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# INTRO PHYS SCI&ENG PHYS 1211

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This 117 page Class Notes was uploaded by Heath Hane III on Saturday September 12, 2015. The Class Notes belongs to PHYS 1211 at University of Georgia taught by Staff in Fall. Since its upload, it has received 71 views. For similar materials see /class/202489/phys-1211-university-of-georgia in Physics 2 at University of Georgia.

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Chapter 21 Superposition The combination of two or more waves is called a superposition of waves Applications of superposition range from musical instruments to the colors of an oil lm to lasers Chapter Goal To understand and use the idea of superposition Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 21 Superposition Topics The Principle of Superposition Standing Waves Transverse Standing Waves Standing Sound Waves and Musical Acoustics Interference in One Dimension The Mathematics of Interference Interference in Two and Three Dimensions Beats Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 21 Reading Quizzes Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley When a wave pulse on a string re ects from a hard boundary how is the re ected pulse related to the incident pulse A Shape unchanged amplitude unchanged B Shape inverted amplitude unchanged C Shape unchanged amplitude reduced D Shape inverted amplitude reduced E Amplitude unchanged speed reduced Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley When a wave pulse on a string re ects from a hard boundary how is the re ected pulse related to the incident pulse A Shape unchanged amplitude unchanged V B Shape inverted amplitude unchanged C Shape unchanged amplitude reduced D Shape inverted amplitude reduced E Amplitude unchanged speed reduced Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley There are some points on a standing wave that never move What are these points called A Harmonics B Normal Modes C Nodes D Antinodes E Interference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley There are some points on a standing wave that never move What are these points called A Harmonics B Normal Modes V C Nodes D Antinodes E Interference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Two sound waves of nearly equal frequencies are played simultaneously What is the name of the acoustic phenomena you hear if you listen to these two waves A Beats B Diffraction C Harmonics D Chords E Interference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Two sound waves of nearly equal frequencies are played simultaneously What is the name ofthe acoustic phenomena you hear if you listen to these two waves A Beats B Diffraction C Harmonics D Chords E Interference Copyright 2008 Pearson Education Inc publishing asPearson AddisonWesley The various possible standing waves on a string are called the A antinodes B resonant nodes C normal modes D incident waves Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The various possible standing waves on a string are called the A antinodes B resonant nodes C normal modes D incident waves Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The frequency of the third harmonic of a string is A onethird the frequency of the fundamental B equal to the frequency of the fundamental C three times the frequency of the fundamental D nine times the frequency of the fundamental Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The frequency ofthe third harmonic of a string is A onethird the frequency of the mdamental B equal to the frequency of the mdamental V C three times the frequency of the fundamental D nine times the frequency of the mdamental Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 21 Basic Content and Examples Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley a b Pitching machines A I The halls collide and bounce Qquot apart Two panicles cannot occupy the same point of space at the same time Alan FIGURE 211 Unlike particles two waves A Loudspeakers B can pass directly through each other V The waves pass through each other J as Al an Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The Principle of Superposition If wave 1 displaces a particle in the medium by D1 and wave 2 simultaneously displaces it by D2 the net displacement of the particle is simply D1 D2 Principle of superposition When two or more waves are simultaneously present at a single point in space the displacement of the medium at that point is the sum of the displacements due to each individual wave Copyright 2008 Pearson Education Inc publishing as PearsonAddison Wesley F u 1 Stan M Lung Wmms FHGURE 213 A vibrating string is an example of a standing wave qu aimn Tn AA i FIGURE 215 Standing waves are often