2 are the phases. The resultant amplitude at any instant of time is given by y = y1 + y2 or, y = a1 cos (w t – f1) + a2 cos (wt – f2) or, y = a1 cos wt cos f1 + a1 sin wt sin f1 + a2 sin wt cos f2 + a2 sin wt sin f2 or, y = (a1 cos f1 + a2 cos f2) cos w t + (a1 sin f1 + a2 sin f2) sin w t  Simple Harmonic Motion 1.9 Now, putting A cos f = a1 cos f1 + a2 cos f2 and A sin f = a1 sin f1 + a2 sin f2 where A and f are two constants and given by _______________________________________ A = ÷(a1 cos f1 + a2 cos f2)2 + (a1 sin f1 + a2 sin f2)2 _________________________ or, A = ÷a12 + a22 + 2a1 a2 cos (f1 – f2) a1 sinf1 + a2 sin f2 and ( f = tan–1 _________________ a1 cos f1 + a2 cos f2 ) \ y = A cos f cos w t + A sin f sin w t or, y = A cos (wt – f)...(1.17) Thus, it is observed that the resultant motion is also a simple harmonic motion having the same component motions Special Cases (a) If f1 = f2 then A = a1 + a2 ______ p (b) If f1 ~ f2 = __ then A = ÷a21 + a22 2 (c) If f1 ~ f2 = p then A = a1 ~ a2 (d) If f1 ~ f2 = p and a1 = a2 then A = 0, i.e., the two SHMs destroy each other. If instead of two SHMs, there remain several motions of different amplitudes and phases but of the same w ( time period T = ___ 2p ) ___ ( ) w or frequency, n = 2p then the resultant motion can be obtained in the same process. In such cases, we will get y = y1 + y2 + y3 +... + yn or, y = a1 cos (w t – f1) + a2 cos (w t – f2) + a3 cos (wt – f3) +... + an cos (w t –fn) or, y = (a1 cos f1 + a2 sin f2 + a3 cos f3 +... + an cos fn) cos w t + (a1 sin f1 + a2 sin f2 sin f3 +... + an sin fn) sin w t or, y = A cos (wt – f)...(1.17) _________________________ ÷( S n n ) (S ) 2 2 where A= ai cos fi + ai sin fi i=1 i=1 n and 1.8.2 n ( ) S ai sin fi f = tan–1 __________ i=1 S ai cos fi i=1 Superposition of Two Mutually Perpendicular SHMs of Frequency n Let us consider two SHMs of amplitudes a1 and a2 having same angular velocity w (= 2p n) which act respec- tively along x- and y-axes. Then their amplitudes can be represented by x = a1 cos (wt – f1) and y = a2 cos (wt – f2) 1.10 Basic Engineering Physics Now, we can write y __ = cos (w t – f2) a2 = cos {w t – f1) + (f1 – f2)} y __ or, a2 = cos (w t – f1) cos (f_______ 1 – f2) – sin (w t – f1) sin (f1 – f2) ÷( ) y __ __ x x2 __ or, a2 = a1 cos (f1 – f2) – 1 – a2 sin (f1 – f2) 1 _______ ÷( ) y __ __ x x2 or, 2 – a ◊ cos (f1 – f2) = – 1 – __2 sin (f1 – f2) a 1 a1 Squaring both sides of the equation, we get y2 __ y2 _____ __ + a22 a21 x2 cos 2 (f1 – f 2 ) – 2xy _____ a1 a2 cos (f 1 – f 2 ) ( ) = 1 – x2 __ a21 sin2 (f1 – f2) __ 2xy x2 __ or, – cos (f – f ) + {cos2 (f1 – f2) + sin2 (f1 – f2)} = sin2 (f1 – f2) a22 a1 a2 1 2 a21 y2 _____ __ 2xy x2 __ or, – cos (f – f ) + = sin2 (f1 – f2) a22 a1 a2 1 2 a12 y2 __ xy _____ x2 __ or, – 2 a1 a2 cos f + = sin2 f...(1.18) a22 a12 where f = f1 – f2 Equation (1.18) is a general equation of an ellipse bounded within a rectangle with sides 2a1 and 2a2. Thus the resultant motion (having combined two mutually perpendicular SHMs) is represented by an ellipse. Let us now consider the special cases of this generalized motion. Case 1 If f1 – f2 = 0 or f = 0 then sin f = 0 and cos f = 1 \ Eq. (1.18) gets reduced to x2 __ ___ y2 xy + – 2 ____ a =0 2 a 1 a2 2 1a2 d 2a2 y or, ( x __ __ a1 – a2 )2 =0 a2 or, y = __ a1 x 2a1 This represents a straight line passing through the origin a2 Fig. 1.3 The resultant vibration of two and making an angle of inclination d = tan–1 __ a1 to the x-axis mutually perpendicular vibrations. It takes place in a straight line (Fig. 1.3). which is inclined at an angle d with p the x-axis and d < __ when f = 0. 