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Physical Properties of Waves

Physical Properties of Waves

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    Phys12 Phys12 Document Transcript

    • Unit 12 Physical Properties of Waves I12.1 Types of waves12.2 Useful quantities in describing waves12.3 Waves on a string12.4 Sound waves12.5 The frequency of a sound wave12.6 Sound intensity12.7 Human perception of sound12.8 The Doppler effect12.9 Superposition and interference12.10 Standing waves12.11 Beats12.1 Types of wavesA disturbance that propagates from one place to another is referred to as a wave.Waves propagate with well-defined speeds determined by the properties of thematerial through which they travel. For example, sound waves have different speedsin different materials. The following table lists a sampling of sound speed in variousmaterials. Material Speed (m/s) Aluminum 6420 Steel 5960 Copper 5010 Plastic 2680 Fresh water (20 oC) 1480 Air (20 oC) 343Waves carry energy and propagate it when the waves travel. There are two typicalwaves, namely, the transverse waves and the longitudinal waves. 1
    • Transverse waves:The displacement of individual particles is perpendicular to the direction ofpropagation of the wave, e.g. holding one end of a string with another end fixed onthe wall. When you swing your hands vertically, the waves propagate horizontallyalong the string and the particles of string moves up and down.Longitudinal waves:The displacement of individual particles is in the same direction as the direction ofpropagation of the waves, e.g. sound waves. The particles of air move back and forthsuch that a series of compression and rarefaction are observed. Note that the particledoes not travel with the wave, but vibrating about its equilibrium position. 2
    • 12.2 Useful quantities in describing wavesWavelength: The distance over which a wave repeats,e.g. the distance between successive crests and thedistance between successive troughs. Wavelength isquite often labeled as λ. The SI unit is, of course,meter, m.Angular wave number: The angular wave number is 2πdefined as k = . The SI unit is radian per meter. λAngular frequency: It is the measure of how manyradians the waves change in one second. It is labeled asω.Frequency: The number of oscillation per unit time, f,where ω = 2πf .Period: The time for one oscillation, it is labeled as T, 1 2πwhere T = = . f ωVelocity: The distance that the wave travels per unit time is referred to as the velocity. 3
    • The wave equation is: y ( x, t ) = y m sin(kx − ωt ) , in general we have y ( x, t ) = y m sin( kx − ωt + φ ) ,where φ is the phase angle. ωA useful relation in waves: v = fλ = . k12.3 Waves on a stringThe speed of a wave is determined by the properties of the medium through which itpropagates. For a string of length L, there are two factors that vary the speed of awave: (i) the tension in the string F, and (ii) the mass of the string. For the secondfactor, we should say it more precisely that the speed of a wave varies with thedensity of the string (mass per length) µ. The definition of µ is m/L. The unit is kg/m.We can obtain the velocity v by dimensional analysis. Let the velocity relates thetension of string F and the mass per unit length µ by F v= µThe proof is simple. Let v = F a µ b and consider the dimensions of the followingquantities. [v] = [L][T]−1 (Unit: ms−1) [F] = [M][L][T]−2 (Unit: kg ms−2) [µ] = [M][L]−1 (Unit: kg m−1)Comparing the dimension on both sides of v = F a µ b , we have three equations [L]: 1=a–b [T]: −1 = − 2a [M]: 0=a+b 1 FAfter solving, we find that a = −b = . Hence, we have v = . 2 µExampleA rope of length L and mass M hangs from a ceiling. If the bottom of therope is given a gentle wiggle, a wave will travel to the top of the rope. As 4
