LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA

Transcription

1 CHAPTER 7 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA ANSWERS TO FOCUS ON CONCEPTS QUESTIONS. (d) I we add pulses and 4 as per the principle o linear superposition, the resultant is a straight horizontal line that extends across the entire graph.. (a) These two pulses combine to produce a peak that is 4 units high and a alley that is units deep. No other combination gies greater alues. 3. (c) The smallest dierence in path lengths or destructie intererence to occur is one-hal a waelength λ. As the requency goes up, the waelength goes down, so the separation between the cellists decreases. 4. Smallest separation.56 m 5. (b) According to Equation 7., the diraction angle θ is related to the waelength and sinθ. λ/ D and is determined by the ratio λ/d. Here the ratio is diameter by λ 0/D0 and is the largest o any o the choices, so it yields the largest diraction angle. 6. θ 30.0 degrees 7. (e) Since the waelength is directly proportional to the speed o the sound wae (see Section 6.), the waelength is greatest in the helium-illed room. The greater the waelength, the greater the diraction angle θ (see Section 7.3). Thus, the greatest diraction occurs in the helium-illed room. 8. (d) The trombones produce 6 beats eery seconds, so the beat requency is 3 Hz. The second trombone can be producing a sound whose requency is either 438 Hz 3 Hz 435 Hz or 438 Hz + 3 Hz 44 Hz. 9. Beat requency 3.0 Hz 0. (d) According to the discussion in Section 7.5, one loop o a transerse standing wae corresponds to one-hal a waelength. The two loops in the top picture mean that the waelength o. m is also the distance between the walls, so. m. The bottom picture contains three loops in a distance o. m, so its waelength is 3. m 0.8 m.

2 Chapter 7 Answers to Focus on Concepts Questions 857. (b) The requency o a standing wae is directly proportional to the speed o the traeling waes that orm it (see Equation 7.3). The speed o the waes, on the other hand, depends on the mass m o the string through the relation F/ ( m/ ), so the smaller the mass, the greater is the speed and, hence, the greater the requency o the standing wae.. (c) For a string with a ixed length, tension, and linear density, the requency increases when the harmonic number n increases rom 4 to 5 (see Equation 7.3). According to λ / (Equation 6.), the waelength decreases when the requency increases. 3. Fundamental requency.50 0 Hz 4. (c) One loop o a longitudinal standing wae corresponds to one-hal a waelength. Since this standing wae has two loops, its waelength is equal to the length o the tube, or 0.80 m. 5. (b) There are two loops in this longitudinal standing wae. This means that the nd harmonic is being generated. According to Equation 7.4, the n th harmonic requency is n n, where is the undamental requency. Since 440 Hz and n, we hae 440 Hz 0 Hz. 6. (c) The standing wae pattern in the drawing corresponds to n 3 (the 3 rd harmonic) or a tube open at only one end. Using Equation 7.5, the length o the tube is n ( 3)( 343 m/s) 0.39 m Hz n 7. Frequency o 3 rd harmonic Hz

3 858 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA CHAPTER 7 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA PROBEMS. REASONING For destructie intererence to occur, the dierence in trael distances or the sound waes must be an integer number o hal waelengths. For larger and larger distances between speaker B and the obserer at C, the dierence in trael distances becomes smaller and smaller. Thus, the largest possible distance between speaker B and the obserer at C occurs when the dierence in trael distances is just one hal waelength. SOUTION Since the triangle ABC in Figure 7.7 is a right triangle, we can apply the Pythagorean theorem to obtain the distance d AC as dierence in trael distances is 5.00 m d λ + BC dbc 5.00 m + dbc. Thereore, the where we hae used Equation 6. to express the waelength λ as λ /. Soling or the distance d BC gies ( 5.00 m ) or ( 5.00 m) dbc dbc dbc dbc d dbc + BC BC ( 5.00 m ) d d BC or ( 5.00 m ) d BC 4 4 ( 343 m/s) 4( 5 Hz) ( 5.00 m) ( 5.00 m) m 343 m/s 5 Hz. REASONING When the dierence in path lengths traeled by the two sound waes is a hal-integer number,,, K o waelengths, the waes are out o phase and destructie intererence occurs at the listener. The smallest separation d between the speakers is when the dierence in path lengths is o a waelength, so d λ. The

4 Chapter 7 Problems 859 waelength is, according to Equation 6., is equal to the speed o sound diided by the requency ; λ /. SOUTION Substituting λ / into d λ gies 343 m/s d λ 45 Hz m 3. SSM REASONING According to the principle o linear superposition, when two or more waes are present simultaneously at the same place, the resultant wae is the sum o the indiidual waes. We will use the act that both pulses moe at a speed o cm/s to locate the pulses at the times t s, s, 3 s, and 4 s and, by applying this principle to the places where the pulses oerlap, determine the shape o the string. SOUTION The shape o the string at each time is shown in the ollowing drawings: 4. REASONING The speakers are ibrating in phase. Thereore, in order or constructie intererence to occur at point C, the dierence in path lengths or the sound rom speakers A and B must be zero or an integer number o waelengths. Since the sound rom speaker A

5 860 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA traels a greater distance in reaching point C than the sound rom speaker B does, the dierence in path lengths cannot be zero. The smallest alue o d must, then, be associated with a path dierence o one waelength. The waelength λ o a sound that has a requency and traels at a speed is λ / (Equation 6.). SOUTION Using the Pythagorean theorem to express the distance between speaker A and point C, we set the dierence in path lengths equal to one waelength λ: d + d d λ or ( )d λ Distance between speaker A and point C Distance between speaker B and point C Substituting λ / (Equation 6.) into this result gies 343 m / s d λ or d 3.3 m ( ) ( )( 50 Hz) 5. SSM REASONING According to the principle o linear superposition, the resultant displacement due to two waes is the sum o the displacements due to each wae. In order to ind the net displacement at the stated time and positions, then, we will calculate the indiidual displacements y and y and then ind their sum. We note that the phase angles are measured in radians rather than degrees, so calculators must be set to the radian mode in order to yield alid results. SOUTION a. At t 4.00 s and x.6 m, the net displacement y o the string is y y + y ( 4.0 mm) sin ( 9.00π rad/s)( 4.00 s) (.5π rad/m)(.6 m) mm ( 35.0 mm) sin (.88π rad/s)( 4.00 s) ( 0.400π rad/m)(.6 m) + + b. The time is still t 4.00 s, but the position is now x.56 m. Thereore, the net displacement y is y y + y ( 4.0 mm) sin ( 9.00π rad/s)( 4.00 s) (.5π rad/m)(.56 m) mm ( 35.0 mm) sin (.88π rad/s)( 4.00 s) ( 0.400π rad/m)(.56 m) + +

