PREFACE ii Pre-U STPM Text Physics Term 3 is written based on the new syllabus prepared by the Malaysian Examinations Council (MEC) that has been implemented since 2012. The book is well designed and organised with the following features to help students understand the concepts taught. STPM Scheme of Assessment Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 15 30 Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 30 1—2 hours assessment Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 1—1 2 hours Central assessment Paper 5 Written practical test 3 compulsory structured questions to be answered. First, Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 00 PRELIMS PHYSICS T3 (2018).indd 5 3/19/18 8:58 AM 19 Physics Term 3 STPM Chapter 19 Oscillations Physics Term 3 STPM 19 R Bilingual Keywords OSCILLATIONS Amplitude: Amplitud Angular frequency: Frekuensi sudut Equilibrium position : Kedudukan keseimbangan Extension: Pemanjangan Force constant: Pemalar daya Characteristics Damped Oscillation Forced Oscillation and Resonance Systems in SHM Kinematics Energy Simple Harmonic Motion 1 Concept Map INTRODUCTION 1. There are various types of periodic motion, examples: the vibration of a guitar string and the vibration of air molecules when sound moves through the air. 2. Atoms in a solid vibrate. In electromagnet waves such as light waves, oscillations occur too. When an alternating current fl ows in a circuit, electrical charge oscillates. 3. In this chapter we will discuss oscillation in mechanical systems such as simple pendulum and a loaded spring. These are examples of simple harmonic motion (SHM). 19 PHYSICS T3 (2018).indd 1 3/19/18 8:56 AM R 2 compulsory structured questions to be 2 questions to be answered out of 3 essay All questions are based on topics 1 to 11. 15 30 15 compulsory multiple-choice questions Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. Written practical test 3 compulsory structured questions to be answered. Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. There are various types of periodic motion, examples: the vibration of a guitar string and the vibration of air molecules when sound moves through the air. Atoms in a solid vibrate. In electromagnet waves such as light waves, oscillations occur too. When an alternating current fl ows in a circuit, electrical charge oscillates. In this chapter we will discuss oscillation in mechanical systems such as simple pendulum and a loaded spring. These are examples of simple harmonic motion (SHM). iv Analysis of STPM Papers (2014 - 2017) Chapter 2014 2015 2016 2017 A B C A B C A B C A B C 19 Oscillations 2 1 2 1 2 1 2 20 Wave Motion 2 2 2 2 21 Sound Waves 3 2 1 2 1 2 1 22 Geometrical Optics 2 1 1 1 2 1 1 1 23 Wave Optics 2 3 1 2 1 3 1 24 Quantum Physics 3 1 1 3 1 3 3 1 25 Nuclear Physics 2 1 2 2 1 2 1 Total 15 2 3 15 2 3 15 2 3 15 2 3 00 PRELIMS PHYSICS T3 (2018).indd 4 3/28/18 12:29 PM 1 1 2 Summary of Key, Quantities and Units QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Amount of matter Amaun jirim n mol Electric current Arus elektrik I A Length Panjang l m Mass Jisim m kg Temperature Suhu T K Time Masa t s Other Quantities Kuantiti Lain Acceleration Pecutan a m s–2 Acceleration of free fall Pecutan jatuh bebas g m s–2 Activity of radioactive source Aktiviti sumber radioaktif A s–1, Bq Amplitude Amplitud A m Angular frequency Frekuensi sudut ω rad s–1 Area Luas A m2 Atomic mass Jisim atom ma kg Atomic number (proton number) Nombor atom (nombor proton) Z Capacitance Kapasitans C F Change of internal energy Perubahan tenaga dalam ∆U J Change carrier density Ketumpatan pembawa cas u Conductivity Kekonduksian σ Ω–1 m–1 Critical angle Sudut genting θc ° Current density Ketumpatan arus J A m–2 Decay constant Pemalar reputan λ s–1 Density Ketumpatan ρ kg m–3 Electric charge Cas elektrik Q, q C Electric fi eld strength Kekuatan medan elektrik E N C–1 Electric fl ux Fluks elektrik fi N C–1 m2 Electric potential Keupayaan elektrik V V Electric potential difference Beza keupayaan elektrik V V Electromotive force Daya gerak elektrik ε, E V Electron mass Jisim electron me kg, u Elementry charge Cas keunsuran e C Emissivity Keterpancaran e Energy Tenaga E, U J Focal length Panjang focus f m Force Daya F N Force constant Pemalar daya k N m–1 Frequency Frekuensi f, v Hz Gravitational fi eld strength Kekuatan medan graviti g N kg–2 Gravitational potential Keupayaan graviti V J kg–1 Half-life Setengah hayat t 1/2 s Heat Haba Q J Heat capacity Muatan haba C J K–1 Image distance Jarak imej v m Impedance Impedans Z Ω 361 Summary.indd 361 3/19/18 9:00 AM 3 compulsory structured questions to be 2 questions to be answered out of 3 essay All questions are based on topics 19 to 25. Written practical test School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 19 Oscillations 2 1 2 1 1 2 3 55 20 Physics Term 3 STPM Chapter 20 Wave Motion 20.2 Waves Intensity Students should be able to: • defi ne intensity and use the relationship I ∝ A2 • describe the variation of intensity with distance of a point source in space Learning Outcome 1. Energy is transferred outward from the source by progressive waves. 2. The intensity, I of a wave is the rate at which energy is transferred by the wave to a unit area perpendicular to the direction of propagation of the wave. 3. For a sinusoidal wave to move in a medium, particles of the medium should perform simple harmonic motion of total energy 1 2 mω2 A2 where ω = angular frequency and A = amplitude of oscillation. 4. Hence, the intensity I of a wave is directly proportional to the square of the amplitude. Intensity I ∝ (Amplitude)2 5. Figure 20.7 shows a point wave source emitting wave in all directions. If P is the energy emitted per second, then the intensity at a distance r from the source S is I = P 4πr2 (area of sphere of radius r is 4πr2 ) I ∝ 1 r2 Quick Check 2 1. Two progressive waves are represented by the equations y1 = 2 sin (5t – 3x) and y2 = 3 sin (5t – 3x) where y1 and y2 are in cm and t in seconds. Compare the intensity of y1 to y2 . A point source emits waves in all directions. The intensity at a distance 2.0 m from the source is 10 W m a distance of (a) 1.0 m, and (b) 3.0 m from the source? 20.3 Principle of Superposition This principle states that the resultant displacement produced by two waves at a point is the vector sum of the individual displacements of the two waves. r I S 2008/P1/Q11, 2016/P3/Q6, 2017/P3/Q3 VIDEO Superposition of Waves 20 PHYSICS T3 (2018).indd 55 3/28/18 12:18 PM This principle states that the resultant displacement produced by two waves at a point is the vector sum of the individual displacements of the two waves. 20.3 Principle of Superposition A point source emits waves in all directions. The intensity at a distance 2.0 m from the This principle states that the resultant displacement produced by two waves at a point is the vector sum of the individual displacements of the two waves. The intensity at a distance 2.0 m from the 20.3 Principle of Superposition This principle states that the resultant displacement produced by two waves at a point is the vector sum of the individual displacements of the two waves. source is 10 W m S (a) 1.0 m, and (b) 3.0 m from the source?
101 Physics Term 3 STPM Chapter 21 Sound Waves Physics Term 3 STPM Chapter 21 Sound Waves (d) When the source approaches the stationary obser the observer is ver, the apparent frequency detected by higher than the actual frequency. This explai approaching siren of ambulance. ns for the higher pitch of the 4. Source moving away from a stationary observer P S us us Figure 21.15 (a) Figure 21.15 shows a sound source which produces sound of frequency f moving away from a stationary observer with a velocity of uS (b) In 1.0 s, . f wavefronts are produced. Distance moved by the wavefront is produced when the source has moved a distance fi rst wavefront is uS Hence, distance between f wavefronts is (v + uS ). (c) The apparent wavelength λ″ = v + uS f The apparent frequency detected by the observer f ″ = v λ″ f ″ = ( v v + uS ) f < f Example 12 The frequency of an ambulance’s siren is 600 Hz. Calculate the apparent frequency heard by a stationary observer (a) when the ambulance approaches him with a speed of 30 m s–1. (b) when the ambulance moves away with a speed of 30 m s–1. (Speed of sound = 330 m s–1) Solution: (a) When the ambulance approaches the observer, the apparent frequency, f ′ = ( v v – uS)f = ( 330 330 – 30) 600 = 660 Hz (b) When the ambulance moves away from the observer, apparent frequency, f ″ = ( v v + uS)f = ( 330 330 + 30) 600 = 550 Hz Exam Tips Apparent frequency increases when source stationary observer, but approaches from the observer. decreases when source moves away 21 PHYSICS T3 (2018).indd 101 3/19/18 9:10 AM Physics Term 3 STPM Chapter 22 Geometrical Optics 152 1 1. v = +5.0 cm Image is enlarged, upright and virtual. 2. v = –13.3 cm Image is virtual, upright and magnifi ed. 3. r = 30.0 cm m = v u , r = 2f 1 v + 1 u = 1 f 4. (a) v = 21.0 cm Image real, inverted and diminished. (b) v = 30.0 cm Image real, inverted and same size as object. (c) v = –30.0 cm Image virtual, upright and magnifi ed. 5. (a) v = –5.83 cm (b) m = –0.42 6. (a) Concave mirror because only concave mirror can produced a magnifi ed image. Image is virtual because it is upright. (b) r = 40.0 cm 1 v + 1 u = 1 f , r = 2f 7. (a) Image is real, hence concave mirror. (b) r = +13.3 cm. v = 10.0 cm, u = 20.0 cm 8. (a) When there is non-parallax between the optical pin and its image, they are at the same position. There is no relative movement between the optical pin and its image when viewed from diff erent angles. (b) (i) u = v = 40.0 cm 1 v + 1 u = 1 f , 1 f = 1 40.0 + 1 40.0 The radius of the bubble is r. An object O is at a distance L from the end T of the tube. Light is refl ected from the surface of the soap bubble. Glass tube Soap bubble L T O r (i) If the distance of the image of O formed by refl ection of light is at a distance s from the end T, derive an expression for s in terms of L O is the centre of curvature of the spherical surface and OP = r the radius of curvature. A point object X is at a distance u from P. The image produced by the refraction of light at the spherical surface is at a distance v from P. n2 O P X n1 (i) Write an expression to relate u, v, r, n1 and n2 . Explain clearly the sign convention you use for r, u, and v. (ii) If r = 20.0 cm, u = 15.0 cm, n1 = 1.0 and n2 = 1.5, calculate the value of v and state whether the image is real or virtual. (iii) An image of X due to refl ection of light from the spherical surface is also formed. Calculate the distance between this image and the image in (c) (ii) above. 22 ANSWERS 22 PHYSICS T3 (2018).indd 152 Example 12 The frequency of an ambulance’s siren is 600 Hz. Calculate the apparent frequency heard by a stationary observer (a) when the ambulance approaches him with a speed of (b) when the ambulance moves away with a speed of 30 m s 30 m s (Speed of sound = 330 m s Solution: (a) When the ambulance approaches the observer, the apparent frequency, (b) When the ambulance moves away from the observer, apparent frequency, f ″ = ( = ( = 550 Hz Image is enlarged, upright and virtual. Image is virtual, upright and magnifi ed. r = 2f Image real, inverted and diminished. Image real, inverted and same size as object. Image virtual, upright and magnifi ed. ANSWERS Physics Term 3 STPM Chapter 23 Wave Optics 189 23 5. In Figure 23.26 (b) X and Y are also sources of light far from the single slit. The angle subtended at the slit, θ is too small for the diffraction patterns to be distinguishable. The two central maxima cannot be detected. The images are not resolved. 6. Rayleigh’s criterion states that for the images to be just resolved the central maximum of one image falls on the fi rst minimum of the other image (Figure 23.27). First minimum First minimum Resultant intensity Figure 23.27 Rayleigh’s criterion 7. Hence for the two images to be resolved, the minimum angle θmin subtended by the sources at the slit equals the angular separation between the central maximum and the fi rst minimum of the single-slit diffraction pattern. sin θmin = λ a where λ is the wavelength and a is the width of the slit. This ratio sin θmin = λ a is known as the resolving power of the aperture. 