2  Simple Harmonic Motion 1.11 p Case 2 If f1 – f2 = p/2 or f = __, then sin f = 1 and 2 cos f = 0. Equation (1.18) reduces to 2a2 x2 __ ___ y2 + =1 a12 a22 This equation represents an ellipse which is sym- metrical about the two axes and a1 and a2 are the semi- 2a1 axes (Fig. 1.4). Case 3 If f1 – f2 = p or f = p then sin f = 0 and Fig. 1.4 The resultant of two mutually perpendicular cos f = – 1 vibrations represents an ellipse. \ Eq. (1.18) gets reduced to y2 __ xy ____ x2 ___ + 2 a1 a2 a 2 = 0 + a22 1 y or, ( x __ __ a1 + a2 ) =0a 2 y = – __ 2 or, a x 1 This equation represents a straight line which passes through the origin making an angle of inclination a2 ( ) d = tan–1 – __ a1 to the x-axis (Fig. 1.5). Case 4 If f1 – f2 = p/2 or f = p/2 and a1 = a2 (=a) (say) then sin f = 1 and cos f = 0. In this case, Eq. (1.18) gets reduced to x2 + y2 = a2 This equation represents a circle which is symetrical about the two axes with a as radius (Fig. 1.6). d 2a2 2a 2a1 2a Fig. 1.5 The resultant of two mutually perpendicular Fig. 1.6 The resultant of two mutually vibrations represents a straight line which perpendicular vibrations represents p p is inclined at an angle d where d > __ when a circle when f = __ and a1 = a2. 2 2 f = p. 1.12 Basic Engineering Physics 1.9 DERIVATION OF THE EQUATION OF MOTION FROM THE LAW OF CONSERVATION OF ENERGY The total (mechanical) energy of a simple harmonic oscillator is given by 1 1 E = __ mv2 + __ mw2y2...(1.19) 2 2 where E is the constant of motion. Now getting differentiated on both sides of Eq. (1.19) with respect to time, we get dv dy mv ___ + mw2y ___ = 0 dt dt or, ( dv ) mv ___ + w2y = 0 dt __ ( ) dy d ___ dt dt + w2y = 0 d2y ___ or, + w2y = 0...(1.20) dt2 Equation (1.20) is the same differential equation as Eq. (1.7). 1.10 TWO VIBRATIONS IN A PLANE (OF COMMENSURATE FREQUENCIES) ACTING AT RIGHT ANGLES TO EACH OTHER Case 1 If the frequencies of the two vibrations be of the ratio 1:2 with initial phase difference f, then we can represent the two vibrations by the following two equations: x = a1 cos (w t) and y = a2 cos (2w t + f) Now, we can write y __ a2 = cos (2 w t) cos f – sin (2 w t) sin f y __ or, a2 = (2 cos 2 w t – 1) cos f – 2 sin w t cos w t sin f _______ ( ) ÷( ) y __ x2 __ x2 x or, a2 = 2 a 2 – 1 cos f – 2 1 – ___2 __ sin f 1 a1 a1 ________ ( ) ÷( ) y __ 2 x2 ____ x2 x or, a2 – – 1 cos f = – 2 1 – ___2 __ sin f a12 a1 a1 Now, having squared both sides of the equation, we get y2 ( ) ( ) ( ) 2 __ __ 2 x2 y ____ 2x2 ___ 2 x2 ___ __ x2 2 – 2 a 2 – 1 cos f + 2 – 1 cos f = 4 1 – 2 2 sin2 f a2 2 a1 a1 a1 a1  Simple Harmonic Motion 1.13 or, ___ 2 a1 a1 ( 4x2 ___ x2 __y y – cos f – 1 + __ 2 a2 a 2 ) + cos f ( ) =0 2...(1.21) The above equation gives the general equa- tion of the resultant motion for any phase dif- ference and amplitudes. If f = 0, Eq. (1.21) gets reduced to 2a2 ( ) 2 2x2 __ ___ y – – 1 =0...