represented as they would be seen in a time lapse photograph D Antinodes 0 Nodes II I 3I I x 0 5 A A E A 2A The nodes and antinodes are spaced A2 apart qu ahm tn M t The Mathematics of Standing Waves A sinusoidal wave traveling to the right along the xaXis with angular frequency a 27g wave number k 2711 and amplitude a is DR asinkx wt An equivalent wave traveling to the left is DL a sinkx out We previously used the symbol A for the wave amplitude but here we will use a lowercase a to represent the amplitude of each individual wave and reserveA for the amplitude of the net wave Tn AA r The Mathematics of Standing Waves According to the principle of superposition the net displacement of the medium when both waves are present is the sum of DR and DL Dx r DR DL asinkx wt asinkx wt We can simplify this by using a trigonometric identity and arrive at Dx t asinkxcoswt coskxsinwt asinkxcoswt l coskxsinwt Where the amplitude function Ax is defined as Dx t Axcoswt The amplitude reaches a maximum value of Amax 2a at points where sin kx 1 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 211 Node spacing on a string QUESTIONS EXAMPLE 211 Node spacing on a string A very long string has a linear density of 50 gm and is stretched with a tension of 80 N 100 Hz waves with amplitudes of 20 mm are generated at the ends of the string a What is the node spacing along the resulting standing wave b What is the maximum displacement of the string r 39 Pear nn F duca nn Tm 39 39 Addi MWe51ey EXANIPLE 211 Node spacing on a string MODEL Two counterpropagating waves of equal frequency create a standing wave qucah39on Iquot mum v x x 1 EXAMPLE 211 Node spacing on a string VISUALIZE The standing wave will look like Figure 215 Copyright 2008 Pearson Education In publishing as Pearson AddsomWesley EXAMPLE 211 Node spacing on a string SOLVE a The speed of the waves on the string is 1 80N 40 V M 00050kgm ms and thus the wavelength is v 40 ms 040 m 40 cm f 100 Hz Thus the spacing between adjacent nodes is A2 20 cm b The maximum displacement at the antinodes is Amax 2a 40 mm Pear nn F duca nn an Addi MWe51ey Standing Waves on a String FIGURE 2110 Reflections at the two boundaries cause a standing wave on the string Wiggle the strinng the middle The reflected waves travel x 0 through each other This creates a standing wave Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley xL Standing Waves on a String For a string of xed length L the boundary conditions can be satis ed only if the wavelength has one of the values 2L A39III m 1234 A standing wave can exist on the string only if its wavelength is one of the values given by Equation 2113 Because if v for a sinusoidal wave the oscillation frequency corresponding to wavelength 1m is V V V 1234 f Aquot 2Lm mZL m Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Standing Waves on a String There are three things to note about the normal modes of a string 1 m is the number of antinodes on the standing wave not the number of nodes You can tell a string s mode of oscillation by counting the number of antinodes 2 The fundamental mode with m 1 has 21 2L not 31 L Only half of a wavelength is contained between the boundaries a direct consequence of the fact that the spacing between nodes is 12 3 The frequencies of the normal modes form a series f1 2f1 3f1 The fundamental frequency f1 can be found as the di erence between the frequencies of any two adjacent modes That isf1 Af fm1 fm Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXANIPLE 214 Cold spots in a microwave oven QUESTION EXAMPLE 214 Cold spots in a microwave oven Cold spots in a microwave oven are found to be 60 cm apart What is the frequency of the microwaves queah39on Iquot mum v x x 1 EXAMPLE 214 Cold spots in a microwave oven MODEL A standing wave is created by microwaves re ecting from the walls Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 214 Cold spots in a microwave oven SOLVE The cold spots are nodes in the microwave standing wave Nodes are spaced AZ apart so the wavelength of the microwave radiation must be A 12 cm 012 m The speed of microwaves is the speed of light v c so the frequency is c 300 X 108 ms z X1IZ H f A OJznl 5 0 z SG 2 k 39 Pear nnF duca nn an 39 39 4quot Wesley Standing Sound Waves 0 A long narrow column of air such as the air in a tube or pipe can support a longitudinal standing sound wave 0 A closed end of a column of air must be a displacement node Thus the boundary conditions nodes at the ends are the same as for a standing wave on a string 0 It is often useful to think of sound as a pressure wave rather than a displacement wave The pressure oscillates around its equilibrium value 0 The nodes and antinodes of the pressure wave are interchanged with those of the displacement wave Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley FIGURE 2115 The m 2 longitudinal standing wave can be represented as a displacement wave or as a pressure wave The closed end is 21 displacement node and a pressure antinode Air molecules undergo longitudinal oscillations This is a displacement antinode and a pressure node 040940340 0 0 40wa C c l t 4 O 4H C N A N A N Q The displacement and pressure nodes p and aminodes are interchanged A N A N A The pressure is oscillating around atmospheric pressure pme Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley FIGURE 2116 The first three standing sound wave modes in columns of air with different boundary conditions a Closedclosed L i H Pressure Displacement quot7 I m 2 7ltgtvltgtv lt W 1113 A 2L m m m1234 v open open or closedclosed tube fquot mg mfl Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley FIGURE 2115 The first three standing sound wave modes in columns of air with different boundary conditions b Openopen Dispiaccmem m l quot m m1234 f mi mf open open or closed closed tube In 2L 1 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley FIGURE 2116 The first three standing sound wave modes in columns of air with different boundary conditions 0 Open closed Displacement m 1 W m3 W W m 5 4L Aquot m m1357 v open closed tube n Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 216 The length of an organ pipe QUESTION EXAMPLE 215 The length of an organ pipe An organ pipe open at both ends sounds its second harmonic at a frequency of 523 Hz Musically this is the note one octave above middle C What is the length of the pipe from the sounding hole to the end Copyright 2008 Pearson Education lnc publishing as Pearson AddisonWesley EXAMPLE 216 The length of an organ pipe MODEL An organ pipe similar to a ute has a sounding hole where compressed air is blown across the edge of the pipe This is one end of an open open tube with the other end at the true end of the pipe Assume a room temperature 20 C speed of sound Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 216 The length of an organ pipe SOLVE The second harmonic is the m 2 mode which for an open open tube has frequency v 2 fz 2L Thus the length of the organ pipe is v 343 ms f2 523Hz 0656 m 656 cm Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 216 The length of an organ pipe ASSESS This is a typical length for an organ pipe Capynghl 2cm PeusmEducahm 1m publishing asPeusanAd seresley EXAMPLE 217 The notes on a clarinet QUESTION EXAMPLE 211 The notes on a clarinet A clarinet is 660 cm long The speed of sound in warm air is 350 ms What are the fre quencies of the lowest note on a clarinet and of the next highest harmonic Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 217 The notes on a clarinet MODEL A clarinet is an openclosed tube because the player s lips and the reed seal the tube at the upper end Copyright 2008 Pemson Education Inc publishing as Pealson AddisonWesley EXAMPLE 217 The notes on a clarinet SOLVE The lowest frequency is the fundamental frequency For an open closed tube the fundamental frequency is 350 f1 L ms 133Hz 4L 40660 In An open closed tube has only odd harmonics The next highest harmonic is m 3 with frequency f3 3f1 399 Hz Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Interference in One Dimension The pattern resulting from the superposition of two waves is often called interference In this section we will look at the interference of two waves traveling in the same direction FIGURE 2117 Two overlapped waves travel along the x axls b Two overlapped sound waves Speaker 2 Speaker 1 Point of detection Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The Mathematics of Interference As two waves of equal amplitude and frequency travel together along the xaXis the net displacement of the medium is D DI D2 asinkx wt lo asinkx2 out 20 asinrbI asincbz We can use a trigonometric identity to write the net displacement as A gt D 2acos7 s1nkxavg mt q omg Where A9 92 91 is the phase difference between the two waves Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The Mathematics of Interference The amplitude has a maximum value A 2a if c0sAa2 1 This occurs when A4 m 277 maximum amplitudeA 2a Where m is an integer Similarly the amplitude is zero if cosAa2 0 which occurs when l m E 277 minimum amplitudeA 0 Aqs CopyIight 2008 PeaIson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2110 Designing an antire ection coating QUESTION EXAMPLE 2110 Designing an antireflection coating Magnesium uoride Mng is used as an antire ection coating on lenses The index of refraction of MgF2 is 139 What is the thinnest film of MgF2 that works as an antire ection coating at 510 nm near the center of the Visible spectrum P at anu ah n Tn Am nWesley EXAMPLE 2110 Designing an antire ection coating MODEL Re ection is