6 Chapter 7 Problems REASONING In Drawing the two speakers are equidistant rom the obserer O. Since each wae traels the same distance in reaching the obserer, the dierence in trael-distances is zero, and constructie intererence will occur or any requency. Dierent requencies will correspond to dierent waelengths, but the path dierence will always be zero. Condensations will always meet condensations and rareactions will always meet rareactions at the obseration point. Since any requency is acceptable in Drawing, our solution will ocus on Drawing. In Drawing, destructie intererence occurs only when the dierence in trael distances or the two waes is an odd integer number n o halwaelengths. Only certain requencies, thereore, will be consistent with this requirement. SOUTION The requency and waelength λ are related by λ / (Equation 6.), where is the speed o the sound. Using this equation together with the requirement or destructie intererence in drawing, we hae + Dierence in trael distances n λ n where n, 3, 5,... Here we hae used the Pythagorean theorem to determine the length o the diagonal o the square. Soling or the requency gies n The problem asks or the minimum requency, so we choose n and obtain 343 m/s 0.75 m ( ) ( ) 550 Hz 7. SSM REASONING The geometry o the positions o the loudspeakers and the listener is shown in the ollowing drawing.

7 86 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The listener at C will hear either a loud sound or no sound, depending upon whether the intererence occurring at C is constructie or destructie. I the listener hears no sound, destructie intererence occurs, so nλ d d n, 3, 5, K () SOUTION Since λ, according to Equation 6., the waelength o the tone is 343 m/s λ 5.00 m 68.6 Hz Speaker B will be closest to Speaker A when n in Equation () aboe, so d nλ 5.00 m + d +.00 m 3.50 m From the igure aboe we hae that, x (.00 m) cos m Then y (.00 m) sin m x + y d (3.50 m) or x (3.50 m) (0.866 m) 3.39 m Thereore, the closest that speaker A can be to speaker B so that the listener hears no sound is x+ x m m 3.89 m. 8. REASONING The two speakers are ibrating exactly out o phase. This means that the conditions or constructie and destructie intererence are opposite o those that apply when the speakers ibrate in phase, as they do in Example in the text. Thus, or two wae sources ibrating exactly out o phase, a dierence in path lengths that is zero or an integer number (,, 3, ) o waelengths leads to destructie intererence; a dierence in path lengths that is a hal-integer number ( ),,,... o waelengths leads to constructie intererence. First, we will determine the waelength being produced by the speakers. Then, we will determine the dierence in path lengths between the speakers and the obserer and compare the dierences to the waelength in order to decide which type o intererence occurs. SOUTION According to Equation 6., the waelength λ is related to the speed and requency o the sound as ollows:

8 Chapter 7 Problems 863 λ 343 m/s m 49 Hz Since ABC in Figure 7.7 is a right triangle, the Pythagorean theorem applies and the dierence Δd in the path lengths is gien by AC BC AB BC BC Δ d d d d + d d We will now apply this expression or parts (a) and (b). a. When d BC.5 m, we hae Δ d dab + dbc dbc.50 m +.5 m.5 m.60 m.60 m m λ, it ollows that the intererence is destructie (the Since speakers ibrate out o phase). b. When d BC.00 m, we hae Δ d dab + dbc dbc.50 m +.00 m.00 m.0 m Since.0 m m λ, it ollows that the intererence is constructie (the speakers ibrate out o phase). 9. REASONING The act that a loud sound is heard implies constructie intererence, which,, 3, K o occurs when the dierence in path lengths is an integer number waelengths. This dierence is 50.5 m 6.0 m 4.5 m. Thereore, constructie intererence occurs when 4.5 m nλ, where n,, 3,. The waelength is equal to the speed o sound diided by the requency ; λ / (Equation 6.). Substituting this relation or λ into 4.5 m nλ, and soling or the requency gies n 4.5 m This relation will allow us to ind the two lowest requencies that the listener perceies as being loud due to constructie intererence. SOUTION The lowest requency occurs when n : ( 343 m/s) n 4 Hz 4.5 m 4.5 m

9 864 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA This requency lies below 0 Hz, so it cannot be heard by the listener. For n and n 3, the requencies are 8 and 4 Hz, which are the two lowest requencies that the listener perceies as being loud. 0. REASONING When the listener is standing midway between the speakers, both sound waes trael the same distance rom the speakers to the listener. Since the speakers are ibrating out o phase, when the diaphragm o one speaker is moing outward (creating a condensation), the diaphragm o the other speaker is moing inward (creating a rareaction). Wheneer a condensation rom one speaker reaches the listener, it is met by a rareaction rom the other, and ice ersa. Thereore, the two sound waes produce destructie intererence, and the listener hears no sound. When the listener begins to moe sideways, the distance between the listener and each speaker is no longer the same. Consequently, the sound waes no longer produce destructie intererence, and the sound intensity begins to increase. When the dierence in path lengths l l traeled by the two sounds is one-hal a waelength, or l l, constructie intererence occurs, and a loud sound will be heard. λ SOUTION The two speakers are ibrating out o phase. Thereore, when the dierence in path lengths l l traeled by the two sounds is one-hal a waelength, or l l, constructie intererence occurs. Note that this λ condition is dierent than that or two speakers ibrating in phase. The requency o the sound is equal to the speed o sound diided by the waelength λ; /λ (Equation 6.). Thus, we hae that l λ or l ( l l ) The distances l and l can be determined by applying the Pythagorean theorem to the right triangles in the drawing: Midpoint 4.00 m l 0.9 m l.50 m.50 m The requency o the sound is l 4.00m +.50 m m 4.68 m l 4.00 m +.50 m 0.9 m 4.04 m 343 m/s 70 Hz ( l l ) 4.68 ( m 4.04 m)

