Superposition and Standing Waves

Transcription

1 8 Superposition and Standing Waes CHPTER OUTLINE 8. Superposition and Intererence 8. Standing Waes 8.3 Standing Waes in a String Fixed at Both Ends 8. Resonance 8.5 Standing Waes in ir Columns 8.6 Standing Waes in Rod and Membranes 8.7 Beats: Intererence in Time 8.8 Nonsinusoidal Wae Patterns NSWERS TO QUESTIONS Q8. No. Waes with all waeorms interere. Waes with other wae shapes are also trains o disturbance that add together when waes rom dierent sources moe through the same medium at the same time. *Q8. (i) I the end is ixed, there is inersion o the pulse upon relection. Thus, when they meet, they cancel and the amplitude is zero. nswer (d). (ii) I the end is ree, there is no inersion on relection. When they meet, the amplitude is (. m ). m. nswer. *Q8.3 In the starting situation, the waes interere constructiely. When the sliding section is moed out by. m, the wae going through it has an extra path length o. m, to show partial intererence. When the slide has come out. m rom the starting coniguration, the extra path length is. m, or destructie intererence. nother. m and we are at r r 3 or partial intererence as beore. t last, another equal step o sliding and one wae traels one waelength arther to interere constructiely. The ranking is then d > a c > b. Q8. No. The total energy o the pair o waes remains the same. Energy missing rom zones o destructie intererence appears in zones o constructie intererence. *Q8.5 nswer (c). The two waes must hae slightly dierent amplitudes at P because o their dierent distances, so they cannot cancel each other exactly. Q8.6 Damping, and non linear eects in the ibration turn the energy o ibration into internal energy. *Q8.7 The strings hae dierent linear densities and are stretched to dierent tensions, so they carry string waes with dierent speeds and ibrate with dierent undamental requencies. They are all equally long, so the string waes hae equal waelengths. They all radiate sound into air, where the sound moes with the same speed or dierent sound waelengths. The answer is and (e). *Q8.8 The undamental requency is described by L, where T μ (i) I L is doubled, then L will be reduced by a actor. nswer (). (ii) I μ is doubled, then μ (iii) I T is doubled, then will be reduced by a actor. nswer (e). T will increase by a actor o. nswer (c) _8_ch8_p73-96.indd 73 /3/7 8::55 PM

2 7 Chapter 8 *Q8.9 nswer (d). The energy has not disappeared, but is still carried by the wae pulses. Each particle o the string still has kinetic energy. This is similar to the motion o a simple pendulum. The pendulum does not stop at its equilibrium position during oscillation likewise the particles o the string do not stop at the equilibrium position o the string when these two waes superimpose. *Q8. The resultant amplitude is greater than either indiidual amplitude, whereer the two waes are nearly enough in phase that cos(φ ) is greater than. This condition is satisied wheneer the absolute alue o the phase dierence φ between the two waes is less than. nswer (d). Q8. What is needed is a tuning ork or other pure-tone generator o the desired requency. Strike the tuning ork and pluck the corresponding string on the piano at the same time. I they are precisely in tune, you will hear a single pitch with no amplitude modulation. I the two pitches are a bit o, you will hear beats. s they ibrate, retune the piano string until the beat requency goes to zero. *Q8. The bow string is pulled away rom equilibrium and released, similar to the way that a guitar string is pulled and released when it is plucked. Thus, standing waes will be excited in the bow string. I the arrow leaes rom the exact center o the string, then a series o odd harmonics will be excited. Een harmonies will not be excited because they hae a node at the point where the string exhibits its maximum displacement. nswer (c). *Q8.3 (a) (c) (d) The tuning ork hits the paper repetitiely to make a sound like a buzzer, and the paper eiciently moes the surrounding air. The tuning ork will ibrate audibly or a shorter time. Instead o just radiating sound ery sotly into the surrounding air, the tuning ork makes the chalkboard ibrate. With its large area this sti sounding board radiates sound into the air with higher power. So it drains away the ork s energy o ibration aster and the ork stops ibrating sooner. The tuning ork in resonance makes the column o air ibrate, especially at the antinode o displacement at the top o the tube. Its area is larger than that o the ork tines, so it radiates louder sound into the enironment. The tuning ork will not ibrate or so long. The tuning ork ordinarily pushes air to the right on one side and simultaneously pushes air to the let a couple o centimeters away, on the ar side o its other time. Its net disturbance or sound radiation is small. The slot in the cardboard admits the back wae rom the ar side o the ork and keeps much o it rom interering destructiely with the sound radiated by the tine in ront. Thus the sound radiated in ront o the screen can become noticeably louder. The ork will ibrate or a shorter time. ll our o these processes exempliy conseration o energy, as the energy o ibration o the ork is transerred aster into energy o ibration o the air. The reduction in the time o audible ork ibration is easy to obsere in case (a), but may be challenging to obsere in the other cases. Q8. Walking makes the person s hand ibrate a little. I the requency o this motion is equal to the natural requency o coee sloshing rom side to side in the cup, then a large amplitude ibration o the coee will build up in resonance. To get o resonance and back to the normal case o a small-amplitude disturbance producing a small amplitude result, the person can walk aster, walk slower, or get a larger or smaller cup. lternatiely, een at resonance he can reduce the amplitude by adding damping, as by stirring high iber quick cooking oatmeal into the hot coee. You do not need a coer on your cup. 379_8_ch8_p73-96.indd 7 /3/7 8::56 PM