    • the wave travels upward does its speed (a) increase, (b) decrease, or (c) stay the same?Answer:Since the tension increases with the height, the speed of the wave increases when itclimbs up the rope. Note also that the tension of the rope increases from almost zeroat the bottom to Mg at the top of rope.12.4 Sound wavesA mechanical model of a sound wave is provided by a slinky. Consider if a slinky isoscillated at one end back and forth horizontally. Longitudinal wave travels inhorizontal direction with some regions are compressed and some regions are morewidely spaced, but these regions are distributed alternatively. If we plot the densityvariation against the displacement x, we observe classical wave shape in the graph.The rarefactions and compressions oscillate in a wave-like fashion. In thecompressions regions, the pressure is high, and in the rarefaction regions, the pressureis low. The speed of sound is determined by the properties of the medium throughwhich it propagates. In air, under normal atmospheric pressure and temperature, thespeed of sound is approximately 343 m/s ≈ 770 mi/h. As the air is heated up to ahigher temperature, the air molecules moves faster and the speed of sound increasesas expected.In a solid, the speed of sound is determined in part by the stiffness of the material.The stiffer the material, the faster the sound wave, just as having more tension in astring causes a faster wave. The speed of sound in steel is greater than that in plastic.And both speeds are much higher than that in air.ExampleYou drop a stone into a well that is 7.35 m deep. How long does it takebefore you hear the splash? 5
    • Answer:The time until the splash is heard is the sum of two time intervals. t1: the time for the stone to drop a distance d and t2: the time for the sound to travel a distance d. 1 2 2d 2(7.35)Since d = gt1 , we obtain t1 = = = 1.22 s . 2 g 9.81 d 7.35To calculate t2, we have d = vt 2 , and t 2 = = = 0.0214 s . v 343Hence, the sum of the two time intervals is (1.22 +0.0214) s = 1.24 s.12.5 The frequency of a sound waveHuman can hear sounds between 20 Hz on the low frequency and 20,000 Hz on thehigh frequency end. Sounds with frequencies above this range are referred asultrasonic, while those with frequencies lower than 20 Hz are classified as infrasonic.12.6 Sound intensityIntensity is a quantitative scale by which loudness may be measured. The intensity isdefined as the amount of energy that passes through a given area in a given time. Thisis illustrated in the figure. If the energy E passes through the area A in the time t the Eintensity, I, of the wave carrying the energy is I = , where E/t is the power. At PRewrite the expression again, we have I = . AThe SI unit is W/m2. An example of intensity of light on the Earth’s upper atmospherecoming from the Sun is about 1380 W/m2. A rock concert has an intensity of 0.1W/m2, while the intensity of a classroom is 0.0000001 W/m2. The threshold ofhearing is 10−12 W/m2.When we listen to a source of sound, such as a person speaking or a radio playing asong, the loudness of the sound decreases as we move away from the source. Thesurface area of a sphere from a distance r is 4π r 2 . The intensity of such sound is P I= . 4πr 2 6
    • 12.7 Human perception of soundWe can detect sounds that are about a million times fainter than a typicalconversation, and listen to sounds that are a million times louder before experiencingpain. We are able to hear sounds over a wide range of frequencies, from 20 Hz to20,000 Hz. Our perception of sound, for example the loudness seems to be “twice asloud” if the intensity of the sound is about 10 times the original one. In the study ofsound, the loudness is measured by a convenient scale, which depends on thelogarithm of intensity. IMathematically, the intensity level β is expressed in the form β = 10 log( ) . The I0intensity level β is dimensionless and the unit is given as decibel (dB), where I0 is theintensity of the faintest sounds that can be heard. Experiments show that the lowestdetectable intensity is I 0 = 10 −12 W / m 2 . The smallest increase in intensity level thatcan be detected by the human ear is about 1 dB. And, the loudness of a sound doubleswith each increase in intensity level of 10 dB. Sound Decibels Ear drum ruptures 160 Jet taking off 140 Loud rock band 120 Heavy traffic 90 Classroom 50 Whisper 20 Threshold of hearing 0ExampleIf a sound has an intensity I = I0, the corresponding intensity level is I0 β = 10 log( ) = 10 log 1 = 0 dB . I0Increasing the intensity by a factor of 10 makes the sound seem twice as loud. Interms of decibels, we have 7
