SUPERPOSITION AND STANDING WAVES 16

Transcription

1 SUPERPOSITION AND STANDING WAVES 6 Q6.. Reason: Where there is a change in ediu in particular a change in the wae speed then reflection can occur. Assess: Light traels at different speeds in water and air, and so soe is reflected at a water-air interface. Q6.. Reason: When a wae encounters a discontinuity in the ediu (sudden change in ediu properties in this case depth) there is partial reflection and partial transission. The incident and reflected waes will hae the sae speed because the ediu is the sae. The reflected and transitted wae will hae a different speed, and this speed depends on the critical property (here the depth of the ediu). If we consider the liiting case where the depth of the water goes to zero, the speed of the waes goes to zero. Fro this we conclude that as the depth of the water decreases, the speed of the water wae decreases. Assess: The aplitude of the reflected and transitted wae will be saller than for the incident wae because the energy of the incident wae is diided between the. The phase of the reflected wae ight change, depending on the relatie agnitude of the critical factor on either side of the discontinuity. Q6.3. Reason: See Figure 6.0 in the text. (a) In Chapter 5 we saw that the speed of a wae on a stretched string is = T / µ. s Since the left side of the string has a lower speed, the linear density ust be greater there. (b) You would start a pulse fro the left side, in the part with the greater linear density, in order to hae the reflection not inerted. Assess: This would be a fairly easy experient to set up at hoe with two different strings tied together. Q6.4. Reason: Knowing that the frequency of a ibrating string is deterined by fn = ( n/( L)) Tµ /, we can conclude that the frequency is directly proportional to the square root of the tension in the string. As a result, if the tension in the string increases, the rate at which it ibrates (the frequency) will also increase. Assess: The next tie you are around a guitar or other string instruent gie this a try. Q6.5. Reason: The frequency of ibration of your ocal cords is related to their linear ass density by f = ( / ( L)) T /. Due to the inflaation, the ocal cords are ore assie and hence hae a greater linear ass density. Since the inflaed ocal cords hae a greater linear ass density, the frequency of ibration of the cords and hence the frequency of the sound generated will decrease. Assess: You are no doubt aware that the frequency of your oice lowers with certain illnesses. Q6.6. Reason: When the frequency is doubled the waelength is haled. This haling of the waelength will increase the nuber of antinodes to six. Assess: Calculating the nuber of antinodes in this situation is easier than calculating the nuber of nodes because there are nodes on each end of the string, so the nuber is not doubled: It goes fro four to seen. Q6.7. Reason: When standing waes are set up in a tube that is open at both ends, the length of the tube is an integral nuber of half waelengths L= λ/. Looking at the figure we see four half waelengths, hence = 4, and the air colun is ibrating in the fourth haronic. Since this is sound, we hae a longitudinal wae and the air olecules are ibrating horizontally, parallel to the tube. Assess: The wae diagra superiposed on top of the open-ended tube sketch is a representation of the pressure at different points in the gas. It is not an air olecule displaceent sketch. 6-

2 6- Chapter 6 Q6.8. Reason: According to Conceptual Exaple 6.7, a flute ay be odeled as an open-open tube. Since the piccolo is considered to be a sall flute, it ay also be odeled by an open-open tube. When standing wae resonance is set up in an open-open tube, the length of the tube is soe integral nuber of half waelengths. L= λ/ or λ = ( L / ) Inserting this expression for the waelength into the relationship for speed, frequency and waelength obtain = fλ = f( L / ) or f = /( L) This last expression for the frequency in ters of the length of the tube allows us to conclude that the frequency in an open-open tube is inersely proportional to the length of the tube. Hence a shorter instruent (saller L), such as the piccolo, will produce higher-frequency sounds. Assess: Haing the ability to inspect a function such as the one preiously discussed for the frequency in order to deterine what will occur is a skill that physics will help you deelop. Q6.9. Reason: The adantage of haing low-frequency organ pipes closed at one end is that they will sound an octae lower without being twice as long as an open-open pipe. Because a pipe closed on one end contains only /4 of a waelength in the lowest ode, the waelength is twice as long as for an open-open pipe of the sae length. If the waelength is twice as long then the frequency is half, and that corresponds to a usical interal of an octae. Assess: Pipes closed on one end also hae a different sound (tibre) than open-open pipes sounding the sae note. Q6.0. Reason: Modeling the flute as an open-open tube, we know that the length is an integral nuber of half waelengths ( L= λ /). Cobining this with the relationship between the speed, frequency, and waelength of the traeling wae ( = fλ ), we get an expression for the frequency of the sound fro the flute as a function of the speed of sound in air and the length of the flute. As the flute wars up, L increases and hence f decreases this will cause the flute to go flat. As the air wars up, the speed of sound in air increases and this will cause the flute to go sharp. Since we are told that the flute goes sharp, the increase in the speed of sound due to the waring of air doinated the increase in length of the flute due to theral expansion. Assess: Haing the ability to inspect a function such as the one preiously discussed for the frequency in order to deterine what will occur is a skill that physics will help you deelop. Q6.. Reason: While the fundaental frequencies of noral oices are below 3000 Hz, there are higher haronics ixed in that gie the oice its characteristic sound and coney inforation. These haronics are lost when the telephone cuts off frequencies aboe 3000 Hz, and it is a bit harder to understand what is said. Assess: The range of huan hearing is usually quoted as 0 Hz 0 khz; a large portion of that range is aboe 3000 Hz. This principle holds for usic as well as spoken word. CDs are designed to reproduce frequencies clearly up to 0 khz (and een a tiny bit higher), een though the usical notes theseles (the fundaental frequencies) are not nearly that high. Those higher haronics gie the sound definition and sharpness; the usic sounds uddled if the high frequencies are cut out. Q6.. Reason: Treating the instruent as an open-closed tube, the frequencies are f = ( /(4 L)). Inspecting this relationship, we expect the pitch of the instruent to be slightly higher since the speed of sound is greater in heliu. Assess: Haing the ability to inspect a function such as the one preiously discussed for the frequency in order to deterine what will occur is a skill that physics will help you deelop. Q6.3. Reason: The sound you hear is the ibration of the glass; it is set into otion by the disturbance of the flowing, sloshing liquid. The disturbances are of any frequencies, but the natural frequency of the glass resonates and is aplified by the glass while other frequencies are quickly daped out. The reason the pitch rises as the glass fills is that the natural frequency of the glass changes; as the glass fills the resonance frequency rises. Assess: This is also the basis of the kitchen orchestra; get a set of glasses and fill the with arying aounts of water, then tap the lightly on the side with a spoon. You can tune the by adding or reoing sall aounts of water. Make a whole octae scale and play your faorite tunes.