3 Superposition and Standing Waes 75 *Q8.5 The tape will reduce the requency o the ork, leaing the string requency unchanged. I the bit o tape is small, the ork must hae started with a requency Hz below that o the string, to end up with a requency 5 Hz below that o the string. The string requency is Hz, answer (d). Q8.6 Beats. The propellers are rotating at slightly dierent requencies. SOLUTIONS TO PROBLEMS Section 8. Superposition and Intererence P8. y y + y 3. cos (. x 6. t) +. sin ( 5. x.t) ealuated at the gien x alues. (a) x., t. y 3. cos (. rad) +. sin ( + 3. rad) 6. 5cm x., t.5 y 3. cos ( + 3. rad) +. sin ( +. rad) 6. cm (c) x.5, t y 3. cos ( +. rad) +. sin ( + 5. rad). 5cm P8. FIG. P8. P8.3 (a) y ( x t ), so wae traels in the +x direction y ( x + t ), so wae traels in the x direction To cancel, y + y : 5 5 ( 3x t) + + ( 3x+ t 6) + ( 3x t) ( 3x+ t 6) 3x t ± ( 3x+ t 6) rom the positie root, 8t 6 t. 75 s (at t. 75 s, the waes cancel eerywhere) (c) rom the negatie root, 6x 6 x. m (at x. m, the waes cancel always) 379_8_ch8_p73-96.indd 75 /3/7 8::57 PM

4 76 Chapter 8 P8. Suppose the waes are sinusoidal. The sum is (. cm) sin ( kx ωt) + (. cm) sin ( kx ω t + 9. ) (. cm) sin ( kx ω t + 5. ) cos 5. So the amplitude is ( 8. cm) cos cm. P8.5 The resultant wae unction has the orm y φ φ kx t + cos sin ω (a) φ ( ) π 5 cos. cos 9. m ω π π π 6 Hz P8.6 (a) Δx m The waelength is 33 ms. m 3 Hz Thus, Δx o a wae,. or Δφ π(. 53) rad For destructie intererence, we want Δx Δx where Δx is a constant in this set up. x 33 Δ (. 66 ) 83 Hz P8.7 We suppose the man s ears are at the same leel as the lower speaker. Sound rom the upper speaker is delayed by traeling the extra distance L + d L. ( He hears a minimum when this is n ) with n,, 3, Then, L ( n ) L + d L L + d + d ( ) n + L ( ) n ( n ) L + L + ( ) ( ) d n L n n,, 3, This will gie us the answer to. The path dierence starts rom nearly zero when the man is ery ar away and increases to d when L. The number o minima he hears is the greatest ( n ) integer solution to d d n greatest integer + continued on next page 379_8_ch8_p73-96.indd 76 /3/7 8::57 PM

5 Superposition and Standing Waes 77 (a) d. m s 33 ms + ( )( ) + 9. He hears two minima. With n, d L ( ) ( ). m 33 m s s 33 s L 98. m ( ) ( ) ( ms) ( ) With n, d L ( 3) 3 ( ) 99. m P8.8 Suppose the man s ears are at the same leel as the lower speaker. Sound rom the upper speaker is delayed by traeling the extra distance Δr L + d L. He hears a minimum when Δr ( n ) with n 3,,, Then, L + d L n L + d n L + L + d n + n L+ L d n n L () Equation gies the distances rom the lower speaker at which the man will hear a minimum. The path dierence Δr starts rom nearly zero when the man is ery ar away and increases to d when L. (a) The number o minima he hears is the greatest integer alue or which L. This is the same as the greatest integer solution to d n, or number o minima heard n greatest intege r d max + From equation, the distances at which minima occur are gien by ( ) ( ) ( )( ) d n L n n where n,,, nmax 379_8_ch8_p73-96.indd 77 /3/7 8::58 PM