    • 10 I 0 β = 10 log( ) = 10 log 10 = 10 dB . I0A further increase in intensity by a factor of 10 double the loudness again. 100 I 0 β = 10 log( ) = 10 log 100 = 20 dB . I0Thus, the loudness of a sound doubles with each increase in intensity level of 10 dB.The smallest increase in intensity level that can be detected by the human ear is about1 dB.ExampleA crying child emits sound with an intensity of 8.0 × 10 −6 W / m 2 . Find(a) the intensity level in decibels for the child’s sounds, and(b) the intensity level for this child and its twin, both crying with identical intensities.Answer: I(a) As the intensity level is given by β = 10 log( ) , we substitute I0 I = 8.0 × 10 −6 W / m 2 and the lowest detectable intensity I 0 = 10 −12 W / m 2 , hence β = 10 log( 8.0 × 10 −6 10 −12 [ ] ) = 10 log(8.0 × 10 −6 ) − log(10 −12 ) = 69 dB .(b) When the twins cry, the intensity will be doubled, I = 2 × (8.0 × 10 −6 W / m 2 ) = 1.6 × 10 −5 W / m 2 . 1.6 × 10 −5 The intensity level is β = 10 log( ) = 72 dB . 10 −12 Or, we can write β = 10 log( 2 × 8.0 × 10 −6 10 −12 [ ] ) = 10 log (2) + log(8.0 × 10 −6 ) − log (10 −12 ) = 72 dB 8
    • N.B. We should note that double the intensity increases the intensity level by 3 dB,since 10 log 2 ≈ 3 . Halved the intensity leads to a decrease of intensity level by 3 dB.Obviously, ten times the intensity of sound gives an increase of 10 dB.ExampleMany animal species use sound waves that are too high or too low for human ears todetect, e.g. bats and blue whales.12.8 The Doppler effectThe relative motion between a source of sound and the receiver gives a change inpitch. This is the Doppler effect. There are two cases for Doppler effect: Movingobserver and moving the source. For example, there is a change in pitch of a trainwhistle or a car horn as the vehicle moves past us. Doppler effect applies to all wavephenomena, not just to sound.ExampleFor light, we observe a change in color, e.g. red-shifted in the color of their light whenthe galaxies are moving away from the Earth. However, some galaxies are movingtoward us, and their light shows a blue shift.12.81 Moving observerA sound wave is emitted from a stationary source. The wave travels in the air withvelocity v, having frequency f and wavelength λ, where v = fλ. For an observer 9
    • moving toward the source with a speed u, the sound seems to have a higher speed, e.g.v + u. As a result, more wavefronts move past the observer in a given time than if theobserver had been at rest. To the observer, the sound has a frequency, f’, that is higherthan the frequency of the source. u u 1+1+ v v + u v = v = (1 + u ) f > f f = = = λ λ λ 1 v v fIf the observer moves away from the source, the sound seems to have a lower speed,e.g. v − u. As a result, less wavefronts move past the observer in a given time than ifthe observer had been at rest. To the observer, the sound has a frequency, f’, that islower than the frequency of the source. u u 1− 1− v v−u v = v = (1 − u ) f < f f = = = λ λ λ 1 v v fCombing the two results, we have f = (1 ± u / v) f ,where plus sign is used when the observer moves toward the source and minus sign isused when the observer moves away from the source.12.82 Moving sourceWhen the source moves, the Doppler effect is not due to the sound wave appearing tohave a higher or lower speed, but a variation in the magnitude of wavelength.Consider, then, a source moving toward the observer with speed u, the sound waveshave one compression and then another compression in time T, where T = 1/f. Thewave travels a distance vT, and the source travels a distance uT. As a result, the newwavelength of the sound waves is vT − uT = (v − u )T , which is shorter than that when 10
