Lecture 30. Chapter 21 Examine two wave superposition (-ωt and +ωt) Examine two wave superposition (-ω 1 t and -ω 2 t)

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1 To do : Lecture 30 Chapter 21 Examine two wave superposition (-ωt and +ωt) Examine two wave superposition (-ω 1 t and -ω 2 t) Review for final (Location: CHEM 1351, 7:45 am ) Tomorrow: Review session, 2103 CH at 12:05 PM Last Assignment HW13, Due Friday, May 7 th, 11:59 PM Physics 207: Lecture 30, Pg 1 Standing waves Waves traveling in opposite direction interfere with each other. If the conditions are right, same k (2π/λ) & ω, the superposition generates a standing wave: D Right (x,t)= a sin( kx-ωt ) D Left (x,t)= a sin( kx+ωt ) Energy flow in a standing wave is stationary, it stands in place. Standing waves have nodes and antinodes D(x,t)= D L (x,t) + D R (x,t) D(x,t)= 2 a sin(kx) cos(ωt) The outer curve is the amplitude function A(x) = ±2 a sin(kx) when ωt = πn n = 0,1,2, k = wave number = 2 / Anti-nodes Nodes Physics 207: Lecture 30, Pg 2 Page 1

2 Standing waves on a string Longest wavelength allowed: ½ of the full wave Fundamental: λ/2 = L λ = 2 L λ m m = 2 m L = = 1,2,3,... Recall v = f λ f v f m v m = 2L Overtones m > 1 m Physics 207: Lecture 30, Pg 3 Vibrating Strings- Superposition Principle Violin, viola, cello, string bass Guitars Ukuleles Mandolins Banjos D(x,0) Antinode D(0,t) Physics 207: Lecture 30, Pg 4 Page 2

3 Standing waves in a pipe Open end: Must be a displacement antinode (pressure minimum) Closed end: Must be a displacement node (pressure maximum) Blue curves are displacement oscillations. Red curves, pressure. Fundamental: λ/2 λ/2 λ/4 Physics 207: Lecture 30, Pg 5 λ m f m m = = = 2 m L m v 2 L 1, 2,3,... Standing waves in a pipe λ = 2 m L m f m m = = m v 2 L 1, 2,3,... λ m f m m = = = 4 m L m v 4 L 1,3,5,... Physics 207: Lecture 30, Pg 6 Page 3

4 Combining Waves Fourier Synthesis Physics 207: Lecture 30, Pg 7 Superposition & Interference Consider two harmonic waves A and B meet at t=0. They have same amplitudes and phase, but ω 2 = 1.15 x ω 1. The displacement versus time for each is shown below: A(ω 1 t) B(ω 2 t) C(t) = A(t) + B(t) CONSTRUCTIVE INTERFERENCE DESTRUCTIVE INTERFERENCE Physics 207: Lecture 30, Pg 8 Page 4

5 Superposition & Interference Consider A + B, [Recall cos u + cos v = 2 cos((u-v)/2) cos((u+v)/2)] y A (x,t)=a cos(k 1 x 2π f 1 t) y B (x,t)=a cos(k 2 x 2π f 2 t) Let x=0, y=y A +y B = 2A cos[2π (f 1 f 2 ) t/2] cos[2π (f 1 + f 2 ) t/2] and f 1 f 2 f beat = = 1 / T beat f average (f 1 + f 2 )/2 A(ω 1 t) B(ω 2 t) t C(t) = A(t) + B(t) T beat Physics 207: Lecture 30, Pg 9 Exercise Superposition The traces below show beats that occur when two different pairs of waves are added (the time axes are the same). For which of the two is the difference in frequency of the original waves greater? A. Pair 1 B. Pair 2 C. The frequency difference was the same for both pairs of waves. D. Need more information. Physics 207: Lecture 30, Pg 10 Page 5

6 Superposition & Interference Consider A + B, [Recall cos u + cos v = 2 cos((u-v)/2) cos((u+v)/2)] y A (x,t)=a cos(k 1 x 2π f 1 t) y B (x,t)=a cos(k 2 x 2π f 2 t) Let x=0, y = y A + y B = 2A cos[2π (f 1 f 2 )t/2] cos[2π (f 1 + f 2 )t/2] and f 1 f 2 f beat = = 1 / T beat f average (f 1 + f 2 )/2 A(ω 1 t) B(ω 2 t) t C(t) = A(t) + B(t) T beat Physics 207: Lecture 30, Pg 11 Review Final is semi cumulative Early material, more qualitative (i.e., conceptual) Later material, more quantitative (but will employ major results from early on) % will be multiple choice Remainder will be short answer with the focus on thermodynamics, heat engines, wave motion and wave superposition Physics 207: Lecture 30, Pg 12 Page 6