3 Superposition and Standing Waes 6-3 Q6.4. Reason: The introduction of heliu into the outh allows haronics of higher frequencies to be excited ore than in the noral oice. The fundaental frequency of the oice is the sae but the quality has changed. Our perception of the quality is a function of which haronics are present. Assess: The fundaental frequency of a coplex tone fro the oice is deterined by the ibration of the ocal cords and depends on the tension and linear ass density of the cords, not the gas in the outh. Q6.5. Reason: It is the haronics and forants that help us understand what is being said in the first place (that is, the haronics of the ocal cords, odified by the forants of the ocal tract, ake an ee sound different fro an oo sound at the sae pitch). The haronics are ultiples of the fundaental, and when the fundaental is high (000 Hz or ore), the haronics are out of the range of hearing for any people. The forants ephasize different haronics and allow us to distinguish different ocalizations, but if the haronics cannot be heard then it would be difficult to tell the difference between an ee sound and an oo sound. Assess: The upper liit on the frequencies people can hear aries fro person to person. Men and people who hae endured lengthy periods of ery loud sounds cannot hear frequencies as high as woen and people who e protected their hearing, on aerage. Q6.6. Reason: The figure below shows a string and a bat ibrating with transerse standing waes in the = and = odes. The figure akes it clear that the sweet spot is one fourth the distance fro the end of the bat. Assess: You can try this by hitting soething with the bat at a potition one fourth the distance fro the end and then at about the idpoint of the bat. Q6.7. Reason: The quality of the sound of your oice depends on the resonant frequencies of the caities (throat, outh, and nose) in your ocal tract. When one of these caities is stuffed up, the quality of your oice will be affected. Assess: In this case the fundaental frequency has not changed but the ability to create and enhance certain haronics has. The haronics play an iportant role in the quality of the sound of your oice. Q6.8. Reason: The pitch ay be the sae, but the ocal tract is considerably different. The ocal tract deterines the forants that ephasize different haronics and allow us to distinguish different ocalizations. Assess: Due to the forants, it is easy to distinguish the oice of a child fro the oice of an adult. Q6.9. Reason: Two points that experience crests at the sae tie would hae their displaceents in phase. The displaceents at points A, B, C, D, and F are all experiencing crests at the sae tie. The displaceent at point E is has a trough at that tie. (a) Displaceents at A and B are in phase; they experience the sae crest at the sae tie. (b) Displaceents at C and D are in phase; they experience different crests at the sae tie. (c) Displaceents at E and F are out of phase; E experiences a trough at the ties that F experiences a crest. Assess: Half a period later the displaceents at points A, B, C, D, and F are all troughs, but they are still in phase. The displaceent at point E will always be exactly out of phase with the displaceents at all the other labeled points. Q6.0. Reason: The sketch below shows the displaceent at t = 0 s and again at t = s. The displaceent at x = 7 c is read fro the sketch to be.0.

4 6-4 Chapter 6 The correct choice is D. Assess: The ability to inspect a figure and deterine what will occur is a skill that physics will help you deelop. Q6.. Reason: The earliest (and only) tie that y will equal at the point x = 3 c is when the top of the triangular peak on the left oes right to the point x = 3 c. That wae pulse is oing at 6 c/s, so it will take 0.5 s to oe 3 c. The correct choice is A. Assess: Because the pulse oing fro right to left is a negatie pulse, it will ake y = at t = 3s, but not a positie +. Q6.. Reason: Sketch (a) below shows each pulse at t = 0 s and again at t =.5 s. Sketch (b) shows the resultant interference as the two pulses pass through each other. Note that at x = 0 c the resultant aplitude is.0. The correct choice is B. Assess: The ability to inspect a figure and deterine what will occur is a skill that physics will help you deelop. Q6.3. Reason: The axiu displaceent will be the su of the contributions fro the two traeling waes at each point and at each tie; howeer, you are not guaranteed to be watching a point where a crest will eet a crest. It is true that you ight be watching an antinode where the axiu displaceent would be A, but it is also possible that you ight be watching a node that doesn t oe at all or any place in between. The correct choice is D. Assess: If you watch the whole string, you will see points whose axiu displaceent is A and other points whose axiu displaceent is 0 and other points in between. Q6.4. Reason: Since the lowest frequency that creates resonance is 0 Hz, we know that the fundaental frequency is f = 0 Hz. Note that the next haronic is three ties as uch. This inforation tells us that the student is using an open-closed tube as reasoned next. Successie frequencies for an open open tube are deterined by f = f where =,, 3,...,

5 Superposition and Standing Waes 6-5 Successie frequencies for an open-closed tube are deterined by f = f where =, 3, 5,..., Since the next frequency is three ties the fundaental frequency, the student is using an open-closed tube. As a result the next frequency will be f 5 = 5 f = 00 Hz. The correct choice is B. Assess: Open-closed tubes hae only the odd nubered haronics. Q6.5. Reason: We know that the speed, frequency, and waelength of a traeling wae are related by f = / λ. As the air and pipe war fro 0 C to 5 C, there is an insignificant expansion of the pipe, hence the resonant waelengths reain the sae. Howeer, the change in teperature of the air is significant copared to the teperature of the air. Since the speed of sound increases with teperature, the frequency will increase. The correct response is A. Assess: Your ability to look at an expression that describes a situation and predict what will happen under certain circustances will be enhanced by your study of physics. Q6.6. Reason: If the lengths of the two strings are the sae, then the fundaental waelength is the sae ( λ = L for a string tied down at both ends). This eliinates choices C and D. We will interpret the higher pitch to ean String has a higher fundaental frequency. The fundaental relationship for periodic waes is = λ f. If the λ s are the sae for both strings but f > fthen that iplies that >. The correct choice is A. Assess: Since the two strings are ade of the sae type of wire, µ is the sae for both. The cause of > is ( Ts) >( T s). Q6.7. Reason: For the lowest standing wae ode on a string, the waelength is twice the length of the string. If this is the case the speed of the disturbance is deterined by = fλ = (0 Hz)(.0 ) = 0 /s The correct choice is B. Assess: For the lowest standing wae ode on a string, one-half a waelength fits into the string. Q6.8. Reason: l new = (4/5) l old. But the properties of the string (T s and µ ) haen t changed, so hasn t changed. That is, =. new old Let s sole the fundaental relationship for the frequency f = / λ. f 4 ( lold ) new new new old 5 f = old old l = old old l = old f new / l / 5 / / = old = (440 Hz) = 550 Hz 4 4 The correct choice is D. Assess: We (especially the usicians aong us) already knew (or suspected) that a shorter string would produce a higher frequency, so we tentatiely eliinated choices A and B. The answer we cae up with jibes well with our intuition. Probles P6.. Prepare: The principle of superposition coes into play wheneer the waes oerlap. The waes are approaching each other at a speed of /s, that is, each part of each wae is oing eery second. Sole: The graph at t = s differs fro the graph at t = 0 s in that the left wae has oed to the right by and the right wae has oed to the left by. This is because the distance coered by the wae pulse in s is

6 6-6 Chapter 6. The snapshot graphs at t = s, 3 s, and 4 s are a superposition of the left- and the right-oing waes. The oerlapping parts of the two waes are shown by the dotted lines. Assess: This is an excellent proble because it allows you to see the progress of each wae and the superposition (addition) of the waes. As tie progresses, you know exactly what has happened to each wae and to the superposition of these waes. P6.. Prepare: The principle of superposition coes into play wheneer the waes oerlap. The waes are approaching each other at a speed of /s, that is, each part of each wae is oing eery second. Sole: The graph at t = s differs fro the graph at t = 0 s in that the left wae has oed to the right by and the right wae has oed to the left by. This is because the distance coered by the wae pulse in s is. The snapshot graphs at t = 4 s and t = 6 s are a superposition of the left- and the right-oing waes. The oerlapping parts of the two are shown by the dotted lines.