    • the source is at rest. The new frequency of the sound waves is obtained by v = f’λ’,that is v 1 ⎛ 1 ⎞ f = = =⎜ ⎟f > f . (v − u )T u ⎝1− u / v ⎠ (1 − )T vWhen the source reverses its direction, the new wavelength of the sound waves,vT + uT = (v + u )T , is longer than that when the source is at rest. v 1 ⎛ 1 ⎞ f = = =⎜ ⎟f < f . (v + u )T u ⎝1+ u / v ⎠ (1 + )T vCombine the two results, we have ⎛ 1 ⎞ f = ⎜ ⎟f , ⎝1∓ u / v ⎠where minus sign is used when the source moves toward the observer, and the plussign when the source moves away from the observer.12.83 General caseDoppler effect for both moving source and observer is concluded in a simple formula: ⎛ 1 ± uo / v ⎞ f = ⎜ ⎜1∓ u / v ⎟ f . ⎟ ⎝ s ⎠ 11
    • ExampleA car moving at 18 m/s sounds its 550 Hz horn. A bicyclist on the sidewalk, movingwith a speed of 7.2 m/s, approaches the car. What frequency is heard by the bicyclist?Answer:As the car (source) and the bicyclist (observer) approach each other, we apply the ⎛ 1 + uo / v ⎞ 1 + 7.2 / 343 ⎞ ⎜formula f = ⎜ ⎟f =⎛ ⎟ ⎜ ⎟ (550 Hz ) = 592.7 Hz . ⎝ 1 − us / v ⎠ ⎝ 1 − 18 / 343 ⎠The right figure shows the Doppler shiftedfrequency versus speed for a 400-Hz soundsource. The upper curve corresponds to amoving source, the lower curve to a movingobserver. Notice that while the two casesgive similar results for low speed, the high-speed behavior is quite different. In fact, theDoppler frequency for the moving sourcegrows without limit for speeds near the speedof sound, while the Doppler frequency forthe moving observer is relatively small.12.84 Supersonic speed and shock wavesWhat happen when the speed of the source exceeds the speed of sound? The equationsderived above are no longer valid. For supersonic speeds, a V-shaped envelope isobserved, all wavefronts bunch are along thisenvelop, which is in three dimensions. This cone iscalled the Mach cone. A shock wave is said to existalong the surface of this cone, because the bunchingof wavefronts causes an abrupt rose and fall of airpressure as the surface passes through any point. TheMach cone angle is given by vt v sin θ = = . vs t vs 12
    • The ratio vs/v is the Mach number. The shock wave generated by a supersonic aircraftor projectile produces a burst of sound, called a sonic boom.12.9 Superposition and interferenceThe combination of two or more waves to form a resultantwave is referred to as superposition. When waves are ofsmall amplitude, they superpose in the simplest of ways –they just add.For example, consider two waves on a string, as shown infigure.ExampleSince two waves add, does the resultant wave y always have a greater amplitude thanthe individual waves y1 and y2?Answer:The wave y is the sum of y1 and y2, but remember that y1 and y2 are sometimespositive and sometimes negative. Thus, if y1 is positive at a given time, for example,and y2 is negative, the sum y1 + y2 can be zero or even negative.As simple as the principle of superposition is, it still leads to interestingconsequences. For example, consider the wave pulse on a string shown in the abovefigure (a). When they combine, the resulting pulse has an amplitude equal to the sumof the amplitudes of the individual pulses. This is referred to as constructiveinterference. When two pulses like those in figure (b) may combine and gives a netdisplacement of zero. That is the pulses momentarily cancel one another. This isdestructive interference. 13
    • It should also be noted that interference is not limited to waves on a string; all wavesexhibit interference effects. In fact, interference is one of the key characteristics thatdefine waves.Suppose the two sources emit waves in phase. At pointA the distance to each source is the same, hence crestmeets crest and constructive interference results. At Bthe distance from source 1 is greater than that fromsource 2 by half a wavelength. The result is crestmeeting trough and destructive interference. Finally, atC the distance from source 1 is one wavelength greaterthan the distance from source 2. Hence, we find constructive interference at C, and thewaves are in phase again at C. If the sources