8. The wavelength λ of light is very small compared to the width of the slit a (λ << a). Since θmin is small, and sin θmin ≈ θmin in radians. Hence resolving power, θmin = λ a 9. Most optical instruments have circular aperture. The resolving power for a circular aperture of diameter a is given as θmin = 1.22 λ a Info Physics Electron microscope has very high resolving power, the angle θmin, is very much smaller than that of optical microscope. Electron microscope uses a beam of electrons to ‘illuminate’ the specimen. The wavelength of an electron beam is 1/104 , the wavelength of light. Hence the value of θmin for an electron microscope is 1/104 that of an optical microscope. Example 12 The diameter of the pupil of a person’s eye is 2 mm. The average wavelength of light is 550 nm. (a) Estimate the resolving power of the eye. (b) Two parallel lines that are separated by a distance of 1 cm are drawn on a piece of card. What is the maximum distance of the card from the eye for the images of the lines to be resolved? 23 PHYSICS T3 (2018).indd 189 3/19/18 10:09 AM iii (a) Figure 21.15 shows a sound source which produces sound of frequency f moving away wavefronts are produced. Distance moved by the fi rst wavefront is wavefront is produced when the source has moved a distance uS v ″ wavefronts are produced. Distance moved by the fi rst wavefront is , is very much smaller than that of optical microscope. Electron microscope uses a beam of electrons to ‘illuminate’ the specimen. The wavelength of an electron beam is that of an optical microscope. The diameter of the pupil of a person’s eye is 2 mm. The average wavelength of light is 550 nm. (b) Two parallel lines that are separated by a distance of 1 cm are drawn on a piece of card. What is the maximum distance of the card from the eye for the images of the lines to be 352 STPM Model Paper (960/3) Paper 3 Kertas 3 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fi fteen questions in this section. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam bahagian ini. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. What is the condition for a simple pendulum to be simple harmonic? Apakah syarat bagi sebuah bandul ringkas melakukan gerakan harmonik ringkas? A The pendulum bob must be heavy. Ladung bandul perlu berat. B The length of the string must be long. Tali bandul perlu panjang. C The amplitude of oscillation must be small. Amplitud ayunan perlu kecil. D The period of oscillation must be short. Tempoh ayunan perlu pendek. 2. A particle is in simple harmonic motion of amplitude A. At time mean position O and moving in the negative t = 0, the particle is at the Suatu zarah adalah dalam gerakan harmonik ringkas dengan amplitud A. Pada masa t = 0, zarah di titik x direction as shown in the diagram below. keseimbangan dan bergerak menuju arah negatif x seperti ditunjukkan dalam rajah di bawah. –A +A O t = 0 Which graph correctly shows the variation of displacement, x with time t? Graf yang manakah menunjukkan dengan betul ubahan sesaran x dengan masa t? A t 0 x C 0 x t B 0 x t D 0 x t 26 STPM Model Paper.indd 352 3/19/18 12:17 PM 1/104 , the wavelength of light. Hence the value of The diameter of the pupil of a person’s eye is 2 mm. The average wavelength of light is 550 nm. (b) Two parallel lines that are separated by a distance of 1 cm are drawn on a piece of card. What is the maximum distance of the card from the eye for the images of the lines to be The diameter of the pupil of a person’s eye is 2 mm. The average wavelength of light is 550 nm. (b) Two parallel lines that are separated by a distance of 1 cm are drawn on a piece of card. What is the maximum distance of the card from the eye for the images of the lines to be Physics Term 3 STPM Chapter 22 Geometrical Optics 147 22 Important Formulae 1. For spherical mirrors: focal length, f = r 2 , r = radius of curvature 2. Spherical mirror equation and thin lens equation: 1 f = 1 u + 1 v Linear magnifi cation, m = hi h0 = v u = v f – 1 3. Refractive index of medium 1 to medium 2, n1 = sin i sin r n1 sin θ1 = n2 sin θ2 4. Refraction at a spherical surface: n1 u + n2 v = |n2 – n1| r 5. Lens maker equation: 1 f = (n – 1) ( 1 r1 + 1 r2 ) and 1 f = ( n2 n1 – 1) ( 1 r1 + 1 r2 ) 22 STPM PRACTICE 1. The radius of curvature of a concave mirror is 10.0 cm. An object is placed 20.0 cm from the mirror. What is the linear magnifi cation C 1 3 D 1 2 2. An object is 60 cm from a lens. An image of the same size as the object is formed on a screen behind the lens. Which is the lens? A diverging lens of focal length 30 cm B diverging lens of focal length 15 cm C converging lens of focal length 15 cm D converging lens of focal length 30 cm The image I2 , is at the pole P of the convex mirror. It acts as an object for the mirror, with u = 0. Hence the image which is a spot of light is at P. The spot of light at P now acts as a real object for the concave lens. Light refl ected by the mirror passes through the lens with u = +20.0 cm f = –20.0 cm and 1 v = 1 –20 – 1 –20 v = –10.0 cm which is at I1 (from (b) above) The refl ected light after passing through lens 2 seems to come from I object for lens 1 with u f 1 v = v The refl ected light after passing lens 2 is focussed by lens 1 at O, the position of the object. Hence, the fi nal image is at the same position as the object. 22 PHYSICS T3 (2018).indd 147 3/28/18 12:25 PM – 1 –20 = –10.0 cm The refl ected light after passing through lens 2 seems to come from I The refl ected light after passing lens 2 is focussed by lens 1 at O, the position of the object. Hence, the fi nal image is at the same position as the object. = –10.0 cm The apparent frequency detected by the observer an expression for formed by refl ection of light is at a distance s from the end T, derive s in terms of (d) When the source the observer is higher approaching siren of ambulance.
iv Analysis of STPM Papers (2014 - 2017) Chapter 2014 2015 2016 2017 A B C A B C A B C A B C 19 Oscillations 2 1 2 1 2 1 2 20 Wave Motion 2 2 2 2 21 Sound Waves 3 2 1 2 1 2 1 22 Geometrical Optics 2 1 1 1 2 1 1 1 23 Wave Optics 2 3 1 2 1 3 1 24 Quantum Physics 3 1 1 3 1 3 3 1 25 Nuclear Physics 2 1 2 2 1 2 1 Total 15 2 3 15 2 3 15 2 3 15 2 3
STPM Scheme of Assessment v Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 960/3 Physics Paper 3 Oscillations and Waves, Optics and Modern Physics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 960/5 Physics Paper 5 Written Physics Practical Written practical test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 To be scaled to 45 (20%) Throughout the three terms School-based assessment
Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 19 OSCILLATIONS 1 19.1 Characteristics of Simple Harmonic Motion 2 19.2 Kinematics of Simple Harmonic Motion 7 19.3 Energy in Simple Harmonic Motion 11 19.4 Systems in Simple Harmonic Motion 15 19.5 Damped Oscillation 28 19.6 Forced Oscillations and Resonance 30 Praktis STPM 19 35 QQ1 41 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 20 WAVE MOTION 46 20.1 Progressive Waves 48 20.2 Wave Intensity 55 20.3 Principle of Superposition 55 20.4 Standing Waves 60 20.5 Electromagnetic Waves 64 Praktis STPM 20 69 QQ2 73 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 21 SOUND WAVES 76 21.1 Propagation of Sound Waves 77 21.2 Sources of Sound 79 21.3 Intensity Level of Sound 94 21.4 Beats 96 21.5 Doppler Effect 100 Praktis STPM 21 105 QQ3 110 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 22 GEOMETRICAL OPTICS 115 22.1 Spherical Mirrors 116 22.2 Refraction at Spherical Surfaces 127 vi 22.3 Thin Lenses 131 Praktis STPM 22 147 QQ4 152 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 23 WAVE OPTICS 157 23.1 Huygens’s Principle 158 23.2 Interference 160 23.3 Two-Slit Interference Pattern 165 23.4 Interference in a Thin Film 179 23.5 Diffraction by a Single Slit 184 23.6 Diffraction Gratings 192 23.7 Polarisation 203 23.8 Optical Waveguides 212 Praktis STPM 23 215 QQ5 221 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 24 QUANTUM PHYSICS 227 24.1 Photons 228 24.2 Wave-particle Duality 241 24.3 Atomic Structure 247 24.4 X-Rays 263 24.5 Nanoscience 275 Praktis STPM 24 276 QQ6 284 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 25 NUCLEAR PHYSICS 292 25.1 Nucleus 293 25.2 Radioactivity 301 25.3 Nuclear Reaction 322 Praktis STPM 25 342 QQ7 346 STPM Model Paper 352 Summary of Key Quantities and Units 361 CONTENTS
19 Physics Term 3 Physics Term 3 STPM Chapter 19 Oscillations 19 CHAPTER Bilingual Keywords OSCILLATIONS 19 Amplitude: Amplitud Angular frequency: Frekuensi sudut Equilibrium position : Kedudukan keseimbangan Extension: Pemanjangan Force constant: Pemalar daya Periodic motion: Gerakan berkala Phase angle: Sudut fasa Phase difference: Beza fasa Simple harmonic motion: Gerakan harmoni ringkas Simple pendulum: Bandul ringkas Characteristics Damped Oscillation Forced Oscillation and Resonance Kinematics Energy Systems in SHM Simple Harmonic Motion 1 Concept Map INTRODUCTION 1. There are various types of periodic motion, examples: the vibration of a guitar string and the vibration of air molecules when sound moves through the air. 2. Atoms in a solid vibrate. In electromagnet waves such as light waves, oscillations occur too. When an alternating current fl ows in a circuit, electrical charge oscillates. 3. In this chapter we will discuss oscillation in mechanical systems such as simple pendulum and a loaded spring. These are examples of simple harmonic motion (SHM).
19 2 Physics Term 3 STPM Chapter 19 Oscillations 19.1 Characteristics of Simple Harmonic Motion Students should be able to: • defi ne simple harmonic motion Learning Outcome 1. Definition: A body is in simple harmonic motion (SHM) if its acceleration, a is directly proportional to its displacement, x from a fi xed point and is always directed towards that point, O. Acceleration a = –ω2 x where ω2 = constant ω is known as the angular frequency of the SHM. The negative sign shows the direction of the acceleration is always directed towards the fi xed point, O as shown is fi gure 19.1(a). 2. Figure 19.1(b) shows the graph of acceleration a against displacement x. 3 Figure 19.2 shows a simple pendulum. The point O, the position of the pendulum when in equilibrium is known as the point of equilibrium. When the bob of the pendulum is at O, it is in equilibrium. 4. The pendulum oscillates between the points X and Y. The maximum distance A from the equilibrium position is known as the amplitude. 5. The period T of an oscillation is the time taken for a complete oscillation, from X to Y and back to X. 6. The frequency f is the number of complete oscillations per second. Frequency f = 1 T , T = period Also, angular frequency, ω = 2πf frequency, f = ω 2π and period, T = 2π ω The unit of frequency is hertz (Hz). Example 1 A body of mass 0.15 kg moves in simple harmonic motion along a straight line. The fi gure shows the relation between the force F on the body and its displacement x from a point O. a ω ω –A A A +A (b) O a –A +A (a) Figure 19.1 X Y O A A Figure 19.2 : Simple pendulum -0.2 0 3.0 -3.0 0.2 x / m F / N
19 3 Physics Term 3 STPM Chapter 19 Oscillations Find (a) the amplitude, (b) the angular frequency, (c) the frequency, (d) the period of the oscillation. Solution: (a) Amplitude = maximum displacement = 0.2 m (b) Using F = ma = – mω2 x –mω2 = Gradient of graph –(0.15) ω2 = 3.0 – 0 –0.2 – 0 = –15 N m–1 Angular frequency, ω = 15 0.15 = 10 s–1 (c) ω = 2πf Frequency, f = ω 2π = 10 2π Hz = 1.59 Hz (d) Period, T = 1 f = 1 1.59 s = 0.628 s Quick Check 1 1. Defi ne simple harmonic motion. 2. A particle moves in simple harmonic motion between the 24.0 cm mark and 36.0 cm mark as shown in the fi gure. It takes 0.50 s to move from the 24.0 cm mark to the 36.0 cm mark. 24.0 28.0 32.0 36.0 (a) Where is the equilibrium position of the particle? (b) What is (i) the amplitude, (ii) the period, (iii) the frequency of the motion. (c) What is the displacement, x of the particle from the equilibrium position when the particle is at (i) the 24.0 cm mark, (ii) the 28.0 cm mark, (iii) the 30.0 cm mark, (iv) the 34.0 cm mark. (d) As the particle moves from the 24.0 cm mark to the 36.0 cm mark, what is the sign (+ or – ) of the acceleration when it is at (i) the 24.0 cm mark, (ii) the 28.0 cm mark, (iii) the 30.0 cm mark, (iv) the 34.0 cm mark.