minimized if the two re ected waves inter fere destructively Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2110 Designing an antire ection coating SOLVE The film thicknesses that cause destructive interference at wavelength A are 1 A d n g m 2 Zn The thinnest film has m 1 Its thickness is A 510nm 4n 4139 m The film thickness is significantly less than the wavelength of vis ible light Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2110 Designing an antire ection coating ASSESS The re ected light is completely eliminated perfect destructive interference only if the two re ected waves have equal amplitudes In practice they don t Nonetheless the re ec tion is reduced from 4 of the incident intensity for bare glass to well under 1 Furthermore the intensity of re ected light is much reduced across most of the visible spectrum 400 700 nm even though the phase difference deviates more and more from Trrad as the wavelength moves away from 510 nm It is the increasing re ection at the ends of the visible Copyright 2008 Pearson Educan39on Inc publishing as Pearson AddisonWesley EXAMPLE 2110 Designing an antire ection coating It is the increasing reflection at the ends of the visible spectrum A z 400 nm and z 700 nm where Ac deviates sig nificantly from 77 rad that gives a reddishpurple tinge to the lenses on cameras and binoculars Homework problems will let you explore situations where only one of the two re ections has a reflection phase shift of 7T rad Copyright 2008 Pearson Education Inc publishing as Pearson Addison Wesley Interference in Two and Three Dimensions Two in phuse sources emit FIGURE 2126 The overlapping ripple Circularm39sphericalwaves patterns of two sources A few points of constructive and destructive interference are noted 0 Points of constructive interference A crest is aligned with a crest or a trough with a trough 0 Points of destructive interference A crest is aligned with a trough of another wave Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Interference in Two and Three Dimensions The mathematical description of interference in two or three dimensions is very similar to that of onedimensional interference The conditions for constructive and destructive interference are Maximum constructive interference Ar Ag 277 A950 l I1 27T m 0 l 2 Perfect destructive interference A A 27TTFA O 277 1 m2 where Ar is the pathlength di erence Copyright 2008 Pearson Education Inc publishing as Peaison AddisonWesley ProblemSolving Strategy Interference of two waves 3 Interference of two waves MODEL Make simplifying assumptions such as assuming waves are circular and of equal amplitude Copyright 2008 Pearson Education Inc publishing as PeaIson AddisonWesley ProblemSolving Strategy Interference of two waves VISUALIIE Draw a picture showing the sources of the waves and the point where the waves interfere Give relevant dimensions Identify the distances r and r2 from the sources to the point Note any phase difference Aqu between the two sources Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley ProblemSolving Strategy Interference of two waves 50va The interference depends on the pathlength difference Ar r2 r1 and the source phase difference Aqbo A Constructive Aqb 277 Aqbo m 277 m 0 1 2 39277 1 m 2 A Destructive Aq 277 A490 For identical sources Aq o 0 the interference is maximum constructive if Ar mA perfect destructive if Ar m Copyiight 2008 Pearson Education Inc publishing as Pearson AddisonWesley ProblemSolving Strategy Interference of two waves ASSESS Check that your result has the correct units is reasonable and answers the question Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2111 Twodimensional interference between two loudspeakers QUESTIONS EXAMPLE 2111 Twodimensional interference between two loudspeakers Two loudspeakers in a plane are 20 m apart and in phase with each other Both emit 700 Hz sound waves into a room where the speed of sound is 341 ms A listener stands 50 min front of the loudspeak ers and 20 m to one side of the center Is the interference at this point maximum constructive perfect destructive or in between How will the situation differ if the loudspeakers are out of phase k 39 Pear nnF duca nn an 39 39 4quot Wesley EXAMPLE 2111 Twodimensional interference between two loudspeakers MODEL The two speakers are sources of in phase circular waves The overlap of these waves causes interference 39 Iquot mum 9 AAA39 x 1 EXAMPLE 2111 Twodimensional interference between two loudspeakers VISUALIZE FIGURE 2129 shows the loudspeakers and defines the distances r1 and r2 to the point of observation The figure includes dimensions and notes that Aobo 0 rad FIGURE 2129 Pictorial representation of the interference between two loudspeakers Copyright 2008 Peatson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2111 