(1.22) a 2 a2 1 Equation (1.22) represents two coincident 2a1 parabolas (Fig. 1.7). p Fig. 1.7 Resultant of two mutually perpendicular vibrations If f = __, Eq. (1.21) gets reduced to 2 with frequency ratio = 1 : 2 and phase difference 2 f = 0 represents coincident parabolas. ( ) 2 2 y 4x ___ ___ x __ – 1 + = 0...(1.23) a21 a12 2 2 Equation (1.23) is a 4th degree equation and represents a curve which has two loops as has been shown in Fig. 1.8. 2a2 Case 2 If the frequencies of two vibrations be of the ratio 1 : 3 with initial phase differ- ence f then we can represent the two vibra- tions by the following two equations: x = a1 cos (w t) 2a1 and = a2 cos (3w t + f) Fig. 1.8 Resultant of two mutually perpendicular vibrations Now, we can write p with frequency ratio 1 : 2 and phase difference f = __ y __ = cos (3w t + f) 2 a2 represents a two loop curve. y or, __a2 = cos 3 w t cos f – sin 3w t sin f y or, __ 3 3 a2 = (4 cos w t – 3 cos w t) cos f – (3 sin w t – 4 sin w t) sin f { ( ) ( )} 1 __ 3 __ or, __ y a2 = ( 4x3 ___ ___ – a31 a1 3x )cos f – 3 1 – x2 2 __ a12 – 4 1 – x2 2 __ a12 sin f 1 __ or, y __ ( 4x3 ___ ___ 1 3x ) x2 __ a2 = a3 – a1 cos f – 1 – a2 1 ( )( ) 2 4x2 ___ a12 – 1 sin f...(1.24) Now, transposing and squaring, we get { ( ) } ( )( ) 2 2 y __ 4x3 ___ ___ 3x x2 4x2 a2 – a3 – a1 cos f = 1 – ___2 ___ –1 sin2 f 1 a1 a12 If f = 0, then Eq. (1.24) gets reduced to 1.14 Basic Engineering Physics { ( )} 2 y __ 4x3 ___ ___ 3x a2 – a3 – a1 =0...(1.25) 1 Equation (1.25) represents two coincident cubic curves (Fig. 1.9). p If f = __, Eq. (1.24) gets reduced to 2 2 y2 ( )( ) x2 4 2 1 – ___2 ___ a1 a21 – 1 – __2 = 0 a2...(1.26) 2a2 2a1 Fig. 1.9 Resultant of two mutually Fig. 1.10 Resultant of two mutually perpendicular perpendicular vibrations with vibrations with frequency ratio 1:3 and frequency ratio 1:3 and phase p phase difference f = __ represents a curve of difference f = 0 represent two 2 coincident cubic curves. three loops. Equation (1.26) is a 6th degree equation and it gives a curve of three loops (Fig. 1.10). If one changes the phase difference gradually, then the shape of the loop gradually changes. For a ratio of frequencies of 1 : n, the curve will have n loops. 1.10.1 Lissajous Figures Definition Lissajous figures are those curves which are generated by superimposing two simple harmonic motion acting at right angles to each other. The constituent simple harmonic motions may have different time periods, different amplitudes and also different initial phases. The size (or dimension) of the resultant curve depends on their amplitudes but the shape of the curve depends on the ratio of their time periods and the initial phase differ- ences. These figures may be experimentally generated by (a) Blackburn’s pendulum, (b) opti- cal method, and (c) cathode ray oscillograph, etc. The diagram of an oscillograph is shown in Fig. 1.11. Fig. 1.11 Cathode-ray oscillograph.  