19 4 Physics Term 3 STPM Chapter 19 Oscillations Displacement and Time 1. From the definition of simple harmonic motion, acceleration a = d2 x dt 2 = –ω2 x is a second order differential equation. Its general solution is x = Asin (ωt + φ) .............................. a or x = Acos (ωt + φ) .............................. b where A = amplitude, (ωt + φ) is known as the phase angle, φ is a constant whose value is determined by the initial conditions. That is the value of x when time t = 0. 2. To show that x = A sin (ωt + φ) is a general solution, we differentiate x = A sin (ωt + φ) with respect to time t. Velocity v = dx dt = ωA cos (ωt + φ) and acceleration a = d2 x dt 2 = –ω2 [A sin (ωt + φ)] = –ω2 x Hence, x = A sin (ωt + φ) is a general solution of the equation d2 x dt 2 = –ω2 x. 3. Similarly, you can show that x = A cos (ωt + φ) is also a general solution of the equation d2 x dt 2 = –ω2 x 4. To determine the value of φ, we should first consider the case of the particle at the equilibrium position, x = 0 at time t = 0. (Figure 19.3) From the equation, x = A sin (ωt + φ) 0 = A sin (0 + φ) sin φ = 0 φ = 0 Hence, displacement, x = A sin ωt. velocity, v = dx dt = ωA cos ωt = ωA sin (ωt + π 2 ) acceleration a = d2 x dt2 = –ω2 A sin ωt 5. Figure 19.4 shows how the displacement x, the velocity v, and the acceleration a varies with time t. Exam Tips For SHM: • Velocity v leads displacement x by π 2 rad. • Acceleration a in antiphase with displacement, x, • phase difference = π radians. –A O A Figure 19.3 Figure 19.4 ω2 A –ω2 A ωA –ωA A –A (a) (b) (c) Acceleration, a Velocity, v Displacement, x Time, t Time, t Time, t
19 5 Physics Term 3 STPM Chapter 19 Oscillations 6. From the equations x = Asin ωt and v = Asin (ωt + π 2 ) the phase angle for x is ωt and the phase angle for v is (ωt + π 2 ). There is a phase difference between v and x of (ωt + π 2 ) – ωt = π 2 radian. The velocity v is said to lead the displacement x by π 2 radian. 7. This phase difference between velocity v and the displacement x can also be seen from the graphs shown in Figures 9.5(a) and 9.5(b). The velocity is maximum at t = 0, and the displacement is maximum T 4 later. T 4 is equivalent to a phase difference of π 2 radian. 8. Comparing the equations displacement, x = Asin ωt, acceleration, a = –ω2 Asin ωt = ω2 Asin (ωt + π) and also from the graphs is Figures 19.4(a) and 19.4(c), there is a phase difference of π radian between the acceleration a and the displacement x. The acceleration a and the displacement x are said to be in antiphase. 9. Using the same method as discussed above, you should be able to show that (a) if at time t = 0, x = A (Figure 19.5(a)) the equation for x is x = Acos ωt (b) if at time t = 0, x = –A (Figure 19.5(b)), the equation for x is x = –Acos ωt Exam Tips Equations for displacement (a) –A O A If at t = 0, x = 0, then x = Asin ωt (b) –A O A If at t = 0, x = A, then x = Acos ωt (c) –A O A If at t = 0, x = –A, then x = –Acos ωt Example 2 A particle performs simple harmonic motion of amplitude 0.05 m and the period of the motion is 12 s. At time t = 0, the particle is at the equilibrium position. Deduce an expression for the displacement x of the particle in terms of the time t. Solution: Since at time t = 0, x = 0, then x = A sin ωt Amplitude A = 0.05 m Period T = 2π ω ω = 2π T = 2π 12 = π 6 –A O A (a) –A O A (b) Figure 19.5 Hence, x = A sin ωt = 0.05 sin π 6 t
19 6 Physics Term 3 STPM Chapter 19 Oscillations Example 3 Figures (a) and (b) show how the displacement x, and the acceleration a of a body in simple harmonic motion varies with time t. a (a) (b) What is the period T of the motion? Solution: From Figure (a), amplitude, A = 2 m From Figure (b) amax =18 m s–2 Using amax = ω2 A ω2 = 18 2 = 9 ω = 3 Period T = 2π ω = 2π 3 = 2.09 s Example 4 A particle P performs simple harmonic motion of amplitude a and period T. The fi gure on the right shows the position of the particle at time t = 0. What is the displacement of the particle from the point O at time t = 5T 8 ? Solution: From the fi gure, at time t = 0, x = a Hence, x = Acos ωt T = 2π ω = a cos ( 2π T t ) ω = 2π T When t = 5T 8 , x = a cos ( 2π T . 5T 8 ) = a cos ( 5π 4 ) = – a 2 The particle is at distance a 2 below the point O. P O a
19 7 Physics Term 3 STPM Chapter 19 Oscillations 19.2 Kinematics of Simple Harmonic Motion Learning Outcomes Students should be able to: • show that x = A sin ωt is a solution of a = −ω2 x • derive and use the formula v = ±ω A2 – x2 • describe, with graphical illustrations, the variation in displacement, velocity and acceleration with time • describe, with graphical illustrations, the variation in velocity and acceleration with displacement Velocity and Displacement 1. From the defi nition of simple harmonic motion, acceleration, a = – ω2 x and a = dv dt = dv dx . dx dt = v dv dx v dv dx = – ω2 x v dv = – ω2 x dx Integrating, ∫v dv = ∫– ω2 x dx v2 2 = – ω2 x2 2 + c v2 = – ω2 x2 + c1 When x = A and velocity v = 0 0 = – ω2 A2 + c1 ∴ c1 = ω2 A2 v2 = – ω2 x2 + ω2 A2 v = ± ω A2 – x2 When x = 0, at the equilibrium position, velocity is maximun, vmax = ωA –A 0 ωA –ωA A O –A x = 0 +A (a) (b) Figure 19.6 2. For each displacement, other than at x = ±A, there are two velocities. The velocity is positive if it is in the positive direction, and negative in the opposite direction (Figure 19.6(a)). 3. Figure 19.6(b) is the velocity-displacement graph. When x = ±A, the velocity = 0. 2009/P1/Q13, 2010/P1/Q10, 2012/P1/Q10, 2013/P3/Q1,Q2, 2014/P3/Q1 2015/P3/Q1, 2017/P3/Q1
19 8 Physics Term 3 STPM Chapter 19 Oscillations Example 5 A body of mass 0.15 kg moves in simple harmonic motion along a straight line. The fi gure shows the relation between the force F and the displacement x. Find (a) the amplitude, (b) the period of the motion, (c) the maximum speed of the body. Solution: (a) Amplitude = maximum displacement = 0.2 m (b) Using F = ma F =–mω2 x Gradient of graph = –mω2 –mω2 = – 3 0.2 = –15 N m–1 ω2 = 15 0.15 = 100 ω = 10 Period T = 2π ω = 2π 10 = π 5 = 0.628 s (c) vmax = ωA = 10 × 0.2 = 2.0 m s–1 Quick Check 2 1. Which of the following graphs best represents the relation between the acceleration a of a body in simple harmonic motion and the displacement x from the equilibrium position? A C B D 2. The acceleration a of a body is related to its displacement x from a fi xed point O as shown in the graph. Which of the following is the graph of speed v against displacement x? A C
19 9 Physics Term 3 STPM Chapter 19 Oscillations B D 3. What is the frequency of a simple harmonic motion represented by the equation: acceleration = –ω2 x where x is the displacement from a fixed point? A ω C 2π ω B 2πω D ω 2π 4. A particle performs simple harmonic motion of amplitude 0.020 m and frequency 2.5 Hz. Its maximum speed is A 0.007 m s–1 B 0.125 m s–1 C 0.157 m s–1 D 0.314 m s–1 5. The maximum speed of a particle which performs simple harmonic motion of amplitude 2.0 × 10–3 m and period 0.10 s is A 3.2 × 10–5 m s–1 B 2.0 × 10–4 m s–1 C 2.0 × 10–2 m s–1 D 1.3 × 10–1 m s–1 6. A particle in simple harmonic motion completes n oscillations per second. What is its angular frequency? A 1 n rad s–1 C 2π n rad s–1 B 2πn rad s–1 D n 2π rad s–1 7. A particle moves with simple harmonic motion along the x-axis according to the equation d2x dt2 + Ax = 0 The period of the motion is A 2π A C 2π 2 A B 2 A π D A 2π 8. The acceleration a of a particle in simple harmonic motion as a function of its displacement x are given in the table below. x/cm –2 –1 0 1 2 a/cm s–2 8 4 0 –4 –8 The period of the motion is A π s C 2 π s B 1 π s D π 2 s 9. When a particle performs simple harmonic motion, the velocity leads the displacement by a phase angle of A π 4 rad C 3π 4 rad B π 2 rad D π rad 10. A particle performs simple harmonic motion about a point O with amplitude a and peroid T. Its displacement from O at time T 8 after passing O is A 1 8 a C 1 2 2 a B 1 2 a D 1 2 a 11. The graphs below show how the displacement x and velocity v of a body vary with time t when it is performing simple harmonic motion. What is the value T? A π 9 s C π 3 s B 2π 9 s D 2π 3 s
19 10 Physics Term 3 STPM Chapter 19 Oscillations 12. The displacement-time graph of a particle in simple harmonic motion is as shown above. Which of the following gives the correct value for the frequency and amplitude? Frequency/kHz Amplitude/μm A 5 2 B 25 2 C 25 4 D 50 2 13. The displacement-time graph of a particle in simple harmonic motion is as shown above. Which of the following is the velocity-time graph? A B C D 14. The figure shows the displacement-time graph of an oscillating particle. P R What does the length PR represent? A Half the period B Twice the frequency C Half the frequency D Twice the period 15. The displacement-time graph of a body performing simple harmonic motion is shown below. At which of the points A, B, C, or D is the body travelling and accelerating in opposite directions. 16. A particle in simple harmonic motion of amplitude A completes N oscillations in time t. What is the expression for the maximum speed vmax of the particle? A vmax = 2πtA N C vmax = 2πNt A B vmax = 2πNA t D vmax = NA 2πt 17. A particle is in simple harmonic motion along the x-axis. The motion of the particle is represented by the equation d2 x dx2 = – kx. Which statement about the motion of the particle is correct? A The angular velocity of the motion equals k . B The velocity of the particle is directly proportional to x2 . C The period of the motion is directly proportional to k. D The acceleration of the particle is always in the negative-x direction.