7 Superposition and Standing Waes 6-7 Assess: This is an excellent proble because it allows you to see the progress of each wae and the superposition (addition) of the waes. As tie progresses, you know exactly what has happened to each wae and to the superposition of these waes. P6.3. Prepare: The principle of superposition coes into play wheneer the waes oerlap. Sole: As graphically illustrated in the following figure, the snapshot graph of Figure P6.3 was taken at t = 4 s.

8 6-8 Chapter 6 Assess: This is an excellent proble because it allows you to see the progress of each wae and the superposition (addition) of the waes. As tie progresses, you know exactly what has happened to each wae and to the superposition of these waes. P6.4. Prepare: Look at the original graph and concentrate on the point x = 3while you entally let the pulse trael to the right at /s. After s the leading edge of the pulse will reach the point of interest. Reeber that since the string is fixed at 5 the pulse will inert on reflection. Sole: Assess: You can also track the trailing edge of the pulse (at the end of the triangular section). It will take 4 s to get to x = 3and 4 ore seconds to get back to the sae point x = 3after reflection. P6.5. Prepare: The pulse will hae to trael to the end of the string (4.0 ), be reflected back to the other end of the string (0.0 ), and then be reflected again and trael to the sae position (6.0 ) in order to hae the sae appearance as at t = 0 s. This eans the total distance traeled is 0.0 and that the pulse is traeling at a speed of 4.0 /s. Sole: The tie required is t= x / x = 0.0 /(4.0 /s) = 5.0 s. Assess: This is basically a straightforward kineatics proble. P6.6. Prepare: Knowing the frequency and waelength of the traeling wae disturbance, we can deterine how fast it is traeling along the string ( = fλ). Standing waes will be established along the entire length of the string once the disturbance has traeled to the end of the string and back. The tie for the traeling wae disturbance to trael to the end of the string and back ay be deterined by t = L/. Sole: The tie for the traeling wae to trael to the end of the string and back is L L (3.0) t =.7 s = fλ = (3.5Hz)(.0) = Assess: This is a reasonable tie for the disturbance to trael 6.0. P6.7. Prepare: Reflections at both ends of the string cause the foration of a standing wae. Figure P6.7 indicates that there are three full waelengths on the.0--long string and that the wae speed is 40 /s. We will use Equation 5.0 to find the frequency of the standing wae. Sole: The waelength of the standing wae is λ = (.0 ) = The frequency of the standing wae is thus 40 /s f = = = 60 Hz λ Assess: The units are correct and this is a reasonable frequency for a ibrating string. 3 P6.8. Prepare: Reflections at the string boundaries cause a standing wae on the string. The oscillating frequency of the wae on the string is 00 Hz.

9 Superposition and Standing Waes 6-9 Sole: Figure P6.8 indicates one and a half full waelengths on the string. Hence λ = (60 c) = 40 c. Thus = λ f = (0.40 )(00 Hz) = 40 /s. Assess: The units are correct and this is a reasonable speed for the wae to be traeling along the string. P6.9. Prepare: We assue that the string is tied down at both ends so there are nodes there. This eans the length of the string is L = λ in the fundaental ode (there are no nodes between the ends). λ = L = (0.89 ) =.78. Sole: Use the fundaental relationship for periodic waes: = λ f = (.78 )(30 Hz) = 53 /s. Assess: Reeber, we are talking about the speed of the wae on the string, not the speed of sound in air. These nubers are reasonable for bass guitar strings. P6.0. Prepare: For a string fixed at both ends, the possible ibration frequencies are related to the tension by f = ( /( L)) T/ Sole: If we hae a new tension T new = T/, we will hae a new frequency: f = ( /( L)) T / = ( /( L)) ( T/)/ = (/ ) f = (/ )(384 Hz) = 7 Hz new new Assess: We could sole this proble by just inspecting the aboe expression for the frequency as a function of the tension. We see the frequency is proportional to the square root of the tension. So if the tension is reduced by a factor of, the frequency will be reduced by a factor of. This will gie a new frequency of f new = f / = 384 Hz/ = 7 Hz. 3 P6.. Prepare: A string fixed at both ends supports standing waes. A standing wae can exist on the string only if its waelength is gien by Equation 6., that is, L λ =, =,, 3... The length L of the string is 40 c. Sole: (a) The three longest waelengths for standing waes will therefore correspond to =,, and 3. Thus, (.40 ) (.40 ) (.40 ) λ= = 4.80 λ = =.40 λ3 = =.60 3 (b) Because the wae speed on the string is unchanged fro one alue to the other, fλ (50.0 Hz)(.40 ) fλ = f3λ3 f3 = = = 75.0 Hz λ.60 3 Assess: The units on each deterination are correct and the alues are reasonable. The axiu waelength of a standing wae in a string is twice the length of the string and all other possible waelengths are fractions of this alue. P6.. Prepare: A string fixed at both ends fors standing waes. The waelengths of standing wae odes of a string of length L are gien by Equation 6., so we can easily deterine the third haronic waelength. With a known frequency of 80 Hz we can find the wae speed using Equation 5.0. Equation 5. will allow us to find the tension in the wire. Mass density µ of the wire is equal to the ratio of its ass and length. Sole: (a) The waelength of the third haronic is calculated as follows: L L.4 λ = λ3 = = = or 0.8 to two significant figures. (b) The speed of the waes on the string is = λ 3 f 3 = (0.807 )(80 Hz) = 45.3 /s. The speed is also gien by = T / µ, so the tension is S kg TS = = = (45.3 /s) = 70.0 N L.