10 Chapter 7 Problems 865. REASONING AND SOUTION Since λ, the waelength o the tone is 343 m/s λ 4.70 m 73.0 Hz The ollowing drawing shows the line between the two speakers and the distances in question. x - x A P Constructie intererence will occur when the dierence in the distances traeled by the two sound waes in reaching point P is an integer number o waelengths. That is, when ( x) x nλ where n is an integer (or zero). Soling or x gies B nλ x () When n 0, x / (7.80 m)/ 3.90 m. This corresponds to the point halway between the two speakers. Clearly in this case, each wae has traeled the same distance and thereore, they will arrie in phase. When n, (7.80 m) (4.70 m) x.55 m Thus, there is a point o constructie intererence.55 m rom speaker A. The points o constructie intererence will occur symmetrically about the center point at /, so there is also a point o constructie intererence.55 m rom speaker B, that is at the point 7.80 m.55 m 6.5 m romspeaker A. When n >, the alues o x obtained rom Equation () will be negatie. These alues correspond to positions o constructie intererence that lie to the let o A or to the right o C. They do not lie on the line between the speakers.. REASONING The diraction angle or the irst minimum or a circular opening is gien by Equation 7.: sinθ.λ / D, where D is the diameter o the opening. SOUTION a. Using Equation 6., we must irst ind the waelength o the.0-khz tone:

11 866 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA λ 343 m/s Hz 0.7 m The diraction angle or a.0-khz tone is, thereore, b. The waelength o a 6.0-kHz tone is θ sin œ. 0.7 m m λ 343 m/s Hz m Thereore, i we wish to generate a 6.0-kHz tone whose diraction angle is as wide as that or the.0-khz tone in part (a), we will need a speaker o diameter D, where D. λ (.)(0.057 m) 0.0 m sinθ sin SSM REASONING Equation 7. speciies the diraction angle θ according to sin θ λ/ D, where λ is the waelength o the sound and D is the width o the opening. The waelength depends on the speed and requency o the sound. Since the requency is the same in winter and summer, only the speed changes with the temperature. We can account or the eect o the temperature on the speed by assuming that the air behaes as an ideal gas, or which the speed o sound is proportional to the square root o the Kelin temperature. SOUTION Equation 7. indicates that sinθ λ D Into this equation, we substitute λ / (Equation 6.), where is the speed o sound and is the requency: λ / sinθ D D Assuming that air behaes as an ideal gas, we can use γ kt/ m (Equation 6.5), where γ is the ratio o the speciic heat capacities at constant pressure and constant olume, k is Boltzmann s constant, T is the Kelin temperature, and m is the aerage mass o the molecules and atoms o which the air is composed: γ kt sinθ D D m Applying this result or each temperature gies

12 Chapter 7 Problems 867 γktsummer γkt sin θsummer and sinθwinter D m D m Diiding the summer-equation by the winter-equation, we ind winter Thus, it ollows that γ ktsummer sinθsummer D m sinθwinter γ ktwinter D m T summer T winter T 3 K sinθ sinθ sin or θ sin summer summer winter summer Twinter 73 K 4. REASONING The diraction angle θ depends upon the waelength λ o the sound wae λ and the width D o the doorway according to sinθ (Equation 7.). We will determine D the waelength rom the speed and requency o the sound wae ia λ (Equation 6.). We note that, in part (a), the requency o the sound wae is gien in kilohertz (khz), so we will use the equialence khz 0 3 Hz. SOUTION λ a. Soling sinθ (Equation 7.) or θ, we obtain D λ θ sin () D Soling λ (Equation 6.) or λ yields λ. Substituting this result into Equation (), we ind that λ θ sin sin D D () Thereore, when the requency o the sound wae is 5.0 khz Hz, the diraction angle is 343 m/s o θ sin sin 5. D ( m Hz) b. When the requency o the sound wae is Hz, the diraction angle is 343 m/s sin sin 63 D ( 0.77 m Hz) θ o

13 868 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 5. REASONING For a rectangular opening ( single slit ) such as a doorway, the diraction λ angle θ at which the irst minimum in the sound intensity occurs is gien by sinθ D (Equation 7.), where λ is the waelength o the sound and D is the width o the opening. This relation can be used to ind the angle proided we realize that the waelength λ is related to the speed o sound and the requency by λ / (Equation 6.). SOUTION a. Substituting λ / into Equation 7. and using D m (only one door is open) gies λ 343 m/s sin θ θ sin ( ) 53.8 D D 607 Hz m b. When both doors are open, D m and the diraction angle is λ 343 m/s sinθ θ sin ( ) 3.8 D D ( 607 Hz)( m) 6. REASONING The diraction angle θ is determined by the ratio o the waelength λ o the sound to the diameter D o the speaker, according to sin θ. λ/d (Equation 7.). The waelength is related to the requency and the speed o the wae by λ / (Equation 6.). SOUTION Substituting Equation 6.into Equation 7., we hae λ sin θ.. D D Since the speed o sound is a constant, this result indicates that the diraction angle θ will be the same or each o the three speakers, proided that the diameter D times the requency has the same alue. Thus, we pair the diameter and the requency as ollows: Diameter Frequency D (0.050 m)( Hz) m/s (0.0 m)( Hz) m/s (0.5 m)( Hz) m/s The common alue o the diraction angle, then, is. 343 m/s θ sin. sin 44 D m / s

14 Chapter 7 Problems REASONING AND SOUTION At 0 C the speed o sound is air is gien as 33 m/s in Table 6. in the text. This corresponds to a waelength o λ / (33 m/s)/( Hz) 0.03 m The diraction angle is gien by Equation 7. as. λ m θ sin sin 50.3 D 0.75 m For an ideal gas, the speed o sound is proportional to the square root o the Kelin temperature, according to Equation 6.5. Thereore, the speed o sound at 9 C is ( 33 m/s) 30 K 73 K 348 m/s The waelength at this temperature is λ (348 m/s)/( Hz) 0.6 m. This gies a diraction angle o θ The change in the diraction angle is thus Δθ REASONING The person does not hear a sound because she is sitting at a position where the irst minimum in the single slit diraction pattern occurs. This position is speciied by the angle θ according to sin θ λ / D (Equation 7.), where λ is the waelength o the sound and D is the width o the slit (or diraction horn). The waelength λ o a sound that has a requency and traels at a speed is λ / (Equation 6.). We will apply these equations to determine the angle α or the requency o 800 Hz. Knowing α, we will apply the equations a second time or the angle α/ and thereby determine the unknown requency. SOUTION The irst diraction minimum or an angle θ is speciied as ollows by Equation 7.: λ λ sin θ or θ sin D D Substituting λ / (Equation 6.) into this result we obtain θ θ sin () or sin () D D Using Equation () to determine the angle α or the requency 343 m / s sin sin 44.9 D (800 Hz)(0.060 m) α 800 Hz, we obtain