7 Exercise Superposition The traces below show beats that occur when two different pairs of waves are added (the time axes are the same). For which of the two is the difference in frequency of the original waves greater? A. Pair 1 B. Pair 2 C. The frequency difference was the same for both pairs of waves. D. Need more information. Physics 207: Lecture 30, Pg 13 Organ Pipe Example A 0.9 m organ pipe (open at both ends) is measured to have it s first harmonic (i.e., its fundamental) at a frequency of 382 Hz. What is the speed of sound (refers to energy transfer) in this pipe? L=0.9 m f = 382 Hz and f λ = v with λ = 2 L / m (m = 1) v = 382 x 2(0.9) m v = 687 m/s Physics 207: Lecture 30, Pg 14 Page 7

8 Standing Wave Question What happens to the fundamental frequency of a pipe, if the air (v =300 m/s) is replaced by helium (v = 900 m/s)? Recall: f λ = v (A) Increases (B) Same (C) Decreases Physics 207: Lecture 30, Pg 15 Chapter 1 Physics 207: Lecture 30, Pg 16 Page 8

9 Important Concepts Physics 207: Lecture 30, Pg 17 Chapter 2 Physics 207: Lecture 30, Pg 18 Page 9

10 Chapter 3 Physics 207: Lecture 30, Pg 19 Chapter 4 Physics 207: Lecture 30, Pg 20 Page 10

11 Chapter 4 Physics 207: Lecture 30, Pg 21 Chapter 5 Physics 207: Lecture 30, Pg 22 Page 11

12 Chapter 5 & 6 Physics 207: Lecture 30, Pg 23 Chapter 6 Chapter 7 Physics 207: Lecture 30, Pg 24 Page 12

13 Chapter 7 Physics 207: Lecture 30, Pg 25 Chapter 7 (Newton s 3 rd Law) & Chapter 8 Physics 207: Lecture 30, Pg 26 Page 13

14 Chapter 9 Chapter 8 Physics 207: Lecture 30, Pg 27 Chapter 9 Physics 207: Lecture 30, Pg 28 Page 14

15 Chapter 10 Physics 207: Lecture 30, Pg 29 Chapter 10 Physics 207: Lecture 30, Pg 30 Page 15

16 Chapter 10 Physics 207: Lecture 30, Pg 31 Chapter 11 Physics 207: Lecture 30, Pg 32 Page 16

17 Chapter 11 Physics 207: Lecture 30, Pg 33 Chapter 12 and Center of Mass Physics 207: Lecture 30, Pg 34 Page 17

18 Chapter 12 Physics 207: Lecture 30, Pg 35 Important Concepts Physics 207: Lecture 30, Pg 36 Page 18

19 Chapter 12 Physics 207: Lecture 30, Pg 37 Angular Momentum Physics 207: Lecture 30, Pg 38 Page 19

20 Hooke s Law Springs and a Restoring Force Key fact: ω = (k / m) ½ is general result where k reflects a constant of the linear restoring force and m is the inertial response (e.g., the physical pendulum where ω = (κ / I) ½ Physics 207: Lecture 30, Pg 39 Simple Harmonic Motion Maximum potential energy Maximum kinetic energy Physics 207: Lecture 30, Pg 40 Page 20

21 Resonance and damping Energy transfer is optimal when the driving force varies at the resonant frequency. Types of motion Undamped Underdamped Critically damped Overdamped Physics 207: Lecture 30, Pg 41 Fluid Flow Physics 207: Lecture 30, Pg 42 Page 21

22 Density and pressure Physics 207: Lecture 30, Pg 43 Response to forces States of Matter and Phase Diagrams Physics 207: Lecture 30, Pg 44 Page 22

23 Ideal gas equation of state Physics 207: Lecture 30, Pg 45 pv diagrams Thermodynamics Physics 207: Lecture 30, Pg 46 Page 23

24 Work, Pressure, Volume, Heat T can change! Physics 207: Lecture 30, Pg 47 Chapter 18 Physics 207: Lecture 30, Pg 48 Page 24

25 Thermal Energy Physics 207: Lecture 30, Pg 49 Relationships Physics 207: Lecture 30, Pg 50 Page 25

26 Chapter 19 Physics 207: Lecture 30, Pg 51 Refrigerators Physics 207: Lecture 30, Pg 52 Page 26

27 Carnot Cycles Physics 207: Lecture 30, Pg 53 Work (by the system) Physics 207: Lecture 30, Pg 54 Page 27

28 Chapter 20 Physics 207: Lecture 30, Pg 55 Displacement versus time and position Physics 207: Lecture 30, Pg 56 Page 28