19 11 Physics Term 3 STPM Chapter 19 Oscillations 18. The graph below shows how the acceleration of an object undergoing simple harmonic motion varies with time. (a) Deduce from the graph (i) the period (ii) the frequency (iii) the angular frequency (iv) the amplitude of the oscillation (b) Sketch a graph to show how the displacement varies with time. 19. The velocity-displacement graph for a body performing simple harmonic motion is as shown below. (a) Deduce from the graph the value of the maximum velocity. (b) Explain why there are two values of velocity for zero displacement. Explain why there are two values of x for zero velocity. 19.3 Energy in Simple Harmonic Motion Students should be able to: • derive and use the expressions for kinetic energy and potential energy. • describe, with graphical illustrations, the variation in kinetic energy and potential energy with time and displacement. Learning Outcomes 1. When there is no loss of energy from a system performing simple harmonic motion, the total energy E which consists of the kinetic energy K and potential energy U is constant. Total energy E = K + U The kinetic energy, K = 1 2 mv 2 v2 = ω2 (A2 – x2 ) K = 1 2 mω2 (A2 – x2 ) 2. The maximum kinetic energy = 1 2 mω2 A2 This occurs when x = 0, at the equilibrium position. 3. At the equilibrium position, the potential energy is assummed zero, U = 0. Hence, the total energy is completely in the form of kinetic energy. From E = K + U At the equilibrium position, U = 0. E = 1 2 mω2 A2 + 0 Hence, total energy E = 1 2 mω 2 A2 A = amplitude 2009/P1/Q11, 2015/P3/Q18
19 12 Physics Term 3 STPM Chapter 19 Oscillations At any other displacement x, E = K + U 1 2 mω2 A2 = 1 2 mω2 (A2 – x2 ) + U ∴ U = 1 2 mω2 A2 – 1 2 mω2 (A2 – x2 ) Potential energy U = 1 2 mω2 x2 When x = ± A, the potential energy is maximum, Umax = 1 2 mω2 A2 4. Figure 19.7 shows how the three forms of energy vary with displacement, x. +A A A –A Figure 19.7 : Variation of energy with displacement Variation of Energy with Time 1. Suppose that the displacement x of a body performing simple harmonic motion is given by x = A sin ωt where t = time 2. Then the variation with respect to time of (a) the potential energy U = 1 2 mω2 x2 U = 1 2 mω2 A2 sin2 ωt (b) the kinetic energy K = 1 2 mω2 (A2 – x2 ) = 1 2 mω2 [A2 – A2 sin2 ωt] = 1 2 mω2 A2 (1 – sin2 ωt) K = 1 2 mω2 A2 cos2 ωt
19 13 Physics Term 3 STPM Chapter 19 Oscillations 3. Figure 19.8 shows the variation of energy with time. A A Figure 19.8 : Variation of energy with time Example 6 A particle of mass 4 kg moves in simple harmonic motion and its potential energy U varies with displacement x as shown in the fi gure. What is the period of oscillation? Solution: From the graph, amplitude = 0.2 m maximum potential energy = 1.0 J 1 2 mω2 A2 = 1.0 1 2 × 4 × (ω2 )(0.2)2 = 1.0 ω2 = 1 0.08 ω = 3.54 T = 2π ω = 2π 3.54 s = 1.78 s Example 7 A particle moves such that its potential energy U varies with the square of its displacement r from a point O, i.e. U ∝ r2 as shown in the graph. Sketch a graph to show how the force F on the particle varies with the displacement r? Solution: Potential energy, U ∝ r2 U = kr2 k = constant Using force F = – dU dr = –2kr Hence, the graph of F against r is shown on the right. 0 1.0 r / m U / J – 0.2 0.2 0 r / m F – 0.2 0.2
19 14 Physics Term 3 STPM Chapter 19 Oscillations Quick Check 3 1. Which of the following graphs shows correctly the relation between the potential energy V, the kinetic energy K, and the total energy T of a body moving in a straight line with simple harmonic motion? A B C D 2. In which group below do all three quantities of a particle in simple harmonic motion remain constant? A Force Total energy Amplitude B Total energy Amplitude Angular frequency C Amplitude Angular Acceleration frequency D Acceleration Force Total energy 3. A body performs simple harmonic motion with period 2.0 s. After being displaced, it is released at time t = 0. Which of the following graphs shows correctly the variations of the kinetic energy K, and the potential energy V of the body with time? A B C D 4. A particle of mass m performs simple harmonic motion with amplitude a. Its maximum kinetic energy is E. Which of the following represents the angular frequency? A 2E ma2 C E ma2 B 2E ma D ma E 5. A body is in simple harmonic motion. Its displacement x at time t from the equilibrium position is given by x = Asin ωt. On the same time-axis, sketch graphs to show (a) the variation of velocity v with time, (b) the variation of kinetic energy with time.
19 15 Physics Term 3 STPM Chapter 19 Oscillations 6. (a) Calculate the gain in potential energy when a mass of 150 g is raised through 1 mm. (Take g = 10 m s–2) A simple pendulum has a bob of mass 150 g. The variation of V, the potential energy, with x, the horizontal displacement of the bob is shown below. The bob is raised vertically through 1.0 mm, with the string taut, and then released. (b) What is the amplitude of oscillation of the pendulum? (c) Sketch labelled graphs to show the variation of (i) the total energy (ii) the kinetic energy with x, as the pendulum oscillates. 19.4 Systems in Simple Harmonic Motion Students should be able to: • derive and use expressions for the periods of oscillations for spring-mass and simple pendulum systems. Learning Outcome Simple Pendulum 1. Figure 19.9 shows a simple pendulum of length l. When the string makes an angle q with the vertical, the forces on the pendulum bob are (a) its weight, mg vertically downwards, and (b) the tension, T in the string. The weight mg has two components • mg cos q in the opposite direction of T, • mg sin q in the negative direction. 2. Using F = ma –mg sin q = ma Acceleration, a =–g sin q Assuming that the angle q is small sin q = q rad = x l Hence, a = –g( x l ) = –( g l ) x = – ω2 x That is the motion of simple harmonic. ω = g l and period, T = 2π ω T = 2π g l O a Figure 19.9 : Simple Pendulum Alternative Method: Using F = m d2 x dt 2 = –mg sin q (x = l q) = d2 (lq) dt 2 = – g sin q Assuming that the angle q is small sin q = q rad Therefore, l d2 q dt 2 = –gq d2 q dt 2 = –( g l )q = –ω2 q ω = g l 2008/P2/Q9, 2010/P1/Q12, 2010/P10, 2011/P2/Q9, 2015/P3/Q16, 2017/P3/Q2
19 16 Physics Term 3 STPM Chapter 19 Oscillations 3. The period T is independent of the mass m of the bob, and the amplitude of oscillation. 4. The fact that its period does not depend on the amplitude of oscillation, the simple pendulum is suitable as a clock. When the amplitude decreases, its period is still the same. 5. A simple pendulum can be used to determine the value of g, the acceleration of free fall. From the equation, T = 2π l g T2 = 4π2 l g 6. The value of g is determined by measuring the period T for pendulum of various lengths l. A graph of T2 against l is then plotted (Figure 19.10). The gradient of the graph = 4π2 g Acceleration due to gravity, g = 4π2 Gradient of graph Figure 19.10 : Graph of T 2 against l Info Physics The pendulum is used to measure time because its period is independent with the amplitude of oscillation. Example 8 When a small sphere is hung from the end of an elastic string, the length of the simple pendulum obtained is 40.0 cm. The time for 20 small oscillations of this pendulum is 26.0 s. The bob is then changed to a sphere of the same size but different mass. The new time for 20 oscillations is 26.4 s. Explain the observation and calculate the ratio of the masses of the bob. Solution: The period of a simple pendulum T = 2π l g When bobs of different mass hang from the elastic string, the extended length of the string differs. Since the time for 20 oscillations increases from 26.0 s to 26.4 s, the new length is longer. Hence, the mass of the new bob is bigger.
19 17 Physics Term 3 STPM Chapter 19 Oscillations For the fi rst sphere, T1 = 26.0 20 = 2π l 1 9.81 l1 = 41.98 cm Extension of string, x1 = (41.98 – 40.00) = 1.98 cm mg = kx m1 g = kx1 .............................. a For the second sphere, T2 = 26.4 20 = 2π l 2 9.81 l2 = 43.29 cm ∴ x2 = 3.29 cm m2 g = kx2 .............................. b a ÷ b m1 m2 = 1.98 3.29 = 0.602 Example 9 (a) What is the frequency of a simple pendulum of length 1.0 m? (b) What is its frequency in a lift (i) which is accelerating upwards at 2.0 m s–2? (ii) which is falling freely? Solution: (a) From T = 2π l g Frequency, f = 1 T = 1 2π g l = 1 2π 9.81 1.0 = 0.499 Hz. (b) (i) When the lift is accelerating upwards with an acceleration a = 2.0 m s–2 g1 = (g + a) = (9.81 + 2.0) = 11.81 m s–2 Frequency, f 1 = 1 2π 11.81 1.0 = 0.547 Hz (ii) When the lift falls freely, a = g g2 = (g – a) = (g – g) = 0 Hence, frequency, f2 = 1 2π 0 1.0 = 0
19 18 Physics Term 3 STPM Chapter 19 Oscillations Quick Check 4 1. When the length of a simple pendulum is doubled, the ratio of the new frequency to the old frequency is A 2 C 1 2 B 2 D 1 2 2. The displacement x of a particle at time t is given by x/m = 5 sin (2t/s). The length of the simple pendulum that has the same period as this particle is A 10.0 m C 2.5 m B 5.0 m D 2.0 m (Take g = 10 m s–2) 3. The acceleration of free fall on the Moon is 1 6 that on the Earth. If the period of a simple pendulum on the Earth’s surface is 1.0 s, what is its period on the Moon? A 1 6 s C 6 s B 1 6 s D 6 s 4. The period of a simple pendulum in a stationary lift is T0 . When the lift descends at a steady speed, the period is T1 , and when the lift accelerates uniformly downwards, the period is T2 . Which of the following is correct? A T0 = T1 < T2 B T0 = T1 > T2 C T0 < T1 < T2 D T0 > T1 > T2 5. A simple pendulum has a period T at a place where the acceleration of free fall is g. At another place, where the acceleration of free fall is g′, its period is T′. What is the ratio of T ′ T ? A g′ g C g′ g B g g′ D g g′ 6. A simple pendulum is suspended from the ceiling of a lift. As the pendulum swings freely, the lift accelerates uniformly upwards. Which of the following is true? A Its frequency remains unchanged. B It oscillates faster. C Its amplitude increases. D It stops oscillating. 7. The displacement x at time t of a particle in simple harmonic motion is given by x = 4.0 cos 2πt, where x is in metre and t in seconds. What is the lowest value of t when the kinetic energy of the particle equal its potential energy? A 0.13 s C 0.25 s B 0.15 s D 0.50 s 8. A student is under the impression that ω, the angular frequency of a simple pendulum is depended only on the length l of the pendulum and the mass m of its bob. Use dimensional analysis to show that this is incorrect. Derive from fi rst principle an expression for the angular frequency ω. 9. (a) Sketch the velocity-time graph of a body performing simple harmonic motion. Assume that at time = 0, the body is at the equilibrium position. (b) Explain how the displacement of the body from the equilibrium position could be deduce from the velocity-time graph. Hence, sketch the displacementtime graph for the body. (c) A simple pendulum is suspended from the ceiling of a room. The table below shows the period t of the pendulum when its length is changed. The length of the pendulum is not measured, but the height x of the pendulum bob from the fl oor is recorded. x/cm 10 40 80 120 160 t/s 3.38 3.20 2.95 2.66 2.34 By a graphical method, determined the value of the (i) acceleration of free fall, and (ii) the height of the ceiling. What precautions you would take when measuring the period T of the pendulum.