15 870 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA where we hae carried an extra signiicant igure or this intermediate result. When the person moes to an angle α /, the unknown requency at which the irst diraction minimum now occurs is dierent and can be ound rom Equation () with θ α /: α 343 m / s 4 sin or.5 0 Hz D Dsin ( α / ) (0.060 m) sin ( 44.9 / ) 9. SSM REASONING The beat requency o two sound waes is the dierence between the two sound requencies. From the graphs, we see that the period o the wae in the upper text igure is 0.00 s, so its requency is / T /(0.00 s) Hz. The requency o the wae in the lower igure is /(0.04 s)4. 0 Hz. SOUTION The beat requency o the two sound waes is beat Hz 4. 0 Hz 8 Hz 0. REASONING The time between successie beats is the period o the beat requency. The period that corresponds to any requency is the reciprocal o that requency. The beat requency itsel is the magnitude o the dierence between the two sound requencies produced by the pianos. Each o the piano requencies can be determined as the speed o sound diided by the waelength. SOUTION According to Equation 0.5, the period T (the time between successie beats) that corresponds to the beat requency beat is T /. The beat requency is the magnitude o the dierence between the two sound requencies A and B, so that beat A. With this substitution, the expression or the period becomes B T beat A B The requencies A and B are gien by Equation 6. as A and B λ λ Substituting these expressions into Equation () gies A A B B beat T 0.5 s 343 m/s 343 m/s A B λ λ m m (). REASONING The beat requency is the dierence between two sound requencies. Thereore, the original requency o the guitar string (beore it was tightened) was either

16 Chapter 7 Problems 87 3 Hz lower than that o the tuning ork (440.0 Hz 3 Hz 337 Hz) or 3 Hz higher (440.0 Hz + 3 Hz 443 Hz): To determine which o these requencies is the correct one (437 or 443 Hz), we will use the inormation that the beat requency decreases when the guitar string is tightened SOUTION When the guitar string is tightened, its requency o ibration (either 437 or 443 Hz) increases. As the drawing below shows, when the 437-Hz requency increases, it becomes closer to Hz, so the beat requency decreases. When the 443-Hz requency increases, it becomes arther rom Hz, so the beat requency increases. Since the problem states that the beat requency decreases, the original requency o the guitar string was 437 Hz Hz Hz 443 Hz 437 Hz } } 3-Hz beat requency 3-Hz beat requency 443 Hz } Beat requency increases } Beat requency decreases 437 Hz Tuning ork Original string Tightened string. REASONING Two ultrasonic sound waes combine and orm a beat requency that is in the range o human hearing. The beat requency is the dierence between the two ultrasonic requencies, that is, the larger minus the smaller o the two requencies. One o the ultrasonic requencies is 70 khz. We can determine the smallest and the largest alue or the other ultrasonic requency by considering the implications o a beat requency o 0 khz on the one hand and 0 Hz on the other hand, since these two alues deine the limits o hearing or a healthy young person. SOUTION et us irst assume that the beat requency beat is 0 khz. Noting that the symbols > and < mean, respectiely, greater than and less than, we hae the two ollowing possibilities or the unknown ultrasonic requency : > 70 khz beat 0 khz 70 khz or 0 khz + 70 khz 90 khz < 70 khz beat 0 khz 70 khz or 70 khz 0 khz 50 khz Next, we assume that the beat requency beat is 0 Hz. Ignoring signiicant igures, we now hae the two ollowing possibilities or the unknown ultrasonic requency :

17 87 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA > 70 khz 3 beat 0 0 khz 70 khz 3 or 0 0 khz + 70 khz khz 3 < 70 khz beat 0 0 khz 70 khz 3 or 70 khz 0 0 khz khz a. From these our alues, we can see that the smallest possible requency or the other ultrasonic wae is 50 khz. b. From these our alues, we can see that the largest possible requency or the other ultrasonic wae is 90 khz. 3. SSM REASONING When two requencies are sounded simultaneously, the beat requency produced is the dierence between the two. Thus, knowing the beat requency between the tuning ork and one lute tone tells us only the dierence between the known requency and the tuning-ork requency. It does not tell us whether the tuning-ork requency is greater or smaller than the known requency. Howeer, two dierent beat requencies and two lute requencies are gien. Consideration o both beat requencies will enable us to ind the tuning-ork requency. SOUTION The act that a -Hz beat requency is heard when the tuning ork is sounded along with the 6-Hz tone implies that the tuning-ork requency is either 63 Hz or 6 Hz. We can eliminate one o these alues by considering the act that a 3-Hz beat requency is heard when the tuning ork is sounded along with the 66-Hz tone. This implies that the tuning-ork requency is either 69 Hz or 63 Hz. Thus, the tuning-ork requency must be 63 Hz. 4. REASONING The beat requency heard by the bystander is the dierence between the requency o the sound that the bystander hears rom the moing car and that rom the (stationary) parked car. The bystander hears a requency o rom the moing horn that is greater than the emitted requency s. This is because o the Doppler eect (Section 6.9). The bystander is the obserer o the sound wae emitted by the horn. Since the horn is moing toward the obserer, more condensations and rareactions o the wae arrie at the obserer s ear per second than would otherwise be the case. More cycles per second means that the obsered requency is greater than the emitted requency. The bystander also hears a requency rom the horn o the stationary car that is equal to the requency s produced by the horn. Since this horn is stationary, there is no Doppler eect. SOUTION According to Equation 6., the requency o that the bystander hears rom the moing horn is

18 Chapter 7 Problems 873 o s s where s is the requency o the sound emitted by the horn, s is the speed o the moing horn, and is the speed o sound. The beat requency heard by the bystander is o s, so we ind that o s s s s s s ( 395 Hz) 4 Hz.0 m/s 343 m/s 5. REASONING When the wae created by the tuning ork is superposed on the sound wae traeling through the seawater, the beat requency beat heard by the underwater swimmer is equal to the dierence between the two requencies: beat wae ork () We know the requency ork o the tuning ork, and we will determine the requency wae o the sound wae rom its waelength λ and speed ia B speed o a sound wae in a liquid is gien by ρ adiabatic bulk modulus o the liquid, and ρ is its density. SOUTION Soling this result into Equation (), we obtain Substituting B waeλ (Equation 6.). The ad (Equation 6.6), where B ad is the waeλ (Equation 6.) or wae yields wae. Substituting λ beat wae ork () ork λ ad (Equation 6.6) into Equation (), we ind that ρ B ad Pa ρ 05 kg/m beat ork ork Hz 8 Hz λ λ 3.35 m 3 6. REASONING AND SOUTION The speed o the speakers is s π r/t π (9.0 m)/(0.0 s).83 m/s