Twodimensional interference between two loudspeakers SOLVE It s not r1 and r2 that matter but the di erence Ar between them From the geometry of the figure we can calculate that r 50 m2 10 m2 510m r2 50 m2 30 m2 583 m Thus the pathlength difference is Ar r2 r1 073 m The wavelength of the sound waves is v 341 ms A 0487 f 700 Hz m r 39 Pear nn F duca nn Tm 39 39 Addi MWe51ey EXAMPLE 2111 Twodimensional interference between two loudspeakers In terms of wavelengths the path length difference is ArA 150 or 3 A t r 2 Because the sources are in phase A O 0 this is the condi tion for destructive interference If the sources were out of phase Acbo 77 rad then the phase difference of the waves at the lis tener would be A 3 A 2w7rA 02w 2 77rad477rad This is an integer multiple of 27739 rad so in this case the interfer ence would be constructive Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2111 Twodimensional interference between two loudspeakers ASSESS Both the path length difference and any inherent phase difference of the sources must be considered when evaluating interference P at anu 2h n Tn Am nWesley Beats FIGURE 2132 Beats are caused by the superposition of two waves of nearly identical frequency The medium oscillates rapidly at frequency fwg 20 2a Loud Soft Loud Soft Loud Soft Loud The amplitude is slowly modulated as 2acoswn1le 1 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Beats 0 With beats the sound intensity rises and falls twice during one cycle of the modulation envelope 0 Each loudsoftloud is one beat so the beat frequency Aw which is the number of beats per second is twice the modulation frequency fmod The beat frequency is wmod 1 ml fbeal 2 meod 2 2 2 277 2 271 277 f1 f2 where to keep fbeat from being negative we will always let f1 be the larger of the two frequencies The beat is simply the difference between the two indiVidual frequenc1es Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 2113 Listening to beats QUESTIONS EXAMPLE 2113 Listening to beats One utist plays a note of 510 Hz while a second plays a note of 512 Hz What frequency do you hear What is the beat frequency r 39 Pear nnF duca nn an 39 39 Addi quotwwcsley EXAMPLE 2113 Listening to beats SOLVE You hear a note with frequency favg 511 Hz The beat frequency is fbeat fl f2 ZHZ You and they would hear two beats per second EXAMPLE 2113 Listening to beats ASSESS If a 510 Hz note and a 512 Hz note were played sepa rately you would not be able to perceive the slight difference in frequency But when the two notes are played together the obvi ous beats tell you that the frequencies are slightly different Musi cians learn to make constant minor adjustments in their tuning as they play in order to eliminate beats between themselves and other players 39 Iquot mlhlihinwn n Chapter 21 Summary Slides Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley General Principles Principle of Superposition The displacement of a medium when more than one wave is present is the sum of the displacements due to each individual wave Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Important Concepts Stand in g waves are due to the superposition of two traveling waves moving in opposite directions Antinodes K Node spacing is The amplitude at position x is Ax 2a sinkx Nodes where a is the ampli m 2 1 tude of each wave O The boundary conditions determine m 2 which standing wave ltgtltgt frequencies and wavelengths are allowed The allowed standing waves are modes of the system Standing waves on a string Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Important Concepts Interference In general the superposition of two or more waves Aquot Tquotdquotquot 11quot mm mmve Interference A 21 I a into a single wave is called interference Maximum constructive interference occurs Where crests are aligned with crests and troughs with troughs These waves are in phase The maximum displacement is A 261 Perfect destructive interference occurs where crests are aligned with troughs These waves are out of phase The amplitude is A 0 Interference depends on the phase difference Ad between the two waves A Constructive Aq 277 Aqu m 277 aquot Nodal lines destructive A l Destructive Aq 277 Adio m 27T interference A 0 Ar is the pathlength difference of the two waves and Aqbo is any phase difference between the sources For identical sources in phase Adm 0 Interference is constructive if the pathlength difference Ar mA Interference is destructive if the path length difference Ar m 2a cos The amplitude at a point where the phase difference is Ad is A Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Applications Boundary conditions Strings electromagnetic waves and sound waves in Closed closed tubes must have nodes at both ends 2L AMI v m m m f 2L f1 