Simple Harmonic Motion 1.15 Figures 1.7, 1.8, 1.9, and 1.10 are examples of Lissaious figures. Uses of Lissajous Figures (a) Lissajous figures can be used for determination of the ratio of the frequencies of two mutually per- pendicular superposing vibrations. (b) They also can be used to determine the unknown frequency of a tuning fork. (c) The nature and shape of a signal can be known with the help of Lissajous’ figures. (d) The amplitude of a signal also can be determined with the help of these figures. Worked-out Examples Example 1.1 The displacement of a body of mass 2 g executing simple harmonic motion is indicated by p p ( ) y = 10 sin __ t + ___ cm. Calculate the (a) amplitude, (b) angular velocity, (c) time period, (d) maximum 3 15 and minimum velocity and maximum and minimum acceleration, (e) epoch, (f) kinetic energy, and (g) poten- tial energy. Is its energy conserved? Sol. The general equation of a simple harmonic oscillator is given y = a sin (w t + d)...(1) And in the present case, it is p p ( y = 10 sin __ t + ___ 3 15 )...(2) Now, comparing eqs. (1) and (2), we get (a) The amplitude, a = 10 cm, p (b) The angular velocity, w = __ rad s–1 3 2p 3 (c) The time period, T = ___ or, T = 2p × __ p=6s w ______ (d) The velocity of the oscillating body at any time is given by v = ±w ÷a2 – y2 ______ Hence, | vmax = ± w ÷a2 – y2 | max = |± w a| [for y = 0 ] p i.e., vmax = aw = 10 × __ = 10.46 cm/s 3 Similarly, the minimum velocity is given by vmin = 0 cm s–1 [for y = a] The acceleration at any time t is given by f = w 2y \ fmax = aw 2 [for y = a] p2 or, 3.14 fmax = 10 × __ = 10 × ____ 9 3 ( ) 2 or, fmax = 10.95 cm s–2 1.16 Basic Engineering Physics And the minimum acceleration fmin is given by fmin = 0 cm s–1 [for y = 0] (e) The phase of the oscillating body is given by, f = (w t + d) Now, we know that epoch is the initial phase at t = 0 p \ from Eq. (2), we get d = + ___ 15 (f) The kinetic energy is given by 1 Ek = __ mw2 (a2 – y2) 2 So, the value of kinetic energy will vary with y. (g) The potential energy is given by 1 Ep = __ mw2y2 2 \ the value of the potential energy varies as y2. At any time, the total energy is given by E = Ep + Ek 1 1 or, E = __ mw2y2 + __ mw2 (a2 – y2) 2 2 1 1 1 or, E = __ mw2y2 + __ mw2a2 – __ mw2y2 2 2 2 1 \ E = __ mw2a2 2 which is independent of the variables t and y. So, it is a constant w.r.t. time. Hence, the total energy of the oscillator is conserved. Example 1.2 A particle is executing SHM. At an instant of time its displacement is 12 cm, velocity is 5 cm s–1 and when its displacement is 5 cm, the velocity is 12 cm s–1. Calculate its (i) amplitude, (ii) fre- quency, and (iii) time period. Sol. The velocity of a particle executing SHM is given by ______ dy v = ___ = w ÷a2 – y2 dt In the first case, ______ v1 = w ÷a2 – y12 –1 where v1 = 5 cm s and y1 = 12_______ cm or, 5 = w ÷a2 – 144...(1) In the second case ______ v2 = w ÷a2 – y22 –1 where v2 = 12 cm s , y2 = 5 cm______ \ 12 = w ÷a2 – 25...(2)  Simple Harmonic Motion 1.17 Dividing (1) by (2), we get _______ ___ ÷a2 – 144 5 _________ = ______ 12 ÷a2– 25 a2 – 144 25 _______ ____ or, = 2 144 a – 25 \ the amplitude is 13 cm. Now, substituting the value of a = 13 cm in Eq. (1), we get ________ 5 = w ÷132 – 144 ___ or, 5 = w ÷25 \ w = 1 rad s–1 w 1 The frequency n = ___ = ___ Hz 2p 2p 1 The time period is T = __ = 2p s n Example 1.3 Show that for a particle executing SHM, its velocity at any instant of time is given by ______ dy ___ = w ÷a2 – y2 dt Sol. The displacement of a particle executing SHM is given by y = a sin (w t + f)...