19 874 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The sound that an obserer hears coming rom the right speaker is Doppler shited to a new requency gien by Equation 6. as OR s 00.0 Hz Hz s/ (.83 m/s )/ ( m/s) The sound that an obserer hears coming rom the let speaker is shited to a new requency gien by Equation 6. as O s 00.0 Hz 99.8 Hz + s/ + (.83 m/s )/ ( m/s) The beat requency heard by the obserer is then Hz 99.8 Hz.7 Hz 7. REASONING The time it takes or a wae to trael the length o the string is t /, where is the speed o the wae. The speed can be obtained since the undamental requency is known, and Equation 7.3 (with n ) gies the undamental requency as /(). The length is not needed, since it can be eliminated algebraically between this expression and the expression or the time. SOUTION Soling Equation 7.3 or the speed gies. With this result or the speed, the time or a wae to trael the length o the string is 3 t.95 0 s 56 Hz 8. REASONING For a ibrating string that is ixed at both ends, there are an integer number o hal-waelengths between the ends. Each hal-waelength gies rise to one loop in the standing wae. From this inormation, we can determine the waelength. Once the waelength λ is known, the speed o the waes ollows rom λ (Equation 6.), since the requency is gien in the statement o the problem. SOUTION a. In the distance o.50 m there are ie loops, so there are ie hal-waelengths. Thus, the waelength o the wae can be obtained by noting that λ 5.50 m or λ.00 m b. The speed o the wae can be obtained directly rom Equation 6.: λ 85.0 Hz.00 m 85.0 m/s

20 Chapter 7 Problems 875 c. The undamental requency arises when there is only one loop in the standing wae. The waelength in that case is ()(λ/).50 m, or λ 5.00 m. The undamental requency o the string can be obtained now rom Equation 6.: 85.0 m/s λ 5.00 m 7.0 Hz 9. SSM REASONING The undamental requency is gien by Equation 7.3 with n : /( ). Since alues or and are gien in the problem statement, we can use this expression to ind the speed o the waes on the cello string. Once the speed is known, the tension F in the cello string can be ound by using Equation 6., F/( m/ ). SOUTION Combining Equations 7.3 and 6. yields F m/ Soling or F, we ind that the tension in the cello string is F 4 ( m/ ) 4(0.800 m) (65.4 Hz) (.56 0 kg/m) 7 N 30. REASONING The harmonic requencies are integer multiples o the undamental requency. Thereore, or wire A (on which there is a second-harmonic standing wae), the undamental requency is one hal o 660 Hz, or 330 Hz. Similarly, or wire B (on which there is a third-harmonic standing wae), the undamental requency is one third o 660 Hz, or 0 Hz. The undamental requency is related to the length o the wire and the speed at which indiidual waes trael back and orth on the wire by /() (Equation 7.3, with n ). This relation will allow us to determine the speed o the wae on each wire. SOUTION Using Equation 7.3 with n, we ind Wire A or. m 330 Hz 790 m/s Wire B. m 0 Hz 530 m/s

21 876 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 3. SSM REASONING According to Equation 7.3, the undamental (n ) requency o a string ixed at both ends is related to the wae speed by /, where is the length o the string. Thus, the speed o the wae is. Combining this with Equation 6., F/( m/ ), we ind, ater some rearranging, that F 4 ( m/ ) Since the strings hae the same tension and the same lengths between their ixed ends, we hae ( m/ ) ( m/ ) E E G G where the symbols E and G represent the E and G strings on the iolin. This equation can be soled or the linear density o the G string. SOUTION The linear density o the string is E E G E E G G ( m/ ) ( m/ ) ( m/ ) Hz kg/m kg/m 96.0 Hz 3. REASONING The series o natural requencies or a wire ixed at both ends is gien by n/ (Equation 7.3), where the harmonic number n takes on the integer alues, n, 3, etc.. This equation can be soled or n. The natural requency n is the lowest requency that the human ear can detect, and the length o the wire is gien. The speed at which waes trael on the wire can be obtained rom the gien alues or the tension and the wire s linear density. SOUTION According to Equation 7.3, the harmonic number n is n n () F The speed is (Equation 6.), where F is the tension and m/ is the linear m/ density. Substituting this expression into Equation () gies n n / kg/m n n m ( 0.0 Hz) ( 7.60 m) F F 33 N m/

22 Chapter 7 Problems REASONING A standing wae is composed o two oppositely traeling waes. The speed F o these waes is gien by (Equation 6.), where F is the tension in the string m/ and m/ is its linear density (mass per unit length). Both F and m/ are gien in the statement o the problem. The waelength λ o the waes can be obtained by isually inspecting the standing wae pattern. The requency o the waes is related to the speed o the waes and their waelength by /λ (Equation 6.). SOUTION a. The speed o the waes is F 80 N 80 m/s m/ kg/m b. Two loops o any standing wae comprise one waelength. Since the string is.8 m long and consists o three loops (see the drawing), the waelength is λ.8 m. m 3 λ.8 m c. The requency o the waes is 80 m/s 50 Hz λ. m 34. REASONING Equation 7.3 (with n ) gies the undamental requency as /(), where is the wire s length and is the wae speed on the wire. The speed is gien by Equation 6. as F, where F is the tension and m is the mass o the wire. m / SOUTION Using Equations 7.3 and 6., we obtain F / F 60 N m m kg 0.4 m 3 ( ) 30 Hz 35. REASONING The undamental requency o the wire is gien by /() (Equation 7.3, with n ), where is the speed at which the waes trael on the wire and is the length o the wire. The speed is related to the tension F in the wire according to (Equation 6.), where m/ is the mass per unit length o the wire. F m/

23 878 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The tension in the wire in Part o the text drawing is less than the tension in Part. The reason is related to Archimedes principle (see Equation.6). This principle indicates that when an object is immersed in a luid, the luid exerts an upward buoyant orce on the object. In Part the upward buoyant orce rom the mercury supports part o the block s weight, thus reducing the amount o the weight that the wire must support. SOUTION Substituting Equation 6. into Equation 7.3, we can obtain the undamental requency o the wire: F () m/ In Part o the text drawing, the tension F balances the weight o the block, keeping it rom alling. The weight o the block is its mass times the acceleration due to graity (see Equation 4.5). The mass, according to Equation. is the density ρ copper times the olume V o the block. Thus, the tension in Part is Part tension F ( mass) g ρcoppervg In Part o the text drawing, the tension is reduced rom this amount by the amount o the upward buoyant orce. According to Archimedes principle, the buoyant orce is the weight o the liquid mercury displaced by the block. Since hal o the block s olume is immersed, the olume o mercury displaced is V. The weight o this mercury is the mass times the acceleration due to graity. Once again, according to Equation., the mass is the density ρ mercury times the olume, which is V. Thus, the tension in Part is Part tension copper mercury F ρ Vg ρ V g With these two alues or the tension we can apply Equation () to both parts o the drawing and obtain ρcoppervg Part m/ Part ρ Vg ρ m/ copper mercury V g Diiding the undamental requency o Part by that o Part gies