wherem 123 The frequencies and wavelengths are the same for a sound wave in an openopen tube which has antinodes at both ends A sound wave in an openclosed tube must have a node at the closed end but an antinode at the open end This leads to 4L V A m m fin m mfl wherem1357 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Applications Beats loud soft loud soft modulations of intensity occur when two waves of slightly different frequency are superimposed I Soft Loud Soft Loud Soft The beat frequency between waves of frequencies f and f2 is fbea1f1fz Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 21 Clicker Questions Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley 1 ms 1 ms 4 A X In I I f 2 0 2 4 6 8 10 12 14 16 18 Approaching waves at I 0 s Two pulses on a string approach each other at speeds of 1 ms What is the shape of the string at t 6 s v K II I u 1 x m u H n x m l2 u z x m 2 H an mu in my qucah39on Iquot Wham v x x 1 1 ms 1 ms 4 A x m 0 2 4 6 8 10 12 14 16 18 Approaching waves at t 0 3 Two pulses on a string approach each other at speeds of 1 ms What is the shape of the string at t 6 s 1 INF r L r Y vA z ml Lv Y vA VIHU l1m a x In 12 14 I x lo I 14 r x In 12 H n s m I I4 1 b 10 M gtlt I I I I I I 0 2 4 6 8 10 12 14 16 18 20 Copyn39ght 2008 Pearson Education Inc publishing as Pearson AddisonWesley Original standing wave ltgtltgt A standing wave on a string vibrates as shown at the top Suppose the tension is quadrupled while the frequency and the length of the string are held constant Which standing wave pattern is produced No standing wave for lhcsc conditions 1 j 1155 12 lt1 x391 j 39lti jrx 1 gt 13 gtCKLICIZJCVQ a b C d C Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Original standing wave ltgtltgt A standing wave on a string vibrates as shown at the top Suppose the tension is quadrupled while the frequency and the length of the string are held constant Which standing wave pattern is produced No standing wave for lhcsc conditions 22 Jr vii 121 111 lt fix391j39gtltfjrxfft j VgtCxx39jzyg r39j 39 a b 1 d C Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley An openopen tube of air supports standing waves at frequencies of 300 Hz and 400 Hz and at no frequencies between these two The second harmonic of this tube has frequency A 800 Hz B 200 Hz C 600 Hz D 400 Hz E 100 HZ Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley An openopen tube of air supports standing waves at frequencies of 300 Hz and 400 Hz and at no frequencies between these two The second harmonic ofthis tube has frequency A 800 Hz 1 B 200 Hz C 600 Hz D 400 Hz E 100 Hz Copyright 2008 Pearson Education Inc publishing asPearson AddisonWesley Two loudspeakers emit waves with I 20 m Speaker 2 is 10 In 1 in front of speaker 1 What if anything must be done to cause 2 constructlve W 1nterference between 1390 m the two waves A Move speaker 1 forward to the right 05 m B Move speaker I backward to the left 10 m C Move speaker 1 forward to the right 10 m D Move speaker I backward to the left 05 m E Nothing The situation shown already causes constructive interference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Two loudspeakers emit waves with I 20 m Speaker 2 is 10 In 1 in front of speaker 1 What if anything must be done to cause 2 constructlve W 1nterference between 1390 m the two waves A Move speaker 1 forward to the right 05 m B Move speaker I backward to the left 10 m C Move speaker 1 forward to the right 10 m DMove speaker I backward to the left 05 m E Nothing The situation shown already causes constructive interference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The interference at point C in the gure at the right is A maximum constructive B destructive but not perfect C constructive but less than maximum D perfect destructive E there is no interference at point C Iquot quotum Mu n At A Ar 7 ISOKhii dp0il1T of comtructivc intcrfcrcncc V 39 Al B A39 u this is u pmm of destruclnc interference O AI A Ar 7 A so 1i Ls I pull of mnxlmcliw inlel39ll n ncc The interference at point C in the gure at the right is A maximum constructive B destructive but not perfect C constructive but less than maximum VDperfect destructive E there is no interference at Al 3 An 7 M will Is u 1mm point ol39dexlruume inlerference Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley These two loudspeakers are in phase They emit equalamplitude sound waves with a wavelength of 10 m At the point indicated is the interference maximum constructive perfect destructive or something in between A10m AZIDm A perfect destructive B maximum constructive C something in between Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley These two loudspeakers are in phase They emit equalamplitude sound waves with a wavelength of 10 m At the point indicated