had been oppositein phase, then A and C would be points of destructiveinterference, and B would be a point of constructiveinterference.Remarks: • The system that when one source emits a crest, the other emits a crest as well is referred to as synchronized system. The sources are said to be in phase. • In general, we can say that constructive and destructive interference occur under the following conditions for two sources that are in phase: i) Constructive interference occurs when the path length from the two sources differs by 0, λ, 2λ, 3λ, …. ii) Destructive interference occurs when the path length from the two sources differs by λ/2, 3λ/2, 5λ/2, ….ExampleTwo speakers separated by a distance of 4.30 m emit soundof frequency 221 Hz. The speakers are in phase with oneanother. A person listens from a location 2.8 m directly infront of one of the speakers. Does the person hearconstructive or destructive interference? 14
    • Answer:The wavelength of sound: λ = v / f = 343 m / s / 221 Hz = 1.55 m .To determine the path difference, d = d2 – d1, we need to find d2 first, and d 2 = D 2 + d12 = (4.30 m) 2 + (2.80 m) 2 = 5.13 m .Now d = 5.13 m – 2.80 m = 2.33 m. The number of wavelength that fit into the path d 2.33 mdifference: = = 1.50 . Since the path difference is 3λ/2 we expect destructive λ 1.55 minterference. In the ideal case, the person would hear no sound. As a practical matter,some sound will be reflected from objects in the vicinity, resulting in a finite soundintensity.12.10 Standing wavesIf you plucked a guitar string, or blown across themouth of a pop bottle to create a tone, you havegenerated standing waves. In general, a standingwave is one that oscillates with time, but remainsin its location. It is in this sense that the wave issaid to be “standing”. In some respects, a standingwave can be considered as resulting fromconstructive interference of a wave with itself. 15
    • 12.10.1 Waves on a stringA string is tied down at both ends. If the string is pluckedin the middle a standing wave results. This is thefundamental mode of oscillation of the string. Thefundamental consists of one-half a wavelength betweenthe two ends of the string. Hence, its wavelength is 2L,or we write λ = 2 L .If the speed of waves on the string is v, it follows that thefrequency of the fundamental, f1, is determined byv = λf 1 = (2 L) f 1 . Therefore, v v f1 = = . λ 2LNote that the fundamental frequency increases with the speed of the waves, anddecreases as the string is lengthened. Other than the fundamental frequency, there arean infinite number of standing wave modes – or harmonics – for any given string.Remarks: • Points on a standing wave that stay fixed are referred to as nodes, N. • Halfway between any two nodes is a point on the wave that has a maximum displacement, is called an anti-node, A.The second harmonics can be constructed by includingone more half wavelength in the standing wave. Thismode has one complete wavelength between the walls. v v f2 = = = 2 f1 . λ LSimilarly, the third harmonic has one-and-a-halfwavelength in the length L, therefore, its frequency 16
    • v v f3 = = = 3 f1 . λ 2 L 3Remark: vIn general, we have f 1 = , f n = nf1 , and λ n = 2 L / n where n = 1, 2, 3, …. That is, 2Lall harmonics are present.12.10.2 Vibrating columns of airIf you blow across the open end of a pop bottle, you hear atone of a certain frequency. If you pour some water into thebottle and repeat the experiment, the sound you hear has ahigher frequency. The standing wave will have an antinode,A, at the top (where the air is moving) and a node, N, at thebottom (where the air cannot move.) The lowest frequencystanding wave has one-quarter of a wavelength fits into thecolumn of air in the bottom. Thus, we have the wave form N-A in the pipe 1 λ=L 4 λ = 4LThe fundamental frequency, f1, is given by v = λf 1 = (4 L) f 1 . Or we can write v f1 = . 4LThe second harmonic is produced by addinghalf a wavelength, i.e. N-A-N-A, therefore, 43λ / 4 = L , and hence λ = L . The frequency 3 v v vis = = 3( ) = 3 f 1 . λ 4 L 4L 3Similarly, the next-higher harmonic isrepresented by N-A-N-A-N-A. Inside the pipe, v v vwe have standing waves 5λ / 4 = L , the frequency is = = 5( ) = 5 f1 . λ 4 L 4L 5 17