10 6-0 Chapter 6 Assess: You ust reeber to use the linear density in SI units of kg/. Also, the speed is the sae for all odes, but you ust use a atching λ and f to calculate the speed. P6.3. Prepare: A string fixed at both ends fors standing waes. Three antinodes eans the string are ibrating as the = 3 standing wae. The waelengths of standing wae odes of a string of length L are gien by Equation 6.. Sole: (a) The frequency is f 3 = 3f, so the fundaental frequency is f = (40 Hz) = 40 Hz. The fifth haronic 3 will hae the frequency f 5 = 5f = 700 Hz. (b) The waelength of the fundaental ode is λ = L =.0. The wae speed on the string is = λ f = (.0 )(40 Hz) = 68 /s. Alternatiely, the waelength of the n = 3 ode is λ 3 = (L) = 0.40, fro 3 which = λ 3 f 3 = (0.40 )(40 Hz) = 68 /s. The wae speed on the string, gien by Equation 5., is = T S TS = = (0.000 kg/)(68 /s) = 56 N µ Assess: You ust reeber to use the linear density in SI units of kg/. Also, the speed is the sae for all odes, but you ust use a atching λ and f to calculate the speed. P6.4. Prepare: A string fixed at both ends fors standing waes. Sole: A siple string sounds the fundaental frequency f = /(L). Initially, when the string is of length L A = 3.8 c, the note has the frequency f A = /(L) A. For a different length, f C = /(L) C. Taking the ratio of each side of these two equations gies f /( L) L f = = L = f L L f A A C A C C /( ) C A C We know that the second frequency is desired to be f B = 53 Hz. The string length ust be 440 Hz L C = (3.8 c) = 7.6 c 53 Hz The question is not how long the string ust be, but where ust the iolinist place his finger. The full string is 3.8 c long, so the iolinist ust place his finger 5. c fro the end. Assess: A fingering distance of 5. c fro the end is reasonable. L A P6.5. Prepare: Reflections at the string boundaries cause a standing wae on a stretched string. The waelengths of standing wae odes of a string of length L are gien by Equation 6., so we can easily deterine the waelength fro the ibrating length of the string, which is.90. With a known frequency of 7.5 Hz we can find the wae speed using Equation 5.0. Equation 5. will now allow us to find the tension in the wire. Mass density µ of the wire is equal to the ratio of its ass and length. Sole: Because the ibrating section of the string is.9 long, the two ends of this ibrating wire are fixed, and the string is ibrating in the fundaental haronic. The waelength is L λ = λ = L = (.90 ) = 3.80 The wae speed along the string is = f λ = (7.5 Hz)(3.80 ) = 04.5 /s. The tension in the wire can be found as follows: T length.00 S ass kg = TS = = = (04.5 /s) = 80 N Assess: You ust reeber to use the linear density in SI units of kg/. Also, the speed is the sae for all odes, but you ust use a atching λ and f to calculate the speed. P6.6. Prepare: Expressions for the fundaental frequency for an open-open tube and an open-closed tube are respectiely gien by

11 Superposition and Standing Waes 6- f L oo = /( ) oo and foc = /(4 L) oc Sole: Soling these for the length of each tube, we hae Length for open-open tube: Loo = /( f) oo = (343 /s)/((0 Hz)) = 8.6 Length for open-closed tube: Loc = /(4 f) oc = (343 /s)/(4(0 Hz)) = 4.3 Assess: When the air in a tube is resonating in the fundaental ode, half of a waelength fits into an open-open tube and one fourth of a waelength fits into an open-closed tube. Since the waelength is the sae ( and f are the sae, so λ is the sae) for each case, the open-closed tube only needs to be half as long as the open-open tube. P6.7. Prepare: We want to use the fundaental relationship for periodic waes. But first, conert the length to SI units (use Table.3) L = 8 ft = 5.49 ft Sole: (a) For an open-closed pipe in the fundaental ode L = λ (one-quarter of a waelength fits in the pipe), λ = 4L = 4(5.49 ) = /s f = = = 6 Hz λ.9 This is below the arbitrary lower liit of the range of huan hearing. (b) We notice that the true alue (7.5 Hz) is different fro the answer we got in part (a). To find the effectie length of the instruent with a fundaental frequency of 7.5 Hz using the open-closed tube odel, we siply do the preious calculations in reerse order. First find λ for the fundaental ode. Now λ = 350 /s.7 f = 7.5 Hz = L = λ = (.7 ) = 3.8 = 0.4 ft 0 ft 4 4 Assess: The effectie length is just oer half of the real length. This would lead us to conclude that the openclosed tube odel is not a ery accurate odel for this situation. A good contrabassoon gies foundation and body to the orchestra. P6.8. Prepare: Please refer to Figure P6.8. We hae an open-open tube that fors standing sound waes. Sole: The gas olecules at the ends of the tube exhibit axiu displaceent, aking antinodes at the ends. There is another antinode in the iddle of the tube. Thus, this is the = ode and the waelength of the standing wae is equal to the length of the tube, that is, λ = Since the frequency f = 500 Hz, the speed of sound in this case is = fλ = (500 Hz)(0.80 ) = 400 /s. Assess: The experient yields a reasonable alue for the speed of sound. P6.9. Prepare: For the open-open tube, the two open ends exhibit antinodes of a standing wae. The possible waelengths for this case are λ = L /, where =,, 3... On the other hand, in the case of an open-closed tube λ = 4 L /, where =, 3, 5... The length of the tube is c. Sole: (a) The three longest waelengths are (.) (.) (.) λ = =.4 λ = =. λ3 = = (b) The three longest waelengths are 4(.) 4(.) 4(.) λ = = 4.84 λ = =.6 λ3 = = Assess: It is clear that the end of the air colun, whether open or closed, changes the possible odes. 4

12 6- Chapter 6 P6.0. Prepare: For an open-closed organ pipe, the length of the pipe is an integral nuber of quarter waelengths L= λ / 4, where =, 3, 5, etc. The expression = fλ connects speed with the waelength and frequency. Sole: For an open-closed pipe with a fundaental frequency of f = 6 Hz, the length of the pipe is L λ 343 /s 4 4 f 4 6 Hz sound = = = = 5.36 Assess: While this is a long pipe, it is a reasonable length for the longest pipe in the rack of pipes. P6.. Prepare: For the open-open tube, the fundaental frequency of the standing wae is f = 500 Hz when the tube is filled with heliu gas at 0 C. We will use Equation 6.6, which gies the allowed waelengths in an open-open tube. The speeds of sound in air and in heliu at 0 C are gien in Table 5.. Sole: For the fundaental ode, = in λ = L /, so using Equation 5.0, Siilarly, when the tube is filled with air, f f heliu heliu = = l air air air air l L f heliu 970 /s L 33/s f 33/s 33/s = = = f = (500 Hz) = 50 Hz 970 /s 970 /s Assess: Note that the length of the tube is one-half the waelength, whether the tube is filled with heliu or air. P6.. Prepare: For an open-closed organ pipe, the length of the pipe is an integral nuber of quarter waelengths L= λ / 4, where =, 3, 5, etc. The expression = fλ connects speed with the waelength and frequency. Sole: A general expression for the allowed frequiences is f = λ = (4 L/ ) = 4L where =, 3, 5, etc For the fundaental (=) we hae 350 / s f = () 58 Hz 4L = 4L = 4(.5 ) = Since all other frequencies are ultiples of this frequency we hae f3 = 3 f = 70 Hz and f5 = 5 f = 9 Hz Assess: These are reasonable frequencies and they are ultiples of the fundaental frequency. P6.3. Prepare: The frequencies at which resonance will occur in an open-open pipe are deterined by f = ( /( L)). Sole: (a) Knowing the speed of sound, the fundaental frequency is deterined by f = ()( /( L)) = (340 /s)/((30.0 )) = 5.67 Hz (b) The lowest frequency we can hear is about 0 Hz. For an open-open tube the frequency of the haronics are related to the fundaental frequency by f, = f hence the lowest haronic that would be audible to the huan ear is = f / f = 0 Hz/(5.67 Hz) = (c) As the air cools the speed of sound will decrease. Inspecting the function gien in the preious Prepare step, we see that this would in turn result in a decrease in frequency. Assess: This is a straightforward exaple of resonance in an open-open pipe. P6.4. Prepare: For an open-closed tube we need Equation 6.7