19 19 Physics Term 3 STPM Chapter 19 Oscillations Mass Spring System-Vertical Spring 2014/P3/Q18, 2015/P3/Q2 1. Figure 19.11 shows a spiral spring extended by e when a mass m hangs in equilibrium. 2. Assuming that the spring obeys Hooke’s law, tension in the spring, T0 = ke = mg .............................. a where k = force constant of spring. The mass is then pulled down a distanceA (A e) from its equilibrium and then released, it oscillates with an amplitude = A. Suppose that at the instant t, the displacement of the mass m from the equilibrium position is x, then using F = ma mg – T1 = ma T1 = k(x + e) mg – k(x + e) = ma ke – k(x + e) = ma Acceleration, a = – k m x ≡ –ω2 x Hence, the mass m is in simple harmonic motion, ω = k m Period, T = 2π ω T = 2π m k = 2π e g since mg = ke 3. During the derivation above, the following assumptions were made: • The spring obeys Hooke’s law, and the elastic limit of the spring is not exceeded. • The amplitude of oscillation Ae, the extension of the spring when the mass m hangs in equilibrium. • The mass of the spring is negligible. 4. If the effective mass s of the spring is not negligible, the formula for the period T is T = 2π m + s k T 2 = 4π2 m k + 4π2 s k The effective mass, s of the spring can be determined experimentally by plotting a graph of T 2 against m (Figure 19.12). From the graph, When m = 0, T2 = 4π2 s k Hence, s = k (value of interception on T2 -axis) 4π2 a Figure 19.11 : Loaded vibrating spring Figure 19.12 INFO Simple Harmonic Motion
19 20 Physics Term 3 STPM Chapter 19 Oscillations Example 10 The period of vertical oscillation of a mass m at the end of a light helical spring of force constant k (Figure (a)) is T1 . (a) Write an expression for T1 in terms of m and k. (b) The spring is cut into two pieces of equal length and one portion is used to support the same mass m (Figure (b)). What is the period T2 in terms of T1 ? (c) Both portions of the spring are used in parallel in Figure (c). What is the period T3 in terms of T1 ? Solution: (a) T1 = 2π m k (b) The spring constant for half the spring, k2 = 2k T2 = 2π m 2k = 1 2 T1 (c) When two halves of the spring are in parallel, k3 = 2k2 = 4k T3 = 2π m 4k = 1 2 T1 Example 11 The fi gure shows a mass m on a pan hanging from a vertical spring. The pan is moving vertically in simple harmonic motion with an amplitude of 10 cm. Above a certain frequency, the mass ceases to remain in contact with the pan. (a) Find the lowest frequency when this happens. (b) At this minimum frequency, at what point in the motion does contact cease? (Take g = 10 m s–2) Solution: (a) Using F = ma, the normal reaction R of the pan on the mass is given by mg – R = ma R = m(g – a) When contact between the mass and the pan ceases, R = 0, then a = g. The maximum acceleration for SHM is amax = ω2 A Hence, ω2 (0.10) = 10 ω = 10 Lowest frequency, f = ω 2π = 10 2π = 1.59 Hz (b) Contact ceases when the pan is at the highest point so that the direction of the acceleration is downwards. (a) (b) (c) a
19 21 Physics Term 3 STPM Chapter 19 Oscillations Quick Check 5 1. When a mass m is suspended from an elastic string of unstretched length a, the string extends to a total length l. The mass is pulled down a small distance b and released. It performs simple harmonic motion, described by the equation x .. + ω2 x = 0, where x is the displacement, and ω2 is A g (l – a) C gb (l – a) B g (l – b) D ga (l – a) 2. a l m x b The original length of an elastic string is a. The string extends to a total length of l when a mass m hangs from it. When the mass is moved down a small distance b [b < (l – a)] and then released, it oscillates. Assuming that the time t is zero at the instant the mass is released, the equation that describes the subsequent motion is A x = b cos ωt C x = (l – a) sin ωt B x = b sin ωt D x = (b + l – a) cos ωt 3. When a mass of 0.10 kg hangs in equilibrium from a light helical spring, the extension of the spring is 0.10 m. The mass is pulled down a distance of 0.02 m and then released. Taking g as 10 m s–2, the equation that relates the distance x of the mass from the equilibrium position at time t after it is released is A x/m = 0.1 cos [0.2 π(t/s)] B x/m = 0.02 sin [0.2 π(t/s)] C x/m = 0.02 cos [0.1 (t/s)] D x/m = 0.02 cos [10(t/s)] 4. A mass of 0.10 kg at the end of a light helical spring performs simple harmonic motion with period of 1.0 s. How much mass must be added for the period of oscillation to be 3.0 s? A 0.10 kg C 0.80 kg B 0.30 kg D 0.90 kg 5. A mass hanging from the free end of a light helical spring performs simple harmonic motion. Which graph represents how T, the tension in the spring varies with x, the displacement of the mass from the equilibrium position? A C B D 6. A mass on the end of a light helical spring is given a vertical displacement of 3.0 cm from its rest position and then released. If the subsequent motion is simple harmonic motion with a period of 2.0 s, through what distance will the bob move in (a) the fi rst 1.0 s, (b) the fi rst 0.75 s. 7. A mass suspended from one end of a helical spring performs vertical simple harmonic motion with an amplitude of 2.0 cm. If three complete oscillations are made in 4.0 s, what is the acceleration of the mass at (a) the equilibrium position, (b) the position of maximum displacement? 8. A mass rests on a horizontal platform which is performing vertical simple harmonic motion with an amplitude 50 mm. Above a certain frequency, the particle ceases to remain in contact with the platform. (a) Find the lowest frequency at which this occurs.
19 22 Physics Term 3 STPM Chapter 19 Oscillations (b) At this minimum frequency, which point in the motion does contact cease? [Take g = 10 m s–2] 9. (a) When a mass M hanging in equilibrium from the free end of a vertical spring, the extension produced is e. The mass is displaced slightly and then released. (i) State two conditions required for the subsequent motion to be simple harmonic. (ii) Show that the period of oscillation is given by T = 2π e g where g = acceleration of free fall. 2014 (b) When the mass M = 0.040 kg, the vertical displacement y of the mass from the equilibrium position varies with time t according to the equation, y = a cos ωt where a = 0.010 m and ω = 20 rad s–1. Calculate (i) the amplitude of oscillation, (ii) the period of oscillation, (iii) the equilibrium extension e, (iv) the force constant of the spring. (v) What is the maximum kinetic energy of the mass? (vi) What is the potential energy of the mass at time t = 0.50 s? Mass Spring System-Horizontal Springs 1. Figure 19.13(a) shows a mass m in equilibrium between two stretched helical springs on a smooth horizontal surface. If k1 and k2 are the force constants of the springs and e1 , e2 their respective equilibrium extensions, then tension T1 = T2 , tension in the other spring. k1 e1 = k2 e2 m (a) Equilibrium position (b) Oscillating k1 T1 T2 T’ 1 T’ 2 k2 m a x Figure 19.13 2. The mass m is displaced a distance A to one side (A < e1 , e2 ) and then released. Figure 19.13 (b) shows the mass when its displacement from the equilibrium position is x. Using F = ma T2 ’ – T1 ’ =ma k2 (e2 – x) – k1 (e1 + x) = ma k2 e2 – k2x – k1 e1 – k1 x = ma (k1 e1 = k2 e2 ) ma = –(k1 + k2 )x Acceleration, a = – (k1 + k2 ) m x = –ω2 x Hence, the motion is simple harmonic. ω = k1 + k2 m T = 2π ω = 2π m k1 + k2
19 23 Physics Term 3 STPM Chapter 19 Oscillations Example 12 (a) Figure (a) on the right shows a body of mass m on a smooth horizontal surface. The body is attached to the free end of a light helical spring which has a force constant k. The body is displaced slightly from the equilibrium position and then released. Show that the subsequent motion of the body is simple harmonic, and deduce the period T of oscillation in terms of m and k. (b) In the Figure (b), the block P has a mass of 0.40 kg at the end of a light helical spring is in equilibrium. A block Q of mass 1.60 kg is placed close to P. The block P is pushed a distance A towards the wall and then released. P subsequently starts to move in simple harmonic motion towards Q. The velocity of P just before colliding with Q is 10.0 m s–1. After collision, P and Q coalesce and move together. If the force constant of the spring is 200 N m–1, calculate (i) the value of A, (ii) the common velocity of P and Q immediately after collision, (iii) the period of oscillation of P and Q, (iv) the amplitude of oscillation of P and Q. Solution: (a) Suppose that at any time t during the oscillation, the displacement of the mass m from the equilibrium position is x. Using F = ma –T1 = ma –kx = ma Acceleration, a = –( k m ) x = –ω2 x Motion is simple harmonic, ω = k m Period, T = 2π ω = 2π m k (b) When Q passes the equilibrium position, its velocity is maximum, vmax = 10.0 m s–1. (i) Using vmax = ωA = k ( m) A A = vmax m k = 10.0 × 0.40 200 = 0.447 m O a m P Q Figure (a) Figure (b)
19 24 Physics Term 3 STPM Chapter 19 Oscillations (ii) By the principle of conservation of linear momentum, (0.40 + 1.60) v = 0.40 × 10.0 v = 2.0 m s–1 (iii) Using T1 = 2π m1 k = 2π 0.40 + 1.60 200 = 0.628 s (iv) Using v’ max = ω’A′ amplitude, A′ = 2.0 10 ω′ = 200 (0.40 + 1.60) = 10 = 0.20 m Quick Check 6 1. A trolley of mass 2 kg with free running wheels is attached to two fi xed points P and Q by two springs under tension as shown below. P Q The trolley is displaced a distance of 0.05 m towards Q by a force of 10 N and then released. The equation of the subsequent motion is x = –ω2 x, where x is the displacement from the equilibrium position. What is the value of ω2 ? A 1.0 rad2 s–2 B 4.0 rad2 s–2 C 100 rad2 s–2 D 400 rad2 s–2 2. A mass m on a smooth horizontal table is attached by two light springs to two fi xed supports as shown below. The mass executes linear simple harmonic motion of amplitude a and frequency f. The energy associated with this simple harmonic motion is A 2πf 2 ma2 B 2πfm2 a2 C 2π2 f 2 ma2 D 4π2 f 2 ma2 3. Three bodies P, Q and R are in simple harmonic motion as shown below. P Q R In which of these systems will the period increase if the mass of the body is increased? A P only B Q only C P and Q only D P, Q and R 4. When a diver stands at the free end of a diving board, the end is depressed by 0.30 m. If the diver leaps up and down at the end of the board, what is the period of the oscillation? Explain what would happen to the period of oscillation when a heavier diver does the same act. 5. A load of weight W hangs from the free end of a vertical helical spring. It moves in simple harmonic motion. The displacement y of the load at time t is given by y = a sin ωt. A student drew the graph shown below to show the variation of the tension F in the spring with time t.