    • Remark: vIn general, we have f1 = , f n = nf1 and λ n = 4 L / n , where n = 1, 3, 5, …. That is 4Lodd harmonics are present.Standing waves in a pipe that is open at both endshave the following modes, as shown in the figure.Remark: vIn general, we have f1 = , f n = nf1 and 2Lλ n = 2 L / n , where n = 1, 2, 3, …. That is allharmonics are present.ExampleAn empty pop bottle is to be used as a musical instrument in aband. In order to be tuned properly the fundamental frequencyof the bottle must be 440.0 Hz. If the bottle is 26.0 cm tall,how high should it be filled with water to produce the desiredfrequency?Answer:Since f1 = v / 4 L , we have 343 m / s L = v / 4 f1 = = 0.195 m . 4(440.0 Hz )The depth of water to be filled: h = H − L = 0.260 m − 0.195 m = 6.5 cm .ExampleIf you fill your lungs with helium and speak you sound something like Donald Duck.From this observation, we can conclude that the speed of sound in helium must be (a)less than (b) the same as, or (c) greater than the speed of sound in air. 18
    • Answer:When we speak with helium our words are higher pitched. Looking at the relation, ve.g. f1 = , the velocity of sound is increased if the length of vocal chords is fixed 2Lwhile the frequency is increased.12.11 BeatsBeats can be considered as the interference pattern in time. To be specific, imagineplucking two guitar strings that have slightly different frequencies. If you listencarefully, you notice that the sound produced by the strings is not constant is time. Infact, the intensity increases and decreases with a definite period. These fluctuations inintensity are the beats, and the frequency of successive maximum intensities is thebeat frequency.Consider two waves, with frequencies f1 = 1/T1 and f2 = 1/T2, interfere at a given,fixed location. At this location, each wave moves up and down to the verticalposition, y, of each wave yields the following: ⎛ 2π ⎞ y1 = A cos⎜ ⎜T t ⎟ = A cos(2πf1t ) ⎟ ⎝ 1 ⎠ ⎛ 2π ⎞ y 2 = A cos⎜ ⎜T t ⎟ = A cos(2πf 2 t ) ⎟ ⎝ 2 ⎠If A = 1,we have the following plots, where y total = y1 + y 2 . Mathematically, we have ytotal = y1 + y2 = A cos(2π f1t ) + A cos(2π f 2 t ) ⎡ ⎛ f − f2 ⎞ ⎤ ⎡ ⎛ f1 + f 2 ⎞ ⎤ ytotal = 2 A cos ⎢ 2π ⎜ 1 ⎟ t ⎥ cos ⎢ 2π ⎜ 2 ⎟ t ⎥ ⎣ ⎝ 2 ⎠ ⎦ ⎣ ⎝ ⎠ ⎦ 19
    • ⎡ ⎛ f − f2 ⎞ ⎤The first part of the ytotal is 2 A cos ⎢2π ⎜ 1 ⎟t ⎥ which gives the slowly-varying ⎣ ⎝ 2 ⎠⎦amplitude of the beats. Since a loud sound is heard whenever this term is 2A or –2A,the beat frequency is f beat = f 1 − f 2 . The rapid oscillations within each beat are due ⎡ ⎛ f + f2 ⎞ ⎤to the second part of ytotal, cos ⎢2π ⎜ 1 ⎟t ⎥ . Now, beats can be understood as ⎣ ⎝ 2 ⎠⎦oscillations at the average frequency, modulated by a slowly varying amplitude.ExampleSuppose two guitar strings have frequencies 438 Hz and 442 Hz. If you sound themsimultaneously you will hear the average frequency, 440 Hz, increasing anddecreasing in loudness with a beat frequency of 4 Hz. Beats can be used to tune amusical instrument to a desired frequency. To tune a guitar string to 440 Hz, forexample, the string can be played simultaneously with a 440 –Hz fork. Listening tothe beats, the tension in the string can be increased or decreased until the beatfrequency becomes vanishingly small.Example (Challenging)An experimental way to tune the pop bottle is tocompare its frequency with that of a 440-Hz tuning fork.Initially, a beat frequency of 4 Hz is heard. As a smallamount of water is added to that already present, thebeat frequency increases steadily to 5 Hz. What were theinitial and final frequencies of the bottle?Answer:Before extra water is added, possible frequency of the bottle is either 436 Hz or 444Hz. After water is added, possible frequency of the bottle is either 435 Hz or 445 Hz.But the frequency of the bottle should be increased as water is added. Hence, thefrequency of the bottle before adding extra water should be 444 Hz. 20