13 Superposition and Standing Waes 6-3 f = 4 L where we are gien that = (for the fundaental frequency) and that f = 00 Hz. We are also gien that = 345 /s. Sole: Sole the equation for L. 350 /s L= = () = 0.44 = 44 c 4 f 4(00 Hz) Assess: Since the 00 Hz is a typical fundaental frequency we don t really expect the length obtained to be the exact length of your ocal tract; but we do note that it is in the right ballpark (put a eter stick next to your stretched neck and guesstiate the distance fro your outh to your diaphrag). P6.5. Prepare: Follow Exaple 6.6 ery closely. Assue the ear canal is an open-closed tube, for which we need Equation 6.7 f = =, 3, 5, L where we are gien that L =.3 c = We assue that = 350 /s in the war ear canal. Take the audible range as 0 Hz 0 khz. Sole: Plug in arious alues of and obtain the corresponding frequencies. 350 /s f = () = 6730 Hz 6700 Hz 4(0.03 ) Higher frequencies are odd ultiples of this fundaental. f 3 = 3(6730 Hz) = 0 00 Hz Already f 3 is out of the audible range. So the only one in the audible range is f = 6700 Hz. P6.6. Prepare: We can deterine the frequency of the first forant fro Figure 6.3. Deeloping an expression for the resonant frequencies of an open-closed pipe, we can deterine the length of the tube. Sole: Notice in the graph that the first 000 Hz is diided into seen equal sections and the desired frequency is /7 of the way to 000 Hz or (/7)(000) = 90 Hz. The relationship between the speed of sound, the frequency of the sound, and the length of the open-closed pipe is 343 / s = fλ = f(4 L/ ) or L= / 4 f = / 4 f = = (90 Hz) Assess: This is a reasonable length for the ocal syste. P6.7. Prepare: First let s agree that the teperature of the air in the ocal tract is a little oer 0 C and hence the speed of sound is 350 /s. The relationship between speed, waelength, and frequency for a traeling wae disturbance in any ediu is = fλ. The frequency of ibration in air is caused by and is the sae as the frequency of ibration of the ocal cords. The length of the ocal tract is an integral nuber of half-waelengths L= λ /. The length of the ocal tract and hence the waelengths that cause standing wae resonance do not change as the dier descends. Howeer, since the speed of the sound waes changes, the frequency will also change. Sole: When a sound of frequency 70 Hz is coing out of the ocal tract, the waelength of standing waes established in the ocal tract associated with this frequency is λ = / f = (350 /s) / (70 Hz) =.96 As the dier descends, the ocal tract does not change length and hence this waelength for standing wae resonance will not change. Howeer since the sound is now traelling through a heliu-oxygen ixture with a speed of 750 /s, the frequency of the sound will change to f = / λ = (750 /s) / (.96 ) = 580 Hz

14 6-4 Chapter 6 Going through the sae procedure for sound at a frequence of 300 Hz we get the frequency in the heliuoxygen ixture to be f = 4900 Hz. Assess: We are aware that the sound should be at a higher frequency and the frequencies obtained hae higher alues. P6.8. Prepare: The two waes will produce destructie interference any tie their path-length difference is a whole nuber of waelengths plus half a waelength. Equation 6.9 describes this situation. Since the frequency of the sound waes is gien to be 686 Hz, we will use Equation 5.0 to find the waelength. Sole: (a) Let d represent the path-length difference. Using = 0 for the sallest d and the condition for destructie interference, we get d = + λ = 343 /s d = = = 0.50 f 686 Hz Assess: This is a reasonable nuber. If you hae two speakers, you can set the up in such a anner as to produce quiet spots. P6.9. Prepare: Interference occurs as a result of the path difference between the path lengths of the sound fro the two speakers. A separation of 0 c between the speakers leads to axiu intensity on the x-axis, but a separation of 30 c leads to zero intensity. Sole: (a) When the waes are in phase and lead to constructie interference, ( d) = λ = 0 c. For destructie interference, ( d) = ( + ) λ = 30 c. Thus, for the sae alue of λ ( d) ( d) = λ = (30 c 0 c) = 40 c (b) If the distance between the speakers continues to increase, the intensity will again be a axiu when the separation between the speakers that produced a axiu has increased by one waelength. That is, when the separation between the speakers is 0 c + 40 c = 60 c. Assess: The distances obtained are reasonable. As a check on our work we ight want to deterine the frequency of sound associated with the waelength. A waelength of 40 c is associated with a frequency of 860 Hz, which is in the audible range. P6.30. Prepare: We assue that the speakers are identical and that they are eitting in phase. Since you don t hear anything, the separation between the two speakers corresponds to the condition of destructie interference. Sole: Equation 6.9 for destructie interference is λ 3λ 5λ d = + λ d =,, Since the waelength is λ = 340 /s.0 f = 70 Hz = three possible alues for d are.0, 3.0, and 5.0. Assess: The units worked out and these are reasonable distances. P6.3. Prepare: Please refer to Figure P6.3. The circular wae fronts eitted by the two sources show that the two sources are in phase. This is because the wae fronts of each source hae oed the sae distance fro their sources. Sole: Let us label the top source as and the botto source as. For the point P, d = 3λ and d = 4λ. Thus, d = d d = 4λ 3λ = λ. This corresponds to constructie interference. 7 For the point Q, d = λ and d = λ, so d = d d = λ 7λ/ = 3 λ/. This corresponds to destructie interference. 5 7 For the point R, d = λ and d = λ d = d d = λ. This corresponds to constructie interference., r r r C/D