29 Sinusoidal Waves (Sound and Electromagnetic) Physics 207: Lecture 30, Pg 57 Doppler effect Physics 207: Lecture 30, Pg 58 Page 29

30 Chapter 21 Physics 207: Lecture 30, Pg 59 Standing Waves Physics 207: Lecture 30, Pg 60 Page 30

31 D(0, t) = 2Acos(2π f 1 f 2 f beat = 1/ T Beats f f 1 2 beat 2 t)cos(2π f + f f + f 1 2 f 2 t) avg = 1/ T avg Physics 207: Lecture 30, Pg 61 Lecture 30 Assignment HW13, Due Friday, May 7 th I hope everyone does well on their finals! Have a great summer! Physics 207: Lecture 30, Pg 62 Page 31

32 Example Interference A speaker sits on a pedestal 2.0 m tall and emits a sine wave at 340 Hz (the speed of sound in air is 340 m/s, so λ = 1.0 m ). Only the direct sound wave and that which reflects off the ground at a position half-way between the speaker and the person (also 2 m tall) makes it to the persons ear. How close to the speaker can the person stand (A to D) so they hear a maximum sound intensity assuming there is no phase change at the ground (this is a bad assumption)? t 0 A B t 0 d C t 1 D h The distances AD and BCD have equal transit times so the sound waves will be in phase. The only need is for AB = λ Physics 207: Lecture 30, Pg 63 Example Interference The geometry dictates everything else. AB = λ AD = BC+CD = BC + (h 2 + (d/2) 2 ) ½ = d AC = AB+BC = λ +BC = (h 2 + d/2 2 ) ½ Eliminating BC gives λ+d = 2 (h 2 + d 2 /4) ½ λ + 2λd + d 2 = 4 h 2 + d 2 t 0 A B t d = 4 h 2 / λ d = 2 h 2 / λ ½ = 7.5 m t C 4.25 D Because the ground is more dense than air there will be a phase change of π and so we really should set AB to λ/2 or 0.5 m. Physics 207: Lecture 30, Pg 64 Page 32

33 Sample Problem The figure shows a snapshot graph D(x, t = 2 s) taken at t = 2 s of a pulse traveling to the left along a string at a speed of 2.0 m/s. Draw the history graph D(x = 2 m, t) of the wave at the position x = 2 m. Physics 207: Lecture 30, Pg 65 Sample Problem History Graph: time (sec) Physics 207: Lecture 30, Pg 66 Page 33

34 Example problem Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a maximum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity decreases until reaching a minimum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s. a. What is the frequency of the sound? b. Draw, as accurately as you can, a wave-front diagram. On your diagram, label the positions of the two speakers, the point at which the intensity is maximum, and the point at which the intensity is minimum. c. Use your wave-front diagram to explain why the intensity is a minimum at a point 3.0 m directly in front of one of the speakers. Physics 207: Lecture 30, Pg 67 Example problem Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a maximum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity decreases until reaching a minimum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s. What is the frequency of the sound? v = f λ but we don t know f or λ DRAW A PICTURE Constructive Interference in-phase Destructive Interference out-of-phase Physics 207: Lecture 30, Pg 68 Page 34

35 Example problem Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a maximum of intensity. v = 340 m/s. PUT IN GEOMETRY v = f λ but we don t know f or λ AC - BC = 0 (0 phase difference) AD - BD = λ/2 (π phase shift) AD = ( ) 1/2 A BD = 3.0 λ = 2(AD-BD) =1.0 m Destructive B Interference out-of-phase 1.8 m 3.0 m Constructive Interference in-phase D C Physics 207: Lecture 30, Pg 69 Example problem Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a maximum of intensity. b. Draw, as accurately as you can, a wave-front diagram. On your diagram, label the positions of the two speakers, the point at which the intensity is maximum, and the point at which the intensity is minimum. c. Use your wave-front diagram to explain why the intensity is a minimum at a point 3.0 m directly in front of one of the speakers. Physics 207: Lecture 30, Pg 70 Page 35