19 25 Physics Term 3 STPM Chapter 19 Oscillations (a) Discuss how the tension F in the spring actually varies with time t. Hence, mention two mistakes in the graph drawn. (b) Sketch the correct graph and mark on your graph the weight W and the period T of oscillation. 6. A mass m is attached to two stretched helical springs of force constants k1 and k2 on a smooth horizontal table as shown in the above fi gure. The mass is slightly displaced to one side and then released. Show that the mass m moves in simple harmonic motion, and deduce the period of oscillation in terms of k1 , k2 and m. P Q 7. The motion of a piston in an engine may be regarded as simple harmonic motion. The mass of the piston is 0.50 kg and the maximum distance moved by the piston is 12 cm. Its frequency is 3 600 revolutions per minute. Calculate (a) the resultant force on the piston at the end of each stroke. (b) the speed of the piston at the middle of the stroke. Relation between SHM and Uniform Circular Motion 1. Figure 19.14 shows a particle moving with uniform angular velocity ω in a circle of radius A. 2. At time t = 0, the particle is at R. At time = t, the particle is at Q. The projection of the particle onto the diameter AB is P. The displacement of P from the centre O of the circle is x = Asin q q = ωt x = A sin ωt 3. As the particle Q moves in the circle, its projection P oscillates between A and B with velocity v = dx dt = ωA cos ωt and acceleration a = d2 x dt 2 = –ω2 A sin ωt = –ω2 x Hence, the motion of P is simple harmonic with • angular frequency, ω = ω, the angular velocity of the circular motion. • period of SHM, T = 2π ω = period of the circular motion. • amplitude of SHM, A = A, radius of the circle. Example 12 (a) A vertical peg is fi xed to the rim of a horizontal turntable of radius r rotating with a constant angular speed ω, as shown in the fi gure. Parallel light is incident on the turntable so that the shadow of the peg is observed on a screen which is normal to the incident light. At time t = 0, q = 0 and the shadow of the peg is at O. At time t, the shadow is at P. B P O Q R A Figure 19.14 : Circular motion and SHM P O
19 26 Physics Term 3 STPM Chapter 19 Oscillations (i) Derive an expression for the distance OP in terms of r, ω and t. (ii) Hence, describe the motion of the shadow on the screen. The radius of the turntable is 15 cm and its angular velocity is 2.5 rad s–1. Calculate, for the motion of the shadow on the screen, (iii) the amplitude. (iv) the period. (v) the speed of the shadow as it passes through O. (vi) the magnitude of the acceleration of the shadow when it is instantaneously at rest. (b) In a harbour, the depth of water is 10 m at 12 noon when the tide lowest. The water is 30 m deep when the tide is at its highest, which occurs at 6.15 pm. A ship which needs a depth of 15 m, is required to enter the dock as soon as possible that afternoon. Calculate the earliest time the ship could dock. Solution: (a) (i) q = ωt OP = AB = r sin q = r sin ωt (ii) The shadow moves in simple harmonic motion. (iii) Amplitude = radius of circle = 15 cm (iv) Period T = 2π ω = 2π 2.5 = 2.51 s (v) The speed at O is the maximum, vmax = ωA = 2.5 × 15 = 37.5 cm s–1 (vi) The acceleration is maximum, amax = ω2 A = (2.5)2 × 15 = 93.8 cm s–2 (b) From the fi gure, assuming that the water level in the sea is in SHM. If A = amplitude, then 2A = 20 m A = 10 m 1 2 period = 6 1 4 h (12 noon to 6.15 pm = 6 1 4 h) Period, T = 12.5 h ω = 2π T = 2π 12.5 = 4π 25 The displacement x of the water level from the equilibrium position (x = 0) is represented by x = –A cos ωt = –10 cos ( 4π 25 t ) When the depth of water = 15 m, x = (15 – 20) m = –5 m P B A O A
19 27 Physics Term 3 STPM Chapter 19 Oscillations Substituting into x = –10 cos ( 4π 25 t ) –5 = –10 cos ( 4π 25 t ) cos ( 4π 25 t) = 0.5 4π 25 t = π 3 t = 25 12 h = 2 h 5 min Hence, the ship can dock at 2.05 pm. Quick Check 7 1. The mean depth of a sea is 3.5 m. The rise and fall of the sea level is assumed to be simple harmonic motion of amplitude 0.5 m and period of 12 hours. If the tide is highest at mid-night (0000 h), which of the following represents the variation of the depth h of the sea with time t? A B C D 2. A particle rotates clockwise in a horizontal circle of radius r with constant angular velocity ω as shown in the figure. The particle is at S at time zero and at P at time t. Q is the projection of the point P onto the diameter through S. Measured with respect to the centre O, the displacement, linear velocity and linear acceleration of Q in the direction OS are y, v, and a respectively. Which of the following sets of expressions is correct? A y = r cos ωt ; v = –rω sin ωt ; a = rω2 cos ωt B y = r cos ωt ; v = –rω sin ωt ; a = –rω2 cos ωt C y = r cos ωt ; v = –rω cos ωt ; a = –rω2 sin ωt D y = r sin ωt ; v = –rω cos ωt ; a = –rω2 sin ωt 3. The motion of the piston in a car engine is almost simple harmonic with amplitude 40 mm and frequency 120 Hz. Calculate (i) the maximum acceleration, (ii) the maximum speed of the piston. 4. A vertical rod is fixed to the rim of a horizontal turntable of diameter 4.0 cm. A horizontal beam of light casts a shadow of the rod on a screen. (a) The turntable rotates at a uniform angular velocity ω. Show that the motion of the shadow of the rod on the screen is simple harmonic. (b) If the turntable rotates at 0.5 revolutions per second, what is the maximum speed of the shadow of the vertical rod and at which point in the motion does it occur? S P Q O S
19 28 Physics Term 3 STPM Chapter 19 Oscillations 19.5 Damped Oscillation Students should be able to: • describe the changes in amplitude and energy for a damped oscillating system • distinguish between under damping, critical damping and over damping Learning Outcomes Free Oscillation 1. When a system oscillates without any external forces acting on it, the system is in free oscillation. 2. When a system is in free oscillation, the amplitude of oscillation remains constant, and the oscillation will not stop. 3. Figure 19.15 shows the displacement-time graph of a free oscillation. Damped Oscillations 1. When the amplitude of oscillation of a system decreases with time, energy is slowly dissipated from the system. The oscillation is damped. 2. A damped oscillation is one where energy is dissipated from the system. 3. In your practical work, you would have noticed that the amplitude of oscillation of a simple pendulum and a loaded spring slowly decreases and fi nally the system stops. These are example of damped oscillations. 4. Figure 19.16 shows an example of a damped oscillation. A disc fi xed to a mass hanging from a helical spring is submerged in a beaker of liquid. 5. When the mass oscillates, damping is caused by the friction between the disc and liquid. This form of damping is known as external damping. 6. Internal damping is caused by energy in the form of heat dissipated from a vibrating spring. 7. The effects of damping are • the amplitude of oscillation decreases to zero. • the frequency of oscillation is lower compared to its frequency when there is no damping. Hence, the period of oscillation is larger. A –A Time Displacement, x 0 Figure 19.15 : Free oscillation Spring Mass Disc Liquid Figure 19.16 : Damped oscillation 2010/P1/Q11, 2011/P1/Q12, 2012/P1/Q11, 2016/P3/Q1
19 29 Physics Term 3 STPM Chapter 19 Oscillations 0 Displacement, x A –A Time T 2T (3) (2) (1) Figure 19.17 : (1) Under damping (2) Critical damping (3) Over damping Exam Tips For under damping 0 x Amplitude 8. Figure 19.17 shows the displacement-time graph of an oscillatory system experiencing under damping (1), critical damping (2) and over damping (3). 9. For under damping, the amplitude of oscillation decreases exponentially with time. 10. When the damping is increased, a critical stage is reached when the object displaced from its equilibrium position takes the shortest time possible to return to its equilibrium position (graph 2). No oscillation occurs. 11. Critical damping is used in a moving coil-ammeter where the pointer is stopped from oscillating. 12. The suspension system of a car is slightly below critical damping to ensure a comfortable ride for passengers when the car travels over a bumpy road. 13. If the suspension system of a car is faulty, damping is insuffi cient, and passengers would be thrown up and down once the car moves over a bump. The vertical oscillation of the car takes some time to stop. 14. If damping is greater than critical, the system is over damped. When the body is displaced, it would take a very long time to return to its equilibrium position (graph (3)). Quick Check 8 1. The amplitude of oscillation of a simple pendulum decreases exponentially with time. How does the total energy of the pendulum vary with time? A It decreases at a steady rate. B It decreases exponentially. C It remains constant. D It oscillates with the same frequency as the pendulum. 2. Two bodies P and Q are displaced a distance x0 from the equilibrium position and then released. The graphs below show how their displacements x vary with time, t. P
19 30 Physics Term 3 STPM Chapter 19 Oscillations Q P and Q are then subjected to a driving force of constant amplitude and of various frequencies f. Which graphs best represent how the amplitudes of P and Q vary with f? A P Q B P Q C P Q D Q P 3. An oscillating system is underdamped. Which is the displacement-time graph of the motion? A Displacement Time B Displacement Time C 0 Displacement Time D 0 Displacement Time 4. (a) Describe the energy transformation in a complete cycle for a simple pendulum under free oscillation. Sketch suitable graphs to support your answer. (b) Explain why the amplitude of a damped oscillation decreases. 5. With the aid of suitable graphs, explain what is meant by (a) free oscillation, (b) under damped oscillation, (c) critically damped oscillation, and (d) over damped oscillation. 19.6 Forced Oscillation and Resonance Students should be able to: • distinguish between free oscillations and forced oscillations • state the conditions for resonance to occur Learning Outcomes 2013/P3/Q16
19 31 Physics Term 3 STPM Chapter 19 Oscillations 1. When a system oscillates freely, it is actually oscillating at its natural frequency, which depends on a number of factors. 2. A simple pendulum has a natural frequency f given by f = 1 2π g l which depends on (i) its length l, and (ii) the acceleration of free fall g. 3. The natural frequency of a loaded spring is given by f = 1 2π k m where k = force constant of spring, and m = mass of load 4. An oscillatory system can be forced to oscillate with a frequency other than its natural frequency by an external periodic force. This is known as forced oscillation. 5. The frequency of a forced oscillation is the same as the frequency of the external driving force. 6. The amplitude of the forced oscillation depends on the frequency of the external driving force in relation to the natural frequency of the oscillatory system. 7. Figure 19.18 shows the arrangement used to investigate the relation between amplitude of forced oscillation and the frequency of the external driving force. 8. B is the driving pendulum. The frequency of the driving force is the natural frequency of B. Driving frequency, f = 1 2π g l B where l B = length of pendulum B. 9. The driven pendulum, pendulum A is attached to the same string. 10. When pendulum B is set oscillating with frequency f = 1 2π g l B , pendulum A is forced into oscillation with the same frequency. 11. The amplitude of oscillation of the driven pendulum A is measured using a mm-scale. 12. The driving frequency f is increased by shortening the length of the driving pendulum B. 13. The graph in Figure 19.19 shows the results obtained. 14. When the damping is light, such as in air, the amplitude of the forced oscillation is maximum when the driving frequency is the same as the natural frequency of the driven pendulum, resonance occurs. 15. If there is no damping, that is the pendulums oscillate in a vacuum, the amplitude of forced oscillation would be much bigger. Figure 19.19 : Forced oscillation f 0 0 Amplitude of forced oscillation Driving frequency, f Resonance frequency No damping Under damping Increased damping 2x0 Driven pendulum mm scale Driving pendulum String A B Figure 19.18 : Forced oscillation Info Physics The collapse of the Tacoma Narrow Bridge that stretched across Puget Sound, USA in 1940 was partly due to forced oscillation. VIDEO Tacoma Narrow Bridge Collapse
19 32 Physics Term 3 STPM Chapter 19 Oscillations 16. On the other hand, if damping is increased, (a) the resonance frequency is slightly lower than the natural frequency of the driver pendulum, and (b) the amplitude of oscillation is smaller. 17. Figure 19.20 shows the phase difference φ between the oscillation of driving pendulum and driven pendulum. At the resonance frequency f o , the phase difference φ is π 2 radians. Example 14 A car is driven at a constant speed over a road on which the surface height varies sinusoidally. The shock absorber which normally damps vertical oscillation is not working. (a) Explain why at a critical speed of the car, the amplitude of vertical oscillation of the car becomes very large. (b) In terms of the quantities listed below, deduce a formula for (i) the natural frequency of vertical oscillation of the car. (ii) the critical speed of the car when the amplitude of vertical oscillation is maximum. Calculate this speed from the data given. Mass of car and passengers, M = 2 000 kg Vertical rise of car when passengers get out, s = 0.10 m Mass of passengers, m = 500 kg Wavelength of road surface corrugations, λ = 20 m (c) Discuss the behaviour of the car when the shock absorber mechanism is working properly. Give suitable sketch graphs. Solution: (a) When the car moves over a road with sinusoidal surface, it is forced into vertical oscillation. The frequency of the forced oscillation is f = v λ where v = speed of car. When the speed v increases, the frequency of the driving force increases. At the critical speed, the frequency of the driving force is the same as the natural frequency of the car suspension system. Resonance occurs and the amplitude of vertical oscillation of the car is maximum. (b) (i) The suspension system of the car is represented by the spring system shown: When the car is stationary (Figure (b)), R = Mg and R = kA (using Hooke’s law) where k = Force constant of spring A = Equilibrium compression of spring Hence, Mg = kA .............................. a A R 1 R Mg Mg +x A + x ( a ) ( b ) ( c ) M M f 0 Phase difference, φ / rad 0 Under damping Increased damping f π _π 2 Figure 19.20 : Phase difference between driving pendulum and driven pendulum
19 33 Physics Term 3 STPM Chapter 19 Oscillations Figure (c) shows the displacement of the car, x when it is oscillating vertically, Using F = Ma Mg – R1 = Ma Mg – k(A + x) = Ma –kx = Ma Acceleration a = – k M x = –ω2 x ω = k M f0 = ω 2π = 1 2π k M mg = ks k = mg s Natural frequency, f0 = 1 2π k M = 1 2π mg Ms (ii) If v0 = critical speed of the car when resonance occurs. Then v0 = f λ Driving frequency, f = v0 λ When resonance occurs Driving frequency = Natural frequency v0 λ = 1 2π mg Ms v0 = λ 2π mg Ms = 20 2π 5 × 102 × 9.81 2.0 × 103 × 1.0 × 10–1 = 15.8 m s–1 (c) When the suspension system is working properly, the vertical oscillation is slightly below critical damping. After going over each hump, the car does not perform under damped oscillation (Figure (i)) but returns to the equilibrium position quickly (Figure (ii)). 0 Time Vertical displacement Time Vertical displacement 0 (i) Suspension system faulty (ii) Suspension system functions properly
19 34 Physics Term 3 STPM Chapter 19 Oscillations Quick Check 9 1. A simple pendulum is driven by a sinusoidal driving force of frequency f. Which graph shows how the amplitude A of the motion varies with f ? A C A A B D A A 2. A simple pendulum is forced to oscillate at different frequencies f in air, and the response is as shown in the fi gure below. Which one of the following graphs best represent the results if the experiment is repeated in a medium where the damping is greater? A B C D 3. The natural frequency of a simple pendulum is f 0. It is driven into oscillation by a periodic force of variable frequency. When resonance occurs, which row about the driver frequency and the amplitude of oscillation of the driven pendulum is correct? Driver frequency Amplitude of driven pendulum A Less than f 0 Minimum B Equals f 0 Maximum C Equals f 0 Minimum D Greater than f 0 Maximum 4. Sketch on the same axes to show how the amplitude of forced oscillations varies with the frequency of the driving force for (a) under damping, (b) moderate damping, (c) over damping. Mark on the graph, the resonance frequency f 0 . What is the effect of damping on the resonance frequency? 5. (a) Explain the term forced oscillation and resonance. (b) State the condition for resonance to occur. 6. (a) What is meant by the natural frequency of an oscillatory system? (b) The suspension system of a car consists of a spring under compression and a shock absorber which damps the vertical oscillations of the car. Sketch graphs to illustrate how the vertical height of the
19 35 Physics Term 3 STPM Chapter 19 Oscillations car above the road varies with time after the car has just passed over a bump if the shock absorber is (i) not functioning, (ii) functioning normally (c) When the driver of mass 80 kg steps into a car of mass 920 kg, the vertical height of the car above the road decreases by 2.0 cm. (i) Explain why when the car is driven over a series of equally spaced bumps, the amplitude of vertical oscillation of the car becomes large for one particular speed. (ii) Calculate the speed of the car if the separation between successive bumps is 15 m. Important Formulae 01. Defi nition of simple harmonic motion: Acceleration, a = –ω2 x 2. Angular frequency, ω = 2πf Period, T = 1 f 3. Variation of displacement x with time t (a) If x = 0, when t = 0, then x = A sin ωt (b) If x = +A, when t = 0, then x = A cos ωt (c) If x = –A, when t = 0, then x = –A cos ωt 4. If x = A sin ωt then v = ω A cos ωt and a = –ω2 A sin ωt 5. Velocity, v = ±ω A2 – x2 When x = 0, v is maximum vmax = ωA a = 0 When x = A, v = 0, a is maximum amax = ω2 A 6. Energy in simple harmonic motion If x = A sin ωt Potential energy, U = 1 2 mx2 = 1 2 mω2 A2 sin2 ωt Kinetic energy, K = 1 2 mω2 (A2 – x2 ) = 1 2 mω2 A2 cos2 ωt Total energy, E = 1 2 mω2 A2 7. Systems in simple harmonic motion Simple pendulum: period, T = 2π l g Mass-spring system: period, T = 2π m k STPM PRACTICE 19 1. The graph shows the variation of displacement x against time t of a simple pendulum. -2.0 2.0 0 12.0 t / s x / cm Which statement is true? A The period is 6.0 s. B The angular frequency is 12π s–1. C The frequency is 1 8 Hz. D The amplitude is 4.0 cm.