15 Superposition and Standing Waes 6-5 P 3λ 4λ λ C 7 Q λ λ 3 λ D 5 R λ 7 λ λ C Assess: When the path difference r is an integral nuber of whole waelengths (cases P and R), constructie interference occurs. When the path difference r is an integral nuber of half waelengths (case Q), destructie interference occurs. P6.3. Prepare: The two speakers are identical, and so they are eitting circular waes in phase. The oerlap of these waes causes interference. An oeriew of the proble follows. Sole: The waelength of the sound waes is Fro the geoetry of the figure, So, d = d d= = 0.47 d λ = 340 /s f = 800 Hz = = d + (.0 ) = (4.0 ) + (.0 ) = 4.47 Because d/λ = 0.47 /0.889 =.5 = 5/ = + or d = + λ, the interference is perfectly destructie. Assess: Destructie interference (for two waes in phase) will occur when the path difference is an integral nuber of half waelengths. P6.33. Prepare: Knowing the following relationships for a ibrating string, = T / µ, = fλ, and L= λ /, we can establish what happens to the frequency as the tension is increased. Knowing the relationship fbeat = f f, we can deterine the frequency of the string with the increased tension. T Sole: Cobining the first three expressions aboe, obtain f =, L which allows us to deterine that the frequency will increase as the tension increases. Using the expression for the beat frequency, we know that the difference frequency between the two frequencies is 3 Hz. Cobining this with our knowledge that the frequency increases, we obtain f = f f 3 Hz = f 00 Hz f = 03 Hz beat Assess: f is larger than f because the increased tension increases the wae speed and hence the frequency. P6.34. Prepare: Knowing the expression for the beat frequency, we can deterine by which the aount the frequency will change. But at this point we don t know if the frequency increases or decreases. Exaining the expression for the frequency of a flute (odeled as an open-open pipe ) as a function of its length, we can establish if the frequency increases or decreases when the tuning joint is reoed.

16 6-6 Chapter 6 Sole: Using the expression for the beat frequency, the flute player s initial frequency is either 53 Hz 4 Hz =57 Hz or 53 Hz 4 Hz =59 Hz. Modeling the flute as an open-open pipe we see that = fλ and L= λ /, which ay be cobined to obtain f = / ( L) This expression allows us to see that as the length increases, the frequency decreases. As a result we know that the initial frequency of sound fro the flute was 57 Hz. Assess: Since she atches the tuning fork s frequency by lengthening her flute, she is increasing the waelength of the standing wae in the flute. A waelength increase eans a decrease of frequency because = fλ. This confirs that the initial frequency was greater than the frequency of the tuning fork. P6.35. Prepare: We ll use the fundaental relationship for periodic waes ( = λ f, where f = 440 Hz) and we ll copute λ fro Equation 6. ( λ = L =.0, where = ) because the proble is about the fundaental frequency. Sole: = λ f = (.0 )(440 Hz) = 880 /s Assess: This speed is oer twice the speed of sound in air, but this string ight be thin (sall µ ) and under a lot of tension in order to ake the wae speed that high. P6.36. Prepare: Since when a sound wae hits the boundary between soft tissue and air, or between soft tissue and bone, ost of the energy is reflected, the situation is like a ibrating string with reflections at both ends. If that is the case, standing wae resonance will be established (and heating will occur) when the thickness of the soft tissue is an integral nuber of waelengths. Sole: The waelength of the ultrasound in the tissue is λ = = = 6 4 / f (350 /s) / Hz) Here we hae used = 350 /s as the speed of sound in soft tissue because the tissue is greater than 0 C. In order for standing wae resonance and hence heating to occur, the thickness of the tissue ust be an integral nuber of whole waelengths or T = λ (where T is the thickness of the soft tissue). Using this we obtain = T λ = = 4 / / Since this nuber is an integral, we conclude that sound at this frequency will produce heating in soft tissue of 0.50 c thickness. Assess: Since the frequency of the ultrasound and the thickness of the soft tissue are reasonable, we expect heating to occur. Since the waelength is so sall, if standing wae resonance does not occur it looks like a little pressure could copress the soft tissue and cause standing wae resonance to occur. P6.37. Prepare: According to Equation 6.6, the standing wae on a guitar string, ibrating at its fundaental frequency, has a waelength λ equal to twice the length L. We will first calculate the frequency of the wae that the string produces using Equations 5. and 5.0. The wae created by the guitar string traels as a sound wae with the sae frequency but with a speed of 343 /s in air. Sole: The wae speed on the stretched string is T 00 N 0.00kg/ S string = = = 447. /s The waelength of the wae on the string is λ = L = (0.80 ) =.60. Thus, the frequency of the wae is string 447. /s f = = = 79.5 Hz λ.60 Finally, the waelength of the sound wae that reaches your ear is sound 343 /s λ air = = =. f 79.5 Hz Assess: This is a reasonable waelength and the units are correct.

17 Superposition and Standing Waes 6-7 P6.38. Prepare: The relationship between the elocity, frequency, and waelength of a traeling wae disturbance is = fλ. The relationship between the elocity, tension, and linear ass density for a traeling wae disturbance in a string is = T / µ. The relationship that ust be satisfied to create standing wae resonance in a stretched tendon is L= λ /. Finally, the relationship between linear and olue ass density is µ = ρa. Sole: Start with = T / µ and insert µ = ρa to obtain = T /( ρ A). Express the waelength as λ = L/. Insert both and λ into = fλ and sole for the frequency. T /( ρa) = f ( L/ ) f = ( /( L)) T /( ρa) where =,, 3, Inserting alues, obtain the fundaental frequency: 3 4 f = (/ ((0.0 ))) 500 N / ((00 kg / )(.00 0 )) = 60 Hz Other possible frequencies are ultiples of the fundaental f = f = 30 Hz, f3 = 3 f = 480 Hz, etc. Assess: These are reasonable frequencies for the ibration of a tendon. P6.39. Prepare: For a string fixed at both ends, successie resonant frequencies occur at Sole: f = f and f( + ) = ( + ) f Inserting gien inforation into the preious expressions, obtain 35 Hz = f and 390 = ( + ) f Diiding the second expression by the first and for, obtain = 5. Knowing that = 5, we can use the first expression to obtain the fundaental frequency. f = 35 Hz/ = 35 Hz/5 = 65 Hz Or using the second expression, we obtain the sae alue. f = 390 Hz/( + ) = 390 Hz/(5 + ) = 65 Hz Assess: It is good practice to look for siple checks on your work. Calculating the fundaental frequency using both expressions is a siple check. P6.40. Prepare: The relationship between the elocity, frequency, and waelength of a traeling wae disturbance is = fλ. The relationship between the elocity, tension and linear ass density for a traeling wae disturbance in a string is = T / µ. The relationship that ust be satisfied to create standing wae resonance in a stretched tendon is L= λ /. Finally, the relationship between linear and olue ass density is µ = ρa. Sole: First deterine the elocity of the traeling wae disturbance of the struggling insect = f λ = f ( L / ) = fl = (00 Hz)(0. ) = 4 /s Next deterine the linear ass density of the web = ρa= ρπd / 4 = ( kg / )( π / 4)(.0 0 ) = kg/ 4. 0 kg/ Finally, deterine the tension in the web 9 6 T = = (4 /s) ( kg/) =.4 0 N Assess: This is a ery sall tension; howeer, it is seeral orders of agnitude greater than the weight of a sall insect that the spider ight wish to capture. P6.4. Prepare: The wae on a stretched string with both ends fixed is a standing wae. We ust distinguish between the sound wae in the air and the wae on the string. The iolin string oscillates at the sae frequency, because each oscillation of the string causes one oscillation of the air. But the waelength of the standing wae on the string is ery different because the wae speed on the string is not the sae as the wae speed in air. Sole: The listener hears a sound wae of waelength λ sound = 40 c = Thus, the frequency is