is the interference maximum constructive perfect destructive or something in between A10m AILOm A perfect destructive Bmaximum constructive C something in between Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley You hear three beats per second When two sound tones are generated The frequency of one tone is known to be 610 Hz The frequency of the other is A 604 Hz B 607 Hz C 613 Hz D 616 Hz E Either b or c Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 16 A Macroscopic Description of Matter Macroscopic systems are characterized as being either solid liquid or gas These are called the phases of matter and in this chapter we ll be interested in when and how a system changes from one phase to another Chapter Goal To learn the characteristics of macroscopic systems Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 16 A Macroscopic Description of Matter Topics Solids Liquids and Gases Atoms and Moles Temperature 0 Phase Changes 0 Ideal Gases Ideal Gas Processes Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 16 Reading Quizzes Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley What is the SI unit of pressure A The Nrn2 Newtonmeter squared B The atmosphere C The psi D The Pascal E The Archimedes Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley What is the SI unit of pressure A The Nm2 Newtonmeter squared B The atmosphere C The psi D The Pascal E The Archimedes Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The ideal gas model is valid if A the gas density and temperature are both lOW B the gas density and temperature are both high C the gas density is low and the temperature is high D the gas density is high and the temperature is low Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The ideal gas model is valid if A the gas density and temperature are both low B the gas density and temperature are both high C the gas density is low and the temperature is high D the gas density is high and the temperature is low Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley An idealgas process in which the volume doesn t change is called A isobaric B isothermal C isochoric D isentropic Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley An idealgas process in which the volume doesn t change is called A isoba c B isothermal C isochoric D isentropic Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Chapter 16 Basic Content and Examples Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Density The ratio of a system s mass to its volume is called the mass density or sometimes simply the density M p 7 mass densuy The SI units of mass density are kgm3 In this chapter we ll use an uppercase M for the system mass and lowercase m for the mass of an atom Cupyngm 2m Fearsun Edumuun Inc publishmg as Fmrsun Addisuanesley TABLE 151 Densities of materials Substance p kgm3 Air at STP 13 Ethyl alcohol 790 Water solid 920 Water liquid 1000 Aluminum 2700 Copper 8920 Gold 19300 Iron 7870 Lead 1 1300 Mercury 1 3600 Silicon 2330 T 0 C p 1 atm Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXANIPLE 161 The mass of a lead pipe QUESTION EXAMPLE 161 The mass of a lead pipe A project on which you are working uses a cylindrical lead pipe with outer and inner diameters of 40 cm and 35 cm respectively and a length of 50 cm What is its mass Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 161 The mass ofa lead pipe SOLVE The mass density of lead is plead 11300 kgm3 The vol ume of a circular cylinder of length l is V 7W2 In this case we need to find the volume of the outer cylinder of radius r2 minus the volume of air in the inner cylinder of radius n The volume of the pipe is V 7TI 221 777121 032 r12 147 X 10 4m3 Hence the pipe s mass is M pleadV 17 kg Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Atoms and Moles The mass of an atom is determined primarily by its most massive constituents the protons and neutrons in its nucleus The sum of the number of protons and neutrons is called the atomic mass number1 The atomic mass scale is established by defining the mass of 12C to be exactly 12 u where u is the symbol for the atomic mass unit The conversion factor between atomic mass units and kilograms is m12C u 166 x 10 27 kg 12 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Atoms and Moles TABLE 162 Some atomic mass numbers Element A 1 H Hydrogen 1 4He Helium 4 12C Carbon 12 14N Nitrogen 14 0 Oxygen 16 20Ne Neon 20 27A Aluminum 27 40Ar Argon 40 2071313 Lead 207 Copyright o 2008 Pearson Education Inc publishing as Pearson AddisonWesley Atoms and Moles By definition one mole of matter be it solid liquid or gas is the amount of substance containing as many basic particles as there are atoms in 12 g of 12C The number ofbasic particles per mole of substance is called Avogadro s number NA 602 X 1023 mol