36 Sample problem A tube, open at both ends, is filled with an unknown gas. The tube is 190 cm in length and 3.0 cm in diameter. By using different tuning forks, it is found that resonances can be excited at frequencies of 315 Hz, 420 Hz, and 525 Hz, and at no frequencies in between these. a. What is the speed of sound in this gas? b. Can you determine the amplitude of the wave? If so, what is it? If not, why not? Physics 207: Lecture 30, Pg 71 Sample problem A tube, open at both ends, is filled with an unknown gas. The tube is 190 cm in length and 3.0 cm in diameter. By using different tuning forks, it is found that resonances can be excited at frequencies of 315 Hz, 420 Hz, and 525 Hz, and at no frequencies in between these. What is the speed of sound in this gas? L=1.9 meters and f m = vm/2l 315 = v m/2l 420 = v (m+1)/2l 525 = v (m+2)/2l v/2l = 105 v = 3.8 x 105 m/s =400 m/s b) Can you determine the amplitude of the wave? If so, what is it? If not, why not? Answer: No, the sound intensity is required and this is not known. Physics 207: Lecture 30, Pg 72 Page 36

37 Sample Problem The picture below shows two pulses approaching each other on a stretched string at time t = 0 s. Both pulses have a speed of 1.0 m/s. Using the empty graph axes below the picture, draw a picture of the string at t = 4 s. Physics 207: Lecture 30, Pg 73 An example A heat engine uses moles of helium as its working substance. The gas follows the thermodynamic cycle shown. a. Fill in the missing table entries b. What is the thermal efficiency of this engine? c. What is the maximum possible thermal efficiency of an engine that operates between T max and T min? B A B B C C A NET Q H 600 J 0 J 0 J 600 J Q L 0 J 0 J W by 0 J -161 J E Th 600 J -243 J A C Physics 207: Lecture 30, Pg 74 Page 37

38 An example A heat engine uses moles of helium as its working substance. The gas follows the thermodynamic cycle shown. a. Fill in the missing table entries b. What is the thermal efficiency of this engine? c. What is the maximum possible thermal efficiency of an engine that operates between T max and T min? A: T= 400 K B: T= 2000 K C: T= 1050 K B A B B C C A NET Q H 600 J 0 J 0 J 600 J Q L 0 J 0 J W by 0 J -161 J E Th 600 J -243 J A C Physics 207: Lecture 30, Pg 75 The Full Cyclic Process A heat engine uses moles of helium as its working substance. The gas follows the thermodynamic cycle shown. a. What is the thermal efficiency of this engine? b. What is the maximum possible thermal efficiency of an engine that operates between T max and T min? (T = pv/nr, pv γ =const.) A: T= 400 K=1.01x / 0.030/8.3 B: T=2000 K p B V B γ = p C V C γ B (p B V B γ / p C ) 1/γ = V C (5 (10-3 ) 5/3 / 1) 3/5 = V C = 2.6 x 10-3 m 3 C: T= 1050 K A C Physics 207: Lecture 30, Pg 76 Page 38

39 The Full Cyclic Process A: T= 400 K=1.01x / 0.030/8.3 B: T=2000 K Q = n C v T = x 1.5 x 8.3 T V C = 2.6 x 10-3 m 3 Q = T C: T= 1050 K W C A (by) = p V = 1.01x x 10-3 J A B B C C A NET Q H 600 J 0 J 0 J 600 J Q L 0 J 0 J 404 J 404 J W by 0 J 355 J -161 J 194 J E Th 600 J -355 J -243 J 0 J B η = W by / Q H = 194/ 600 = 0.32 η Carnot = 1- T L / T H = 1-400/2000 =0.80 A C Physics 207: Lecture 30, Pg 77 An example A monatomic gas is compressed isothermally to 1/8 of its original volume. Do each of the following quantities change? If so, does the quantity increase or decrease, and by what factor? If not, why not? a. The rms speed v rms b. The temperature c. The mean free path d. The molar heat capacity C V Physics 207: Lecture 30, Pg 78 Page 39

40 An example A creative chemist creates a small molecule which resembles a freely moving bead on a wire (rotaxanes are an example). The wire is fixed and the bead does not rotate. If the mass of the bead is kg, what is the rms speed of the bead at 300 K? Below is a rotaxane model with three beads on a short wire Physics 207: Lecture 30, Pg 79 An example A creative chemist creates a small molecule which resembles a freely moving bead on a wire (rotaxanes are an example). Here the wire is a loop and rigidly fixed and the bead does not rotate. If the mass of the bead is kg, what is the rms speed of the bead at 300 K? Classically there is ½ k B T of thermal energy per degree of freedom. Here there is only one so: ½ mv rms2 = ½ k Boltzmann T v rms =(k Boltzmann T/m) ½ = 640 m/s Physics 207: Lecture 30, Pg 80 Page 40