18 6-8 Chapter 6 f λ sound = = = sound 343 /s Hz Bowing a string produces sound at the string s fundaental frequency, so the waelength of the string is λ = λ = L= 0.60 = λ f = (0.60 )(857.5 Hz) = 54.5 /s string string string The tension in the string is found using Equation 5. as follows: string = TS TS = ( string) = (0.00kg/)(54.5 /s) = 70 N µ Assess: This is a relatiely large tension, but still not unusually large. P6.4. Prepare: The wae on a stretched string with both ends fixed is a standing wae. For ibration at its fundaental frequency, λ = L (see Equation 6.6). Sole: The waelength of the wae reaching your ear is 39. c = So the frequency of the sound wae is f λ 344 /s 0.39 air = = = This is also the frequency eitted by the wae on the string. Thus, Hz T 50 N = = = λ = = λ = = λ λ λ kg/ string S Hz L c 8 c to two significant figures. Assess: Since 8 c is less than the length of a iolin, it is a reasonable length for the ibrating section of the string. P6.43. Prepare: For the stretched wire ibrating at its fundaental frequency, the waelength of the standing wae fro Equation 6. is λ = L. Fro Equation 5., the wae speed is equal to T / µ, where S 3 3 = ass/length = kg/0.90 = kg/. The tension T S in the wire equals the weight of the sculpture or Mg. Sole: Thus, The wae speed on the steel wire is s wire = = wire = fλ = f( L) = (80 Hz)( 0.90 ) = 44 /s T Mg µ µ Assess: A ass of kg for the sculpture is reasonable. 3 wire ( kg/)(44 /s) M = = = kg g 9.8 /s P6.44. Prepare: For a string fixed at both ends, successie resonant frequencies occur at f = f and f( + ) = ( + ) f Sole: (a) Inserting gien inforation into the aboe expressions, obtain

19 Superposition and Standing Waes Hz = f and 3 Hz = ( + ) f Diiding the second expression by the first and for, obtain = 3. Knowing that = 3, we can use the first expression to obtain the fundaental frequency: f = 4 Hz/ = 4 Hz/3 = 8 Hz. Or using the second expression, we obtain the sae alue: f = 33 Hz/( + ) = 3 Hz/(3 + ) = 8.0 Hz. (b) Knowing the fundaental frequency and that the waelength is twice the length of the string for this case we can deterine the wae speed in the string: = fλ = (8 Hz)(.6 ) =.8 /s. (c) When the string is ibrating at 3 Hz, it is ibrating in the = 4 ode, or the fourth haronic, and four half waelengths or two waelengths will fit into the string as shown in the following figure. Assess: It is good practice to look for siple checks on your work. Calculating the fundaental frequency using both expressions is a siple check. P6.45. Prepare: The stretched string with both ends fixed fors standing waes. The astronauts hae created a stretched string whose ibrating length is L =.0. The weight of the hanging ass creates a tension T S = Mg in the string, where M =.0 kg. Sole: The wae speed on the string fro Equation 5. is T Mg µ µ S = = where µ = ( kg)/(.5 ) = kg/ is the linear density. The astronauts then obsere standing waes at frequencies of 64 Hz and 80 Hz. The first is not the fundaental frequency of the string because 80 Hz 64 Hz. But we can easily show that both are ultiples of 6 Hz: 64 Hz = 4 f and 80 Hz = 5 f. Both frequencies are also ultiples of 8 Hz. But 8 Hz cannot be the fundaental frequency because, if it were, there would be a standing wae resonance at 9(8 Hz) = 7 Hz. So the fundaental frequency is f = 6 Hz. The fundaental waelength fro Equation 6. is λ = L = 4.0. Thus, the wae speed on the string is = λf= 64.0 /s. Now we can find g on Planet X: = Mg µ kg/ g = = (64 /s) = 8. /s M.0 kg Assess: Coparing this to the acceleration due to graity on Earth, it is a reasonable alue. P6.46. Prepare: The stretched bungee cord that fors a standing wae with two antinodes is ibrating at the second haronic frequency.

20 6-0 Chapter 6 Sole: Because the ibrating cord has two antinodes, Equation 6. tells us that λ = L =.80. The wae speed on the cord is cord = fλ = (0 Hz)(.80 ) = 36 /s The tension T S in the cord is equal to k L, where k is the bungee s spring constant and L is the 0.60 that the bungee has been stretched. Thus, using Equation 5., S cord = = T k L µ µ cord kg (36 /s) k = = = 40 N/ L.0 (0.60 ) Assess: This is a reasonable alue for the spring constant of a bungee cord and the units are correct. P6.47. Prepare This will be a two-step proble. We ll first copute the wae speed fro the fundaental relationship for periodic waes, = λ f, and then use that in Equation 5. ( = T s/ µ ), which we will sole fort s. We will need to copute µ fro = L /, so we will also need to copute (the ass) fro the density and olue of the cable. Since the cable is ibrating in its fundaental ode and the cable is tied down at both ends, L = λ/for (the ode) =. We are gien L = 4, T (the period) = 0.40 s, and r = d/ =.5 c/ = The 3 density of steel is ρ steel = 7900 kg/. Sole: First, copute µ. rv rπ ( rl ) 3 µ ρπr (7900 kg/ ) π(0.05 ) 3.88 kg/ = = = = = = L L L Second, find fro the fundaental relationship for periodic waes. λ L 8 = = = = 70 /s T T 0.40 s Last, sole Equation 5. for T S and plug in the two preious results. T S = = (70 /s) (3.88 kg/) = 9000 N = 9 kn Assess: The answer is large, but not unexpected for a steel cable holding up a bridge. Notice in the calculation of µ that L canceled out; this is reasonable because µ should be the sae, assuing a unifor cable, regardless of the length. Note that the units work out in the last equation. P6.48. Prepare: According to Figure P6.48, a half waelength is 400 k. Fro the study of standing wae resonance we know that a standing wae goes fro axiu displaceent to noral water leel in one fourth of a period, which in this case is 3 hours. Finally we know that the elocity, waelength, and frequency of a traeling wae disturbance are related by = fλ. Sole: According to Figure P6.48, a half waelength is 400 k, so the waelength is λ =800 k. Knowing that a standing wae goes fro axiu displaceent to noral water leel in one fourth of a period, which in this 4 case is 3 hours, we get the period to be τ = h = s or a frequency of 4 5 f = / τ = / s =.3 0 Hz. Using the relationship = fλ, we obtain a wae speed of 5 5 = fλ = (.3 0 Hz)(8.0 0 ) = 8 /s. Assess: The waelength is large and the frequency is sall, howeer, they result in a reasonable wae elocity. P6.49. Prepare: Please refer to Figure P6.49. The steel wire is under tension and it ibrates with three antinodes. At the new tension it will ibrate with two antinodes. Sole: When the spring is stretched 8.0 c, the standing wae on the wire has three antinodes. This eans λ = and the tension T s in the wire is T s = k(0.080 ), where k is the spring constant. For this tension, 3 3 L