(1) The velocity at any instant of time is given by, dy ___ = aw cos (w t + f)...(2) dt ______________ dy or, ___ = aw ÷1 – sin2 (w t + f) dt __________________ dy or, dt ÷ ___ = w a2 – {a sin (w t + f)}2 ______ dy or, ___ = w ÷a2 – y2 [by Eq. (1)] dt Example 1.4 The motion of a particle executing SHM is given by y = a sin w t. If it has a speed u when the displacement is y1, and a speed v when the displacement is y2, show that the amplitude of the motion is [ ] v2y21 – u2y22 __21 a = __________ v2 – u2 Sol. We have, y = a sin w t dy1 ______ \ u = ___ = w ÷a2 – y12...(1) dt dy ______ v = ___ = w ÷a2 – y22 2 and...(2) dt Now, squaring both the equations, and dividing, we have 1.18 Basic Engineering Physics a2 – y12 u2 _______ __ = v2 a2 – y22 or, u2a2 – u2y22 = v2a2 – v2y12 or, a2 [u2 – v2] = u2y22 – v2y12 u2 y22 – v2y21 or, a2 = ___________ u2 – v2 or, [ a = __________ v2 – u2 ] v2y21 – u2y22 __21 ______ Example 1.5 Show that for a particle executing SHM, the instantaneous velocity is w ÷a2 – y2 and the instantaneous acceleration is – w2y where w is the angular frequency, a is the amplitude and y is the instan- taneous displacement. Sol. The displacement of a particle executing SHM is given by y = a cos (w t + d)...(1) The instantaneous velocity is dy v = ___ = – w a sin (w t + d)...(2) dt From Eq. (1) y cos (w t + d) = __ a ______ ÷ ______________ y2 \ sin (w t + d) = ÷1 – cos2 (w t + d) = 1 – __2 a ______ ÷ y2 or, – aw sin (w t + d) = – aw 1 – __2 a ______ aw or, v = – __ (÷a – y2 ) ____ 2 [by Eq.(2)] ÷a2 _______ \ v = ± w ÷a2 – y2) Now, differentiating Eq. (2), we get dv f = ___ = – w2 a cos (w t + d) dt or, f = – w2y [by Eq. (1)] Example 1.6 Calculate the time period of the liquid column of length l in a U-tube, if it is depressed in one arm by x, d is the density of the liquid and A is the cross-sectional area of the arm of the U-tube. [WBUT 2007] Sol. The U-tube as has been described in the question has been shown in Fig. 1.12.  Simple Harmonic Motion 1.19 The liquid column of a liquid with density d has been shown in the diagram. The height of the liquid column in each arm is equal to l. The liquid column has been depressed in the left arm through a depth of x. So, in the right arm it has risen through a height of x. So the difference in the levels of the two arms is 2x. The liquid column of length 2x in the right arm will try to bring back the two levels to their initial values. The force acting in this process will be F = (2x) Adg This force (F) acts on the liquid of both columns with length l. d2x \ (2l Ad) ___2 = – (2x) Adg [here, m = 2l Ad] dt 2 xg or, ___x = – ___ d x 2 l 2x dt g x or, () x + __ x = 0 l l g \ [ x + w2x = 0 where w2 = __ l ] This is the equation of a simple harmonic oscillator. Fig. 1.12 A U-tube with vibrating liquid \ its time-period T is given column. __ ÷ 2p l T = ___ = 2p __ w g Example 1.7 9 kg of mercury is poured into a glass U-tube with a uniform internal diameter of 1.2 cm. It oscillates freely about its equilibrium position. Calculate the time period of oscillation of the mercury column. Sol. Let l be the height of the mercury column in each arm and m be the mass of liquid of the two col- umns. Then m = 2l Ad where A is the area of crossection and d is the density m \ l = ____ 2Ad The mass m = 9 kg, d = 13.6 × 103 kg m–3 ( ) 1.2 2 A = 3.14 × ___ cm2 = 1.1304 × 10–4 m2 2 9 ____________________________ \ l= = 2.927 m 2 × (1.1304 × 10–4) × (13.6 × 103) \ the time period of oscillation T is given by __ ÷ l T = 2p __g _____ ÷ 2.927 or, T = 2p × _____ 9.8 \ T = 3.43 s 1.20 Basic Engineering Physics Example 1.8 Two particles execute SHM with same amplitude and frequency along two parallel straight lines. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. What is the phase difference between them? Sol. Let the equation of motions of the two particles be as follows: y = a sin w t...(1) and y = a sin (w t + f)...(2) a when y = __ for the first particle, we get 2 a __ = a sin w t [from Eq. (1)] 2 1 fi sin w t = __...(3) 2 a Again, when y = __ for the second particle, we get [from Eq. (2)] 2 a __ = a sin (w t + f) 2 1 __ or, = sin w t cos f + cos w t sin f 2 ____ ÷ 1 __ __ 1 1 or, = cos f + 1– __ sin f [by Eq. (3)] 2 2 4 __ or, 1 = cos f + ÷3 sin f __ or, 1 – cos f = ÷3 sin f or, 1 – 2 cos f + cos2 f = 3 sin2 f or, 1 – 2 cos f + cos2 f = 3 – 3 cos2 f or, 4 cos 2 f – 2 cos f – 2 = 0 or, 2 cos2 f – cos f – 1 = 0 or, (2 cos f + 1) (cos f – 1) = 0 Now, (cos f – 1) = 0 will give the value of f which equals to zero. So, this value of f cannot be accepted as the particles will have the same phase. Hence, (2 cos f + 1) = 0 1 or, cos f = – __ = cos 120° 2 Therefore, the phase difference between the two oscillators is 120°. Example 1.9 The displacement of a moving particle at any time t is given by y = a sin w t + b cos w t Show that the motion is simple harmonic. Sol. The equation representing the displacement of the particle is given by y = a sin w t + b cos w t...(1) dy ___ or, = aw cos w t – bw sin w t dt  Simple Harmonic Motion 1.21 d2y ___ or, 2 = – aw 2 sin w t – bw2 cos w t dt d2y ___ or, 2 = – w 2 (a sin w t + b cos w t) dt d2y ___ 2 = – w 2y [by Eq. (1)] dt d2y ___ or, + w 2y = 0...(2) dt2 Equation (2) is the standard differential equation of a simple harmonic oscillator. Hence, the given equation [i.e., Eq. (1)] represents a simple harmonic motion. Example 1.10 Calculate the displacement to amplitude ratio for a SHM when the kinetic energy is 90% of the total energy. Sol. Let m be the mass of the oscillator, a be the amplitude and w be the angular frequency. So, the total energy of the oscillator is given by 1 E = __ mw 2a2 2 Let y1 be the displacement when its kinetic energy Ek is 90% of the total energy. Now, the potential energy is given by 1 Ep = __ mw 2y21 2 1 __ E mw 2y21 y21 10 2 ___p = ________ \ fi __2 = ____ E __ 1 a 100 mw 2 a2 2 ___ \ y1 : a = ÷0.1 = 0.316 Review Exercises Part 1: Multiple Choice Questions 1. The SI unit of the force constant of a spring is given by (a) Nm (b) Nm–2 (c) Nm–1 (d) N 2. Which of the following is not essential for simple harmonic motion? (a) Inertia (b) Gravity (c) Restoring force (d) Elasticity 3. The velocity of a particle executing simple harmonic motion is minimum at a point where displace- ment is (a) zero (b) maximum (c) midway between zero and maximum (d) None of these 1.22 Basic Engineering Physics 4. If two SHMs of the same amplitude, time period and phase act at right angles to each other, then the resultant vibration is (a) elliptical (b) circular (c) str