l The number of moles in a substance containing N basic particles is Capynght 2m PearsanEducahm Inc publishing asPearsanAddlsanWesley Atoms and Moles If the atomic mass is speci ed in kilograms the number of atoms in a system of mass M can be found from The molar mass of a substance is the mass in grams of 1 mol of substance The molar mass Which we ll designate Mmol has units gmol The number of moles in a system of mass M consisting of atoms or molecules With molar mass Mmol is M in grams quot Z Mimi Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 162 Moles of oxygen QUESTION EXAMPLE 162 Moles of oxygen 100 g of oxygen gas is how many moles of oxygen 1de anon Tn EXAMPLE 162 Moles of oxygen SOLVE We can do the calculation two ways First let s determine the number of molecules in 100 g of oxygen The diatomic oxygen molecule 02 has molecular mass m 32 L1 Converting this to kg we get the mass of one molecule 166 x 10 27 kg m32uX1 53l X10 26kg 11 Thus the number of molecules in 100 g 010 kg is M 0100 kg 24 N 6 188X 10 m 531 x 10 kg Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 162 Moles of oxygen Knowing the number of molecules gives us the number of moles N n 313 mol A Alternatively we can use Equation 165 to find M in grams 100 g n 313 mol Mme 32 gmol Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Tem perature The Celsius temperature scale is defined by setting Tc0 for the freezing point of pure water and Tc100 for the boiling point The Kelvin temperature scale has the same unit size as Celsius with the zero point at absolute zero The conversion between the Celsius scale and the Kelvin scale 1s TK TC The Fahrenheit scale still widely used in the United States is defined by its relation to the Celsius scale as follows 9 TF 2 gTC 32 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Temperature TABLE 164 Temperatures measured with different scales Temperature T C T K T F Melting point of iron 1538 1811 2800 Boiling point of water 100 373 212 Normal body temperature 370 310 986 Room temperature 20 293 68 Freezing point of water 0 273 32 Boiling point of nitrogen 196 77 321 Absolute zero 273 0 460 Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Phase Changes 0 The temperature at which a solid becomes a liquid or if the thermal energy is reduced a liquid becomes a solid is called the melting point or the freezing point Melting and freezing are phase changes 0 The temperature at which a gas becomes a liquid or if the thermal energy is increased a liquid becomes a gas is called the condensation point or the boiling point Condensing and boiling are phase changes 0 The phase change in which a solid becomes a gas is called sublimation Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley Ideal Gases The idealgas model is one in which we model atoms in a gas as being hard spheres Such hard spheres y through space and occasionally interact by bouncing off each other in perfectly elastic collisions Experiments show that the idealgas model is quite good for gases if two conditions are met 1 The density is low ie the atoms occupy a volume much smaller than that of the container and 2 The temperature is well above the condensation point Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley The IdealGas Law The pressure p the volume V the number of moles n and the temperature T of an ideal gas are related by the ideal gas law as follows pV nRT idealgas law where R is the universal gas constant R 831 Jmol K The ideal gas law may also be written as pV NkBT idealgas law where N is the number of molecules in the gas rather than the number of moles n The Boltzmann s constant is kB 138 X 23 Copyright 200 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 163 Calculating a gas pressure QUESTION EXAMPLE 163 Calculating a gas pressure 100 g of oxygen gas is distilled into an evacuated 600 cm3 eon tainer What is the gas pressure at a temperature of 150 C Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 163 Calculating a gas pressure MODEL The gas can be treated as an ideal gas Oxygen is a diatomic gas of 02 molecules Copyright 2008 Pearson Education Inc publishing as Pearson AddisonWesley EXAMPLE 163 Calculating a gas pressure SOLVE From the idealgas law the pressure is p nRTV In Example 162 we calculated the number of moles in 100 g of 02 and found 17 313 mol Gas problems typically involve several conversions to get quantities into the proper units and this exam ple is no exception The SI units of V and T are m3 and K respec tively thus m 3 4 V 600 cm3 600 x 10 m3 100 cm T 150 273 K 423K With this information the pressure is nRT 313 mol831 JmolK423 K V 600 x 10 4m3 p 183 X107Pa 181 atm Copyright 2008 Pearson Education lnc publishing as Pearson AddisonWesley

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