19 36 Physics Term 3 STPM Chapter 19 Oscillations 2. The displacement of a particle moving in simple harmonic motion is given by x = 4 sin 5t where x is in centimetres and the time t in seconds. If the period of the motion is T, the acceleration of the particle at t = T 6 is A + 21.7 m s–2 C –17.3 cm s–2 B + 5.41 m s–2 D –86.6 cm s–2 3. The displacement x of an object of mass 0.50 kg which performs simple harmonic motion varies with time t according to the equation x = 4.0 cos 50 πt where x is in m, and t is in s. What is the kinetic energy of the object when x = 3.0 m? A 1.75 J C 43.8 J B 8.75 J D 87.5 J 4. The variation of displacement x with time t for two oscillating systems P and Q are shown below. 0 P Q t x The systems P and Q are then forced into oscillation by a periodic external force of variable frequency f. Which graph shows correctly the variation of the amplitude of oscillation A of P and Q with the frequency f? A C P Q A ƒ P Q A ƒ B D P Q A ƒ P Q A ƒ 5. The displacement-time graph of a particle performing simple harmonic motion is shown below. -5.0 3.0 5.0 0 2.0 4.0 t / s t 1 x / m At the instant t 1 , the displacement is 3.0 cm. What is the displacement 1.0 s after t 1 ? A – 3.0 cm C 2.5 cm B 0 D 3.0 cm 6. The displacement,x of a particle in simple harmonic motion varies with time,t as shown in the graph. -4.0 4.0 0 4.0 8.0 12.0 t / s x / m Which equation shows the variation of x with t? A x = 4.0 sin 8t B x = 4.0 sin (8t + π 2 ) C x = 4.0 sin ( π 4 t + π 2 ) D x = 4.0 sin ( π 4 t – π 2 ) 7. Which statement about a critically damped oscillatory system is true? A Its period of oscillation is large. B Its amplitude of oscillation decreases exponentially. C Energy of the system decreases linearly with time. D After being displaced, it quickly returns to the equilibrium position. 8. A simple pendulum oscillates freely between the points P and Q. P O Q Which statement is correct? A The total energy of the system minimum at P and Q. B The kinetic energy increases from O to Q. C The potential energy of the system is a minimum at O. D The total energy equals the kinetic energy at Q.
19 37 Physics Term 3 STPM Chapter 19 Oscillations 9. The length of a simple pendulum is l and the mass of the pendulum bob is m. The simple pendulum is in simple harmonic motion. At time t, the angular displacement is q. The angular acceleration d2 θ dt 2 is A –( g l )q C –(m l )q B –( l g )q D –( l m)q 10. An object performs underdamped oscillation. Which is the displacement-time graph of the motion? A C 0 t x 0 t x B D 0 t x 0 t x 11. A simple pendulum bob moves in front of a horizontal scale between the 200 mm and 300 mm marks. A camera is used to take a picture of the bob. The photograph shows that while the shutter of the camera was opened, the bob moved from the 250 mm mark to the 275 mm mark. The period of the pendulum is 2 s, the shutter remained open for A 1 2 s C 1 4 s B 1 3 s D 1 6 s 12. The displacement of a bob of a simple pendulum at time t is given by x = 0.08 sin 32t where x is in metres and t in seconds. What is the magnitude of the maximum velocity of the bob? A 0.08 m s-1 C 82 m s-1 B 2.6 m s-1 D 400 m s-1 13. The rise and fall of sea water is simple harmonic. The depth varies between 1.0 m at low tide and 3.0 m at high tide. The time between successive low tides is 12 hours. A boat, which requires a minimum depth of water of 1.5 m approaches the harbour at low tide. How long will the boat have to wait before entering? A 1.0 hour C 2.0 hours B 1.5 hours D 2.5 hours 14. The variation of displacement y with time t for three particles P, Q, and R is shown by the graph below. 0 R Q P y t Which statement is not true of the motion of the particles? A Particle Q and R are not oscillatory. B Particle Q will stop in the shortest time. C The energy of particle P is lost at the fastest rate. D The amplitude of particle P decreases exponentially. 15. Which statement is incorrect about a particle in simple harmonic motion? A Its total energy is constant. B A restoring force acts on the particle. C Its acceleration is the maximum at the mean position. D Its kinetic energy is zero at the mean position. 16. A particle performs simple harmonic motion between the points P and Q with O as the equilibrium position. The period is T and A is the amplitude of oscillation. The position of a point X between O and Q is x = + 0.30A. What is the time taken from the particle to move from Q to X? A 0.10 T C 0.27 T B 0.20 T D 0.30 T 17. A particle of mass m performs simple harmonic motion of amplitude A and the angular frequency is ω. The maximum restoring force is A mωA C mω2 A B 2πωmA D 4π2 ω2 mA
19 38 Physics Term 3 STPM Chapter 19 Oscillations 18. A mass m hangs from the end of a vertical spring of force constant, k. The mass vibrates with an amplitude of A. The maximum speed of the mass is given by the expression A A m k C m k A B A k m D k m A 19. (a) Differentiate between critical damping and over damping. (b) The diagram shows four simple pendulums W, P, Q and R hanging from a horizontal wire. W has a heavy bob, while the bobs of P, Q and R are light. W P Q R W is set into oscillation. Describe and explain what happens to , and . (c) A loving father pushes his child on a swing to set the swing into oscillation. Explain how oscillation of large amplitude can be achieved. 20. A block of mass 2.0 kg on a smooth horizontal surface is attached to two identical springs as shown in the figure. The extension in each spring is 6.0 cm. The force constant of the combined springs is 30 N m-1. The mass is displaced 5.0 cm to the right and released. It performs simple harmonic motion. Spring Block Spring Determine (a) the period of the simple harmonic motion. (b) the maximum velocity of the block, and (c) the potential energy of the system when the displacement of the mass is 2.5 cm. 21. (a) Define simple harmonic motion. (b) The displacement x from the equilibrium position of a mass undergoing simple harmonic motion is given by x = 2.5 cos (2t + φ), where x is in metres and t in seconds. At time t = 0, the velocity of the mass is –1.0 m s–1. Calculate (i) the maximum speed, (ii) the maximum acceleration, and (iii) the phase angle φ. (c) A mass hanging from a vertical spring is displaced from the equilibrium position and then released. It then moves in simple harmonic motion. (i) Find in terms of the period T, the shortest time after the mass is released when the kinetic energy of the system equals the potential energy. (ii) Deduce the subsequent time when the kinetic energy of the system equals its potential energy again. 22. (a) The figure below shows three different situations of a mass m which is attached to a helical spring on a smooth surface. The force constant of the spring is k. In (a) the spring is neither stretched nor compressed and the mass is in static equilibrium. In (b) the mass is at the maximum distance from the equilibrium position O. In (c) the mass is in oscillation and its displacement from O is x. m A (a) (b) (c) O O O m m x (i) Explain in terms of forces on the mass, the static equilibrium of the mass. (ii) Write expressions for the restoring force on the mass in situation (b) and (c). (iii) Hence show that the mass moves in simple harmonic motion, and deduce the expression for the angular frequency of the motion. (b) Copper atoms vibrate with a frequency of 3.2 × 1012 Hz. Assuming that the vibration of the copper atoms is identical to the mass-spring system mentioned in (a) above. Find the interatomic force constant between two copper atoms. (Molar mass of copper = 63.5 g)
19 39 Physics Term 3 STPM Chapter 19 Oscillations 23. (a) A particle is in simple harmonic motion with amplitude 8.0 cm and angular frequency 20.0 rad s-1. At time t = 0, the displacement of the particle from the equilibrium position is +4.0 cm. Derive an expression to represent the variation of the displacement x with time t. (b) A mass of 0.80 kg on a smooth table is attached to one end of a horizontal spring of force constant 400 N m-1. The mass performs simple harmonic motion with amplitude of 8.0 cm. Calculate (i) the frequency of the simple harmonic motion, (ii) the speed of the mass when it is 3.0 cm from the equilibrium position, (iii) the kinetic energy of the mass when it is 3.0 cm from the equilibrium position. (c) Discuss the change in the motion of the mass if the surface of the table is rough. (d) With the mass on the rough surface, it is driven into oscillation by an external periodic force. Discuss whether there is any change in the amplitude of oscillation when resonance occurs. 24. A body moves in simple harmonic motion along a straight line. On the same axes, sketch graphs to show how the kinetic energy and potential energy of the body vary with the displacement from the equilibrium position. Explain how the principle of conservation of energy is applied. 25. State the necessary conditions for the following bodies to execute simple harmonic motion. (a) A mass suspended from an inelastic string. (b) A mass suspended from the free end of a helical vertical spring. Describe how any one of the systems can be used to determine the value of g, the acceleration of free fall. 26. It is often mentioned that a mass at the end of a light helical spring when slightly displaced and then released will perform simple harmonic motion. (a) Is it necessary that the displacement must be small? (b) How is the period of oscillation affected when the mass is doubled? (c) Explain why the amplitude of oscillation decreases. 27. (a) Define simple harmonic motion. (b) A particle performs simple harmonic motion along a straight line. Discuss whether each of the following is necessary for the motion to be simple harmonic. (i) The displacement must be small. (ii) The magnitude of the acceleration is directly proportional to the displacement from the point of equilibrium. (iii) The acceleration must be less than the acceleration of free fall. (iv) The period is constant. (v) The acceleration is always negative. (vi) The force on the mass is always directed towards the equilibrium position. (vii) The amplitude varies sinusoidally with time. 28. (a) State the necessary conditions for a mass suspended at the end of a vertical light helical spring to perform simple harmonic motion. (b) Two identical helical springs of the same force constant k are joined (i) in series, and (ii) in parallel, and support a mass m as shown in the figure above. Compare the periods of vertical oscillation of the systems. (c) A student records the following readings using a light helical spring subjected to various loads. Load m/g 50 100 150 200 250 Extension l/ mm 45 88 132 179 224 Time for 10 oscillations/s 4.2 6.0 7.3 8.5 9.4