24 Chapter 7 Problems 879 ρ Vg ρ V g m/ ρ Vg m/ copper mercury ρ, Part copper, Part copper copper ρ ρ mercury kg/m kg/m 8890 kg/m REASONING The requencies n o the standing waes allowed on a string ixed at both ends are gien by Equation 7.3 as n n, where n is an integer that speciies the harmonic number, is the speed o the traeling waes that make up the standing waes, and is the length o the string. The speed is related to the tension F in the string and the F linear density m/ ia (Equation 6.). Thereore, the requencies o the m/ standing waes can be written as F m/ n F n n n m/ The tension F in each string is proided by the weight W (either W A or W B ) that hangs rom n W the right end, so F W. Thus, the expression or n becomes n m/. We will use this relation to ind the weight W B. A W A B W B SOUTION String A has one loop so n, and the requency A wae is WA m/. String B has two loops so n, and the requency B WB m/. We are gien that the two requencies are equal, so A o this standing wae is B o this standing

25 880 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA Soling this expression or W B gies W B W A m / m / A 4 W A 4 B W B 44 N N 37. SSM REASONING We can ind the extra length that the D-tuner adds to the E-string by calculating the length o the D-string and then subtracting rom it the length o the E string. For standing waes on a string that is ixed at both ends, Equation 7.3 gies the requencies as n n( / ). The ratio o the undamental requency o the D-string to that o the E-string is D /( D) E /( ) E E D This expression can be soled or the length D o the D-string in terms o quantities gien in the problem statement. SOUTION The length o the D-string is D E 4. Hz E (0.68 m) m D 36.7 Hz The length o the E-string is extended by the D-tuner by an amount D E m 0.68 m m 38. REASONING The beat requency is equal to the higher requency o the shorter string minus the lower requency o the longer string. The reason the longer string has the lower requency can be seen rom the drawing, where it is eident that both strings are ibrating at their undamental requencies. The undamental requency o ibration (n ) or a string ixed at each end is gien by /() (Equation 7.3). Since the speed is the same or both strings (see the ollowing paragraph), but the length is greater or the longer string, the longer string ibrates at the lower requency. The waes on the longer string hae the same speed as those on the shorter string. The speed o a transerse wae on a string is gien by 0.57 cm

26 Chapter 7 Problems 88 F/ m/ (Equation 6.), where F is the tension in the string and m/ is the mass per unit length (or linear density). Since F and m/ are the same or both strings, the speed o the waes is the same. SOUTION The beat requency is the requency o the shorter string minus the requency o the longer string; shorter longer. We are gien that shorter 5 Hz. According to Equation 7.3 with n, we hae longer /( longer ), where longer is the length o the longer string. According to the drawing, we hae longer m. Thus, longer m longer ( + ) Since the speed o the waes on the longer string is the same as those on the shorter string, 4.8 m/s. The length o the shorter string can be obtained directly rom Equation 7.3: 4.8 m/s m 5 Hz Substituting this number back into the expression or longer yields longer 4.8 m/s Hz m m m ( + ) ( + ) The beat requency is shorter longer 5 Hz Hz 3 Hz. 39. SSM REASONING The beat requency produced when the piano and the other instrument sound the note (three octaes higher than middle C) is beat 0, where is the requency o the piano and 0 is the requency o the other instrument ( Hz ). We can ind by considering the temperature eects and the mechanical eects that occur when the temperature drops rom 5.0 C to 0.0 C. SOUTION The undamental requency 0 o the wire at 5.0 C is related to the tension F 0 in the wire by 0 F0 /( m/ ) () 0 0 where Equations 7.3 and 6. hae been combined. The amount Δ by which the piano wire attempts to contract is (see Equation.) Δ α 0 Δ T, where α is the coeicient o linear expansion o the wire, 0 is its length at 5.0 C, and ΔT is the amount by which the temperature drops. Since the wire is preented

27 88 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA rom contracting, there must be a stretching orce exerted at each end o the wire. According to Equation 0.7, the magnitude o this orce is Δ Δ F Y A 0 where Y is the Young's modulus o the wire, and A is its cross-sectional area. Combining this relation with Equation., we hae α 0 ΔT Δ F Y A α( ΔT) YA 0 Thus, the requency at the lower temperature is ( F0 +ΔF)/( m/ ) F0 + α ΔT YA /( m/ ) () Using Equations () and (), we ind that the requency is /( / ) α F + α Δ T YA m F + ΔT YA F F0 /( m/ ) 0 ( 093 Hz) 05 Hz N + ( 0 /C )(5.0 C )(.0 0 N/m ) ( m ) 88.0 N Thereore, the beat requency is 05 Hz 093 Hz Hz. 40. REASONING AND SOUTION We are gien that. a. The length o the unretted string is 0 /( 0 ) and the length o the string when it is pushed against ret is /( ). The distance between the rets is j j ( 0.68 m)( 0.056) m. b. The requencies corresponding to the sixth and seenth rets are 6 7. The distance between ret 6 and ret 7 is 7 0 and 6 0

28 Chapter 7 Problems ( 0.68 m)( ) m ( ) 4. SSM REASONING Equation 7.5 (with n ) gies the undamental requency as /(4), where is the length o the auditory canal and is the speed o sound. SOUTION Using Equation 7.5, we obtain 343 m/s Hz m 4. REASONING The undamental (n ) requency or a tube open at both ends is /() (Equation 7.4). The undamental requency o a tube open at only one end is /(4) (Equation 7.5). Since we know the undamental requency o the tube open at only one end, we can use these relations to determine the other undamental requency. SOUTION a. The ratio o the undamental requency o a tube open at both ends to that o a tube open at only one end is open at both ends open at only one end 4 Thus, ( ) ( ) 30.8Hz 6.6Hz open at both ends open at only one end b. The length o the tube can be ound rom either Equation 7.4 or 7.5. Soling /() (Equation 7.4) or reeals that 343 m/s 6.6Hz m