41 An example A small speaker is placed in front of a block of mass 4 kg. The mass is attached to a Hooke s Law spring with spring constant 100 N/m. The mass and speaker have a mechanical energy of 200 J and are undergoing one dimensional simple harmonic motion. The speaker emits a 200 Hz tone. For a person is standing directly in front of the speaker, what range of frequencies does he/she hear? The closest the speaker gets to the person is 1.0 m. By how much does the sound intensity vary in terms of the ratio of the loudest to the softest sounds? Physics 207: Lecture 30, Pg 81 An example A small speaker is placed in front of a block of mass 4 kg. The mass is attached to a Hooke s Law spring with spring constant 100 N/m. The mass and speaker have a mechanical energy of 200 J and are undergoing one dimensional simple harmonic motion. The speaker emits a 200 Hz tone. (v sound = 340 m/s) For a person is standing directly in front of the speaker, what range of frequencies does he/she hear? E mech =200 J = ½ mv max2 = ½ ka 2 v max =10 m/s Now use expression for Doppler shift where v source = + 10 m/s and 10 m/s f observer fsource = v 1 s v Physics 207: Lecture 30, Pg 82 Page 41

42 An example A small speaker is placed in front of a block of mass 4 kg. The mass is attached to a Hooke s Law spring with spring constant 100 N/m. The mass and speaker have a mechanical energy of 200 J and are undergoing one dimensional simple harmonic motion. The speaker emits a 200 Hz tone. (v sound = 340 m/s) For a person is standing directly in front of the speaker, what range of frequencies does he/she hear? E mech =200 J = ½ mv max2 = ½ ka 2 v max =10 m/s Now use expression for Doppler shift where v source = + 10 m/s and 10 m/s f observer fsource = v 1 s v Physics 207: Lecture 30, Pg 83 An example A small speaker is placed in front of a block of mass 4 kg. The mass is attached to a Hooke s Law spring with spring constant 100 N/m. The mass and speaker have a mechanical energy of 200 J and are undergoing one dimensional simple harmonic motion. The speaker emits a 200 Hz tone. (v sound = 340 m/s) The closest the speaker gets to the person is 1.0 m. By how much does the sound intensity vary in terms of the ratio of the loudest to the softest sounds? E mech =200 J = ½ mv max2 = ½ ka 2 A = 2.0 m Distance varies from 1.0 m to 1.0+2A or 5 meters where P is the power emitted Ratio I loud /I soft = (P/4π r loud2 )/(P/4π r soft2 ) = 5 2 /1 2 = 25 Physics 207: Lecture 30, Pg 84 Page 42

43 An example problem In musical instruments the sound is based on the number and relative strengths of the harmonics including the fundamental frequency of the note. Figure 1a depicts the first three harmonics of a note. The sum of the first two harmonics is shown in Fig. 1b, and the sum of the first 3 harmonics is shown in Fig. 1c. Which of the waves shown has the shortest period? a. 1 st Harmonic b. 2 nd Harmonic c. 3 rd Harmonic d. Figure 1c At the 2 nd position(1 st is at t=0) where the three curves intersect in Fig. 1a, the curves are all: a. in phase b. out of phase c. at zero displacement d. at maximum displacement a displacement b displacement c displacement time time time Figure Physics 1: 207: Wave Lecture superposition 30, Pg 85 An example problem The frequency of the waveform shown in Fig. 1c is a. the same as that of the fundamental b. the same as that of the 2 nd harmonic c. the same as that of the 3 rd harmonic d. sum of the periods of the 1 st, 2 nd & 3 rd harmonics Which of the following graphs most accurately reflects the relative amplitudes of the harmonics shown in Fig. 1? a b c amplitude amplitude harmonic harmonic amplitude d amplitude harmonic harmonic a displacement b displacement c displacement time time time Figure Physics 1: 207: Wave Lecture superposition 30, Pg 86 Page 43

44 Get the beat You are studying an unusual pair of sea creatures that use low frequency sound to interact. Each animal produces a perfect sine wave. Your oscilloscope output appears below. What are the respective frequencies? 2 displacement (meters) time (seconds) Physics 207: Lecture 30, Pg 87 Get the beat You are studying an unusual pair of sea creatures that use low frequency sound to interact. Each animal produces a perfect sine wave. Your oscilloscope output appears below. What are the respective frequencies? f 1 f 2 = 1/ T beat = 1/ 5 (f 1 + f 2 ) /2 =1/ T avg = 1/1 2 f 1 = 2.2 Hz f 1 = 1.1 Hz so f 2 = 0.9 Hz 2 displacement (meters) time (seconds) Physics 207: Lecture 30, Pg 88 Page 44

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