21 Superposition and Standing Waes 6- T µ f λ s s wire = 3 = T 3 k(0.08 ) f = µ L We will let the stretching of the spring be x when the standing wae on the wire displays two antinodes. This eans λ = L and T s = kx. For the tension T s, T s T s k x wire = fλ = f = µ µ L µ The frequency f is the sae in the aboe two situations because the wire is drien by the sae oscillating agnetic field. Now, equating the two frequency equations, k x 3 k(0.080 ) = x = 0.8 = 8 c L L Assess: If a spring is stretched by 8.0 c, stretching it another 8.0 c is a reasonable aount. P6.50. Prepare: The standing wae is a cobination of two traeling waes, which are reflecting fro the ends of the tub. A preliinary calculation shows that f = 0/0 s = 0.50 Hz. Sole: (a) The description leads us to beliee that this is the fundaental ode, =, with only one node. The waelength will be twice the length of the tub: λ = L = (.4 ) =.8. (b) Use the fundaental relationship for periodic waes: = λ f = (.8 )(0.50 Hz) =.4 /s. Assess: These nubers all see to be realistic. The frequency of the earthquake waes could be 0.5 Hz, the tub could easily be.4, and.4 /s is a reasonable speed. P6.5. Prepare: Please refer to Figure P6.5. There are reflectors at both ends, so the electroagnetic standing wae acts just like the standing wae on a string that is tied at both ends. The electroagnetic waes of all frequencies trael with the speed of light c. Sole: (a) The frequencies of the standing waes, using Equation 6., are 8 light c /s 9 f = = = (.5 0 Hz) =.5GHz L L (0.0 ) The generator can produce standing waes at any frequency between 0 GHz and 0 GHz. These are f (GHz) (b) There are 7 different standing wae frequencies. Een-nubered alues of create a node at the center, and odd-nubered alues of create an antinode at the center. So the frequencies where the idpoint is an antinode are 0.5, 3.5, 6.5, and 9.5 GHz. Assess: All of the frequencies are in the appropriate range and a quick sketch will erify that antinodes occur at the idpoint of the caity for odd alues of. P6.5. Prepare: Fro Equations 6.6 and 6.7, the fundaental waelength of an open-open tube is L and that of an open-closed tube is 4L. Sole: We are gien that

22 6- Chapter 6 f f f = = air air open-closed 3 open-open 3 open-open 3 l = open-closed l open-open 4L = open-closed Lopen-open L L (78.0 c) open-open open-closed = = = Assess: A length of 3 c for the open-closed pipe is reasonable. 3.0 c 3 P6.53. Prepare: Since light is reflected at both ends of a laser, there is a node at both ends of the caity and an integral nuber of waelengths ust fit into the caity. The speed of light, frequency, and waelength are related by c= fλ. Finally, the length of the caity is related to the waelength of the light by L= λ /. Sole: The inforation that the laser is oscillating in the =00,000 ode tells us that 00,000 halfwaelengths or 50,000 waelengths fit into the Laser caity. The length of the caity is L= λ = = 6 50,000(0.6 0 ) Knowing the waelength and speed of light, we can deterine its frequency. f = c λ = = In a tie of one second, light will trael a distance of / ( /s) / Hz d = ct = = 8 8 ( /s)(.00 s) For a round trip, the light ust trael L =.06. As a result the nuber of round trips the light will ake in.00 s is n= d L = = 8 8 / ( ) / round trips Assess: The one thing that is easy to check is the length of the laser caity. A alue of is reasonable. P6.54. Prepare: Please refer to Figure P6.54. Particles of the ediu at the nodes of a standing wae hae zero displaceent. The cork dust settles at the nodes of the sound wae where there is no otion of the air olecules. The separation between the centers of two adjacent piles is λ. Thus,3 c/3 = λ/ λ = 8.0 c. Sole: Because the piston is drien at a frequency of 400 Hz, the speed of the sound wae in oxygen fro Equation 5.0 is = fλ = (400 Hz)(0.80 ) = 38 /s Assess: A speed of 38 /s in oxygen is close to the speed of sound in air, which is 343 /s at 0 C. P6.55. Prepare: Please refer to Figure P6.55. The nodes of a standing wae are spaced λ/ apart. The waelength of the th ode of an open-open tube fro Equation 6.6 is λ = L/. Or, equialently, the length of the tube that generates the th ode is L = (λ/). Here λ is the sae for all odes because the frequency of the tuning fork is unchanged. Sole: Increasing the length of the tube to go fro ode to ode + requires a length change: L = ( + )(λ/) λ/ = λ/ That is, lengthening the tube by λ/ adds an additional antinode and creates the next standing wae. This is consistent with the idea that the nodes of a standing wae are spaced λ/ apart. This tube is first increased by L = 56.7 c 4.5 c = 4. c, then by L = 70.9 c 56.7 c = 4. c. Thus λ/ = 4. c and λ = 8.4 c = Therefore, the frequency of the tuning fork, using Equation 5.0, is

23 Superposition and Standing Waes /s f = = = 0 Hz λ 0.84 Assess: This is a reasonable alue for the frequency of a tuning for k in the audible range and the units are correct. P6.56. Prepare: The open-closed tube fors standing waes. When the air colun length L is the proper length for a 580 Hz standing wae, a standing wae resonance will be created and the sound will be loud. Fro Equation 6.7, the standing wae frequencies of an open-closed tube are f = (/(4L)), where is the speed of sound in air and is an odd integer: =, 3, 5... The frequency is fixed at 580 Hz, but as the length L changes, 580 Hz standing waes will occur for different alues of. Sole: The length that causes the th standing wae ode to be at 580 Hz is (343 /s) L = (4)(580 Hz) We can place the alues of L, and corresponding alues of h = L, in a table: M L h = L = 85. c = 55.6 c = 6. c h can t be negatie So water heights of 6. c, 55.6 c, and 85. c will cause a standing wae resonance at 580 Hz. The figure shows the = 3 standing wae at h = 55.6 c, or at 56 c (to two significant figures). Assess: Standing wae resonance will occur in the air colun when the length of the air colun is an integral nuber of quarter waelengths for the frequency associated with the tunning fork. You should also note how uch a good sketch can help you understand and sole the proble. P6.57. Prepare: A stretched wire, which is fixed at both ends, fors a standing wae whose fundaental frequency f wire is the sae as the fundaental frequency f open-closed of the open-closed tube. The two frequencies are the sae because the oscillations in the wire drie oscillations of the air in the tube. Sole: The fundaental frequency in the wire fro Equation 6.5 is f T 440 N wire S wire = = = = 469 Hz Lwire Lwire (.0 ) (0.000 kg/0.50 ) The fundaental frequency in the open-closed tube fro Equation 6.7 is