29 884 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 43. REASONING The requency o a pipe open at both ends is gien by Equation 7.4 as n n, where n is an integer speciying the harmonic number, is the speed o sound, and is the length o the pipe. SOUTION Soling the equation aboe or, and recognizing that n 3 or the third harmonic, we hae 343 m/s n 3.96 m ( 6 Hz) n 44. REASONING AND SOUTION We know that /. For 0.0 Hz For 0.0 khz (343 m/s)/[(0.0 Hz)] 8. 6 m (343 m/s)/[( Hz)] m 45. REASONING The undamental requency ends, is gien by Equation 7.4 with n : A o air column A, which is open at both A, where is the speed o sound A in air and A is the length o the air column. Similarly, the undamental requency B o air column B, which is open at only one end, can be expressed using Equation 7.5 with n : B. These two relations will allow us to determine the length o air column B. 4 B SOUTION A B, so that Since the undamental requencies o the two air columns are the same A 4 B A B or B A ( 0.70 m ) 0.35 m 46. REASONING For a tube open at both ends, the sequence o natural requencies n is gien by n n (Equation 7.4), where n,, 3, 4,... The speed o sound is, and the length o the tube is. For the undamental requency n. This equation can be soled or the lengths o the piccolo and the lute, and the desired ratio obtained rom the results.

30 Chapter 7 Problems 885 SOUTION Soling the expression or the undamental requency (Equation 7.4 with n ) or the length, we obtain or Diiding piccolo by lute gies piccolo piccolo lute lute ( ) ( ) 6.6 Hz Hz piccolo ( ) lute SSM REASONING The natural requencies o a tube open at only one end are gien by Equation 7.5 as n n, where n is any odd integer (n, 3, 5, ), is the speed o 4 sound, and is the length o the tube. We can use this relation to ind the alue or n or the 450-Hz sound and to determine the length o the pipe. SOUTION a. The requency n o the 450-Hz sound is gien by 450 Hz n. ikewise, the 4 requency o the next higher harmonic is 750 Hz ( n + ), because n is an odd 4 integer and this means that the alue o n or the next higher harmonic must be n +. Taking the ratio o these two relations gies ( n + ) 750 Hz 4 n Hz n n 4 Soling this equation or n gies n 3. b. Soling the equation 450 Hz n or and using n 3, we ind that the length o 4 the tube is 343 m/s n m 4 n 4( 450 Hz) 48. REASONING According to Equation 6., the requency o the ibrations in the rod is related to the waelength λ and speed o the sound waes by (6.) λ

31 886 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA The distance between adjacent antinodes o a longitudinal standing wae is equal to hal a waelength λ. This rod has antinodes at the ends only, so the length o the rod is equal to hal a waelength. Put another way, the waelength is twice the length o the rod: λ () The speed o sound in the rod depends upon the alues o Young s modulus Y and the density ρ o aluminum ia Equation 6.7: Y (6.7) ρ SOUTION Substituting Equation () into Equation 6. yields Substituting Equation 6.7 into Equation (), we obtain λ () 0 Y N/m ρ. m kg/m 00 Hz 49. SSM REASONING The well is open at the top and closed at the bottom, so it can be approximated as a column o air that is open at only one end. According to Equation 7.5, the natural requencies or such an air column are n n where n, 3, 5, K 4 The depth o the well can be calculated rom the speed o sound, 343 m/s, and a knowledge o the natural requencies n. SOUTION We know that two o the natural requencies are 4 and 70.0 Hz. The ratio o these two requencies is Hz 5 4 Hz 3 Thereore, the alue o n or each requency is n 3 or the 4-Hz sound, and n 5 or the 70.0-Hz sound. Using n 3, or example, the depth o the well is 3 n m/s Hz 6. m 50. REASONING The pressure P at a depth h in a static luid such as the mercury column is gien by P Patm + ρgh (Equation.4), where P atm Pa is the air pressure at the surace o the luid, ρ is the density o the luid, and g is the magnitude o the acceleration due to graity. Because the air-illed portion o the tube is open at one end, it

32 Chapter 7 Problems 887 can hae standing waes with natural requencies gien by n n (Equation 7.5), 4 where n can take on only odd integral alues (n, 3, 5, ), and is the speed o sound in air. The mercury decreases the eectie length o the air-illed portion o the tube rom its initial length m to its inal length. The third harmonic 3,0 o the original tube is ound by choosing n 3 in Equation 7.5, and the undamental requency o the shortened tube is ound by choosing n. We note that the height h o the mercury column is equal to the dierence between the original and inal lengths o the air in the tube: h 0. SOUTION Substituting h 0 into P Patm + ρgh (Equation.4), we obtain atm 0 P P + ρg () The third harmonic requency o the tube at its initial length 0 and the undamental requency o the tube at its inal length are equal, so rom we ind that n n (Equation 7.5), 4 3, or or Substituting Equation () into Equation (), we obtain atm + ρ 0 atm + ρ atm + ρ 3 0 P P g P g P g Thereore, the pressure at the bottom o the mercury column is () P.0 0 Pa kg/m 9.80 m/s 0.75 m.68 0 Pa 5. REASONING We will make use o the series o natural requencies (including the irst n/ (Equation 7.4), or a tube open at both ends. oertone requency) gien by n In this expression, n takes on the integer alues,, 3, etc. and has the alue o n or the irst oertone requency that is gien. We can sole Equation 7.4 or, but must deal with the act that no alue is gien or the speed o sound in gas B. To obtain the necessary alue, we will use the act that both gases are ideal gases and utilize the speed gien or gas A and the masses o the two types o molecules. SOUTION According to Equation 7.4 as applied to gas B, we hae n ( ) n B /, which can be soled or to show that n () B n

33 888 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA Since both gases are ideal gases, the speed is gien by γ kt/ m (Equation 6.5), where γ is the ratio o the speciic heat capacities at constant pressure and constant olume, k is Boltzmann s constant, T is the Kelin temperature, and m is the mass o a molecule o the gas. Noting that γ and T are the same or each gas, we can apply this expression to both gases: A γkt and B m Diiding the expression or gas B by that or gas A gies A γkt m γ kt m m m B B A or A B A A γ kt B B m m Substituting this result or B into Equation (), we ind that m B A B 6 nb na ma 59 m/s kg m m 386 Hz kg n n 5. REASONING AND SOUTION The original tube has a undamental gien by /(4), so that its length is /(4 ). The cut tube that has one end closed has a length o c /(4 c ), while the cut tube that has both ends open has a length o /( o ). We know that c + 0. Substituting the expressions or the lengths and soling or gies o c ( 45 Hz)( 675 Hz) 6 Hz Hz + 45 Hz c o 53. REASONING According to Equation 7.3, the length o the string is related to its third harmonic (n 3) requency 3 and the speed o the waes on the string by or The speed o the waes is ound rom Equation 6.: 3 () F (6.) m Here, F is the tension in the string and the ratio m/ is its linear density. SOUTION Substituting Equation 6. into Equation (), then, gies the length o the string as:

34 Chapter 7 Problems F () 3 3 m Although appears on both sides o Equation (), no urther algebra is required. This is because appears in the ratio m/ on the right side. This ratio is the linear density o the string, which has a known alue o kg/m. Thereore, the length o the string is 3 F N m 30 Hz kg/m m 54. REASONING The speed o sound is related to its requency and waelength λ by λ (Equation 6.). At the end o the tube where the tuning ork is, there is an antinode, because the gas molecules there are ree to ibrate. At the plunger, there is a node, because the gas molecules there are not ree to ibrate. Since there is an antinode at one end o the tube and a node at the other, the smallest alue o occurs when the length o the tube is one quarter o a waelength. SOUTION Since the smallest alue or is a quarter o a waelength, we hae 4 λ or λ 4. According to Equation 6., the speed o sound is λ Hz m 5 m/s 55. SSM REASONING When constructie intererence occurs again at point C, the path length dierence is two waelengths, or Δs λ 3.0 m. Thereore, we can write the expression or the path length dierence as This expression can be soled or s AB. AC BC AB BC BC s s s + s s 3.0 m SOUTION Soling or s AB, we ind that s AB (3.0 m +.40 m) (.40 m) 5.06 m 56. REASONING et A be length o the irst pipe, and B be the inal length o the second pipe. The length Δ remoed rom the second pipe, then, is Δ () A Both pipes are open at one end only, so their lengths are related to their undamental requencies and the speed o sound in air by (Equation 7.5 with n ), or 4 B

35 890 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA () 4 The beat requency beat that occurs when both pipes are ibrating at their undamental requencies is the dierence between the higher requency,b o the second pipe and the lower requency,a o the irst pipe: beat,b (3),A SOUTION Substituting Equation () into Equation () yields Δ A B 4,A 4,B 4,A,B Soling Equation (3) or the unknown requency o the shortened pipe, we ind that,b,a + (5) beat Substituting Equation (5) into Equation (4), we obtain 343 m/s Δ Hz 56 Hz+ Hz,A,B,A,A beat 0.05 m Thereore, the second pipe has been shortened by cutting o 00 cm ( 0.05 m ).5 cm m 57. REASONING For a tube open at only one end, the series o natural requencies is gien by n / 4 (Equation 7.5), where n has the alues, 3, 5, etc., is the speed o sound, n and is the tube length. We will apply this expression to both the air-illed and the heliumilled tube in order to determine the desired ratio. SOUTION According to Equation 7.5, we hae n n 4 4 air n, air and n, helium helium Diiding the expression or helium by the expression or air, we ind that nhelium 3 n, helium 4 helium.00 0 m/s.9 n n, air air air 343 m/s 4 (4) 58. REASONING According to the principle o linear superposition, when two or more waes are present simultaneously at the same place, the resultant wae is the sum o the indiidual waes. We will use the act that the pulses moe at a speed o cm/s to locate the pulses at

36 Chapter 7 Problems 89 the times t s, s, 3 s, and 4 s and, by applying this principle to the places where the pulses oerlap, determine the shape o the string. SOUTION The shape o the string at each time is shown in the ollowing drawings: t s t s t 3 s t 4 s SSM REASONING For standing waes on a string that is clamped at both ends, Equations 7.3 and 6. indicate that the standing wae requencies are F n n where m/ Combining these two expressions, we hae, with n or the undamental requency, F m/ This expression can be used to ind the ratio o the two undamental requencies. SOUTION The ratio o the two undamental requencies is

37 89 THE PRINCIPE OF INEAR SUPERPOSITION AND INTERFERENCE PHENOMENA Since Fnew 4F, we hae old old new Fold m/ Fnew m/ F F old new F 4 F 4 (55.0 Hz) ().0 0 Hz new old new old old old Fold Fold 60. REASONING Each string has a node at each end, so the requency o ibration is gien by Equation 7.3 as n n/(), where n,, 3, The speed o the wae can be determined rom Equation 6. as F / ( m/ ). We will use these two relations to ind the lowest requency that permits standing waes in both strings with a node at the junction. SOUTION Since the requency o the let string is equal to the requency o the right string, we can write n let right ( m/ ) ( m/ ) Substituting in the data gien in the problem yields n let let F let n right F right 90.0 N n 90.0 N right kg/m.50 0 kg/m 3.75m.5m This expression gies n let 6n right. etting n let 6 and n right, the requency o the let string (which is also equal to the requency o the right string) is 6 ( 6) 90.0 N kg/m 3.75 m 45.0 Hz 6. REASONING When the dierence l l in path lengths traeled by the two sound waes is a hal-integer number,,, K o waelengths, destructie intererence occurs at the listener. When the dierence in path lengths is zero or an integer number,, 3, K o waelengths, constructie intererence occurs. Thereore, we will diide the distance l l by the waelength o the sound to determine i constructie or destructie

38 Chapter 7 Problems 893 intererence occurs. The waelength is, according to Equation 6., the speed o sound diided by the requency ; λ /. SOUTION a. The distances l and l can be determined by applying the Pythagorean theorem to the two right triangles in the drawing: l.00 m +.83 m.85 m.83 m.87 m.00 m l l l.00 m +.87 m.500 m P Thereore, l l 0.35 m. The 343 m/s waelength o the sound is λ 0.34 m. Diiding the distance l l by 466 Hz the waelength λ gies the number o waelengths in this distance: l l 0.35 m Number o waelengths.5 λ 0.33 m Since the number o waelengths is a hal-integer number occurs at the listener., destructie intererence 343 m/s b. The waelength o the sound is now λ 0.35 m. Diiding the distance 977 Hz l l by the waelength λ gies the number o waelengths in that distance: l l 0.35 m Number o waelengths λ 0.35 m Since the number o waelengths is an integer number ( ), constructie intererence occurs at the listener. 6. REASONING The blows o the carpenters hammers all at requencies and, which are related to the period T between blows by T (Equation 0.5). The beat requency beat o their hammer blows is the requency at which the blows all at the same instant. The time between these simultaneous blows is the period T beat o the beats. SOUTION