19 40 Physics Term 3 STPM Chapter 19 Oscillations Plot a suitable graph to deduce the value of the acceleration due to gravity. 29. (a) (i) What is meant by torque produced by a force? (ii) Write an expression to relate torque on a rigid body of moment of inertia I about an axis and the angular acceleration d2 θ dt 2 . (b) The figure shows a light rigid rod of length 3l hinged freely at O. A mass m1 is fixed at Q with distance l from O and another mass m2 is fixed at P, the end of the rod. The system oscillates with small amplitude in a vertical plane. (i) Deduce an expression for the moment of inertia I of the system about O in terms of m1 , m2 and l. (ii) By considering the torque produced when the rod makes an angle θ with the vertical, show that the motion of the rod is simple harmonic. Hence, deduce an expression for the period of oscillation in terms of m1 , m2 , l and g, the acceleration of free fall. (iii) If m1 = 2 kg, m2 = 5 kg and l = 30 cm, find the frequency of oscillation. (iv) The mass m1 is shifted to a distance of 2l from O, but m2 remains at P. Explain any change to the frequency of oscillation. 30. (a) Explain what is meant by the amplitude and angular frequency of a simple harmonic motion. (b) The figure below shows a turntable of radius r rotating about its centre O with uniform angular velocity ω. A peg is fixed at the rim of the turntable. A parallel beam of light casts the shadow of the peg on a screen. O P ω At time t = 0, the shadow of the peg is at P. Show that as the turntable rotates, the shadow performs simple harmonic motion. (c) A block on a smooth horizontal surface is attached to one end of a horizontal helical spring with spring constant 1 500 N m–1. The other end of the spring is fixed to a wall. When slightly displaced and then released, the block performs simple harmonic motion of frequency 5.0 Hz and amplitude 6.0 cm. (i) What is the maximum velocity and maximum kinetic energy of the block? (ii) During its motion, as the block passes through the equilibrium position, it breaks into two equal parts. One part remains attached to the spring and the other part is removed. Calculate the amplitude and frequency of the subsequent simple harmonic motion. 31. (a) Sketch the displacement-time graph of a slightly damped oscillation. (b) Spring Support Mass, m Liquid The above figure shows a mass m hanging from a light helical spring of force constant k in a liquid. The mass is displaced a distance A downwards and then released. It performs damped oscillation. The damping force of the liquid on the mass is bx• , where b is a constant, and x is the vertical displacement of the mass at time t. (i) Use Newton’s second law of motion to deduce a differential equation for the damped oscillation. (ii) Given that m = 0.50 kg and k = 12.0 N m–1, if the frequency of the damped oscillation is the same as the natural frequency, calculate the period of the oscillation. P Q O
19 41 Physics Term 3 STPM Chapter 19 Oscillations (iii) The displacement y of the damped harmonic motion is given as y = A e–bt/2m cos t where A = 10.0 cm and b = 0.060 kg s–1. Estimate the total energy loss after 10 complete oscillations. 32. (a) Sketch graphs to show how the sharpness of resonance in an oscillatory system depends on the degree of damping. (b) A loaded test tube fl oats vertically in a liquid. When the test tube is pushed down slightly and then released, it oscillates vertically. Prove that its motion is simple harmonic. Deduce an expression for the frequency of oscillation. Identify the symbols used. (c) A loaded test tube of mass 0.025 kg fl oats vertically at one end of a ripple tank. The diameter of the test tube is 1.20 cm and the density of water is 1.00 × 103 kg m–3. The frequency of a constant amplitude wave generator at the other end is varied until the vertical oscillation of the test tube is maximum. If the speed of the ripples at resonance is 7.50 cm s–1, calculate the wavelength of the waves. 33. (a) Give an example of a free oscillation. Explain why in practice a free oscillation cannot have a constant amplitude. Spring Variable frequency vibrator Mass Figure (a) The Figure (a) shows a mass which can be made to oscillate vertically by an electrically operated variable frequency vibrator. The vibrator itself vibrates at a constant amplitude. As the frequency is varied, the amplitude of vertical oscillation of the mass is as shown by the graph below. 0 5 10 15 20 Frequency / Hz Amplitude of vertical oscillation Figure (b) (b) Name the phenomenon which is illustrated in Figure (b). (c) When the mass is oscillating at maximum amplitude, calculate (i) the angular frequency, (ii) the period of oscillation. (d) Sketch graphs to show the changes in the graph in Figure (b) (i) when a light card of area larger than the cross sectional area of the mass is fi xed with its plane horizontal to the bottom of the mass shown in Figure (a), (ii) when another identical mass is attached to the mass in Figure (a). (e) Give one application of the phenomenon illustrated in Figure (b).
19 42 Physics Term 3 STPM Chapter 19 Oscillations 1 1. A body is in simple harmonic motion if its acceleration is directly proportional to its displacement from a fi xed point and is always directed towards that point. 2. (a) Equilibrium position: 30.0 cm mark. (b) (i) Amplitude, A = 1 2 (36.0 – 24.0) cm = 6.0 cm (ii) Period, T = 2(0.50) s = 1.0 s (iii) Frequency, f = 1 T = 1 1.00 Hz = 1.0 Hz (c) (i) x = – 6.0 cm (ii) x = – 2.0 cm (iii) x = 0 (iv) x = +4.0 cm (d) (i) + (ii) + (iii) 0 (iv) + 2 1. B 2. C 3. D 4. D 5. D 6. B 7. A 8. A 9. B 10. D 11. D 12. D 13. A 14. A 15. D 16. A: T = 2π ω = N t ω = 2πt N vmax = ωA = 2πtA N 17. A 18. (a) (i) 0.04 s (iii) 157 rad s–1 (ii) 25 Hz (iv) 2 mm (b) Displacement and acceleration in anti phase. 19. (a) vmax = 0.75 m s–1 (b) Velocities in opposite directions (c) Displacement x = +A, and x = –A 3 1. B 2. B 3. A 4. A 5. 6. (a) 1.50 × 10–3 J ∆U = mgh (b) 40 mm (c) –A A 4 1. C 2. C 3. C 4. A 5. D 6. B 7. A: 1 2 mω2 (A2 − x2 ) + 1 2 mω2 x2 x = A 2 = 4.0 2 m x = 4.0 2 = 4.0 cos 2πt 2πt = π 4 hence t = 0.13 s 8. ω = g l 9. (a) x = A sin ωt v = ωA cos ωt (b) Displacement = area under velocity-time graph (x – t graph : see above) (c) (i) g = 10.0 m s–2 (ii) height = 2.95 m T = 2π H – x g ; T2 = 4π2 H g – 4π2 x g where T2 = 0, x = l (height) g = – 4π2 Gradient 5 1. A 2. A 3. D 4. C 5. A 6. (a) 6.0 cm (b) 5.12 cm x = A cos ωt ANSWERS
19 43 Physics Term 3 STPM Chapter 19 Oscillations 7. (a) 0 a = –ω2 A (b) 0.444 m s–2 8. (a) 2.25 Hz amax = ω2 A = g (b) At the highest point 9. (b) (i) 0.010 m (ii) 0.314 s (iii) 2.45 cm T = 2π e g (iv) 16.0 N m–1 mg = ke 6 1. C 2. C 3. C 4. 1.10 s 5. (a) F = W + kA sin ωt Mistakes : • Period T marked wrongly. • Negative value for F. (b) 6. T = 2π k1 + k2 m 7. (a) 4 260 N F = m(ω2 A) (b) 22.6 m s–1 v = ωA 7 1. C 2. B 3. (i) 2.27 × 104 m s–2 (ii) 30.2 m s–1 4. (b) 6.28 cm s–1 At the mid-point. 8 1. B 2. C 3. D 4. (a) On the same axes, sketch graph of (i) kinetic energy against displacement (ii) potential energy against displacement (iii) total energy against displacement 9 1. B 2. A 3. B 4. Refer to page 31 5. (a) (b) Refer to page 31 6. (b) (i) (ii) 0 Height Time (c) (i) Car forced into vertical oscillation. Frequency of driving force f = v λ λ = Distance between bumps When force = natural frequency, frequency resonance occurs. (ii) 15.0 m s–1 f = v λ = 1 2π k m mg = ke STPM Practice 19 1. C: Period = 8.0 s, f = 1 8 Hz 2. D: x = 4 sin 5t, ω = 5, T = 2π 5 When t = T 6 = π 15 , a = –ω2 (4 sin 5t) = –(52 )[4 sin 5( π 15 )] = –86.6 cm s–2 3. C: Kinetic energy, K = 1 2 mv2 = 1 2 m ω2 (A2 – x2 ) = 1 2 (0.50)(52 )( 42 – 32 ) J = 43.8 J 4. D: Damping is greater for Q. Peak is lower, and at lower frequency. 5. A: 1.0 s = 1 2 T 6. C: T = 2π ω = 8.0 s, ω = π 4 , A = 4.0 m x = A cos ωt = A sin (ωt + π 2 ) = 4.0 sin ( π 4 t + π 2 ) 7. D: System unable to oscillate when critically damped 8. C 9. A: d2 q dt 2 = – ω2 θ (ω = g l ) = –( g l )θ 10. B 11. D 12. B : vmax= ωA = (32)(0.08)m s–1 = 2.6 m s–1 13. C 14. C : Energy of P decreases exponentially. 15. D
19 44 Physics Term 3 STPM Chapter 19 Oscillations 16. B : x = A cos ωt 0.30 A = A cos(2π T )t Solving, t = 0.20T 17. C : Fmax = mamax = mω2 A 18. A 19. (b) P, Q, R: forced into oscillation Q: Largest amplitude Reason: same length as W. Resonance occurs. (c) Push in the direction of motion with frequency equals the frequency of the swing. 20. (a) 1.62 s (b) 0.19 m s-1 (c) 9.38 × 10-3 J 21. (a) A body is in simple harmonic motion if its acceleration is directly proportional to its displacement,x from the equilibrium position and directed towards the equilibrium position. (b) (i) x = 2.5 cos (2t + φ) v = –(2.5)(2) sin (2t + φ) = –5.0 sin (2t + φ) Maximum speed = 5.0 m s–1 (ii) Acceleration = – 10.0 cos (2t + φ) Maximum acceleration = 10.0 m s–2 (iii) At time t = 0, velocity = –1.0 m s–1 –5.0 sin [2(0) + φ] = –1.0 sin φ = 0.20 φ = 11.5° (c) (i) If the mass is displaced and then released, the displacement x of the mass at time t is x = A cos ωt, and v = –ωA sin ωt Kinetic energy = potential energy 1 2 mω2 (A2 – x2 ) = 1 2 mω2 x2 x = A 2 = 0.707A When x = A cos ωt = 0.707A ωt = π 4 ( 2π T )t = π 4 t = T 8 (ii) 0 0.707A T / 8 3T / 8 -0.707A t x From the above figure, subsequent time when K = U is when t = 3T 8 . 22. (a) (i) Vertical: Normal reaction of the surface on the mass (upwards) = weight of the mass (downwards). Resultant force = 0. (ii) In (b) Restoring force = –kA In (c) Restoring force = –kx (iii) F = ma = –kx a = –( k m)x = – ω2 x, hence SHM ω = k m (b) Mass of 6.02 × 1023 copper atoms = 0.0635 kg Mass of copper atom, m = 0.0635 6.02 × 1023 kg Frequency, f = ω 2π = 3.2 × 1012 Hz 1 2π k m = 3.2 × 1012 Hz Force constant, k = 4π2 (3.2 × 1012)2 ( 0.0635 6.02 × 1023 ) N m–1 = 42.6 N m-1 23. (a) x = 8 sin (20t + π 6 ) where x is in cm and t in s. (b) (i) f = 1 2π k m = 3.56 Hz (ii) v = ω = A2 – x2 = 1.66 m s-1 (iii) K = 1 2 mv2 = 1.10 J (c) Motion of mass is damped, energy is dissipated as heat due to friction between the mass and the rough surface. Amplitude of oscillation decreases exponentially until the mass stops. (d) Amplitude remains constant. When there is resonance, energy is supplied to the system by an external source. 24. Total energy = K + U = 1 2 mω2 A2 = constant 25. (a) Small amplitude, θ < 100 . Oscillation in one plane. (b) Spring obeys Hooke’s Law. Amplitude e, equilibrium extension. 26. (a) No, as long as amplitude e, equilibrium extension. (b) T1 = 2 T T = 2π m k (c) Energy lost to surroundings.