6 78 Chapter 8 P8.9 (a) φ. rad cm 5. cm 3. rad s. s) 36. rad ( )( ) ( )( φ 5. rad cm 5. cm. rad s. s) 5. rad ( )( ) ( )( Δφ 9. radians Δφ. x 3. t [ 5. x. t] 5. x+ 8. t t t. s, the requirement is Δφ 5. x+ 8. (. ) ( n+ ) π or any integer n. For x < 3., 5.x + 6. is positie, so we hae 5. x+ 6. ( n+ ) π,or ( n + ) π x The smallest positie alue o x occurs or n and is ( + ) π x π. 58 cm 5. 3 ms *P8. (a) First we calculate the waelength: 6. m. 5 Hz Then we note that the path dierence equals 9. m. m Point is one-hal waelength arther rom one speaker than rom the other. The waes rom the two sources interere destructiely, so the receier records a minimum in sound intensity. We choose the origin at the midpoint between the speakers. I the receier is located at point (x, y), then we must sole: ( x+ 5. ) + y ( x 5. ) + y Then, ( x+ 5. ) + y ( x 5. ) + y + Square both sides and simpliy to get:. x ( x 5. ) + y Upon squaring again, this reduces to: x. x+ ( x 5. ) + y 6. Substituting 6. m, and reducing, 9. x 6. y or x y The point should moe along the hyperbola 9x 6y. (c) Yes. Far rom the origin the equation might as well be 9x 6y, so the point can moe along the straight line through the origin with slope.75 or the straight line through the origin with slope _8_ch8_p73-96.indd 78 /3/7 8::59 PM

7 Superposition and Standing Waes 79 Section 8. Standing Waes P8. y (. 5 m) sin (. x) cos( t) sin kxcosωt π π Thereore, k. rad m 5. 7 m. rad m ω rad s and ω π so 3. 8 Hz π π rad The speed o waes in the medium is π π ω rad s k. rad m 5 ms P8. From y sin kxcosω t we ind y x k cos kxcosω t y t ωsin kxsinωt y y k sin kxcosω t ω sin kxcosωt x t ( ) Substitution into the wae equation gies k sin kx cosωt ω sin kx cosωt This is satisied, proided that ω. But this is true, because k π π ω k P8.3 The acing speakers produce a standing wae in the space between them, with the spacing between nodes being d NN 33 ms 8 s ( ). m I the speakers ibrate in phase, the point halway between them is an antinode o pressure at a distance rom either speaker o 5. m. 65. Then there is a node at m a node at. 58 m. m. 33 m a node at. 33 m. m. 89 m a node at. 58 m+. m. 73 m a node at. 73 m+. m. 97 m and a node at. 97 m+. m. 6 m rom either speaker. 379_8_ch8_p73-96.indd 79 /3/7 8:5: PM

8 8 Chapter 8 *P8. (a) y t y t 5 ms 6 x 6 x y t ms y t 5 ms 6 x 6 x y t ms 6 x In any one picture, the distance rom one positie-going zero crossing to the next is m. (c) 5 Hz. The oscillation at any point starts to repeat ater a period o ms, and T. (d) m. By comparison with the wae unction y ( sin kx)cos ω t, we identiy k π, and then compute π k. (e) 5 Hz. By comparison with the wae unction y ( sin kx)cos ω t, we identiy ω π π. [ ] [ ] P8.5 y 3. sin π ( x+. 6t) cm; y 3. sin π ( x. 6t) cm y y+ y 3. sin( πx) cos ( 6. πt) + 3. sin( πx) cos (. 6πt) y ( 6. cm) sin( π x) cos ( 6. πt) [ ] (a) We can take cos (. 6πt) to get the maximum y. t x. 5 cm, y max ( 6. cm) sin ( 5. π ). cm t x. 5 cm, y max ( 6. cm) sin ( 5. π ) 6. cm (c) Now take cos (. 6πt) to get y max : t x 5. cm, y max ( 6. cm) sin (. 5π )( ) 6. cm cm continued on next page 379_8_ch8_p73-96.indd 8 /5/7 6:7:33 PM

9 Superposition and Standing Waes 8 n (d) The antinodes occur when x ( n 35,,, ) π But k π so. cm and x. 5 cm as in 3 x 5. cm as in (c) 5 x 3 5. cm *P8.6 (a) The resultant wae is y φ kx+ φ sin cos ωt The oscillation o the sin(kx + φ ) actor means that this wae shows alternating nodes and antinodes. It is a standing wae. φ The nodes are located at kx+ nπ so nπ φ x k k which means that each node is shited φ to the let by the phase dierence between the traeling waes. k n The separation o nodes is Δx π φ ( n+ ) k k π k φ π Δx k k The nodes are still separated by hal a waelength. (c) s noted in part (a), the nodes are all shited by the distance φ k to the let. Section 8.3 Standing Waes in a String Fixed at Both Ends P8.7 L 3. m; μ 9. 3 kg m; T. N; where T μ 7. so m s L. 786 Hz. 57 Hz Hz 3. Hz 3 P8.8 The tension in the string is T ( )( ) kg 9. 8 m s 39. N Its linear density is μ 3 m 8 kg 6. L 5m T and the wae speed on the string is μ In its undamental mode o ibration, we hae L ( 5 m) m ms Thus, m 3 kg m 39. N ms 3.6 kg m 5. 7 Hz 379_8_ch8_p73-96.indd 8 /3/7 8:5: PM

10 8 Chapter 8 P8.9 (a) Let n be the number o nodes in the standing wae resulting rom the 5.-kg mass. Then n + is the number o nodes or the standing wae resulting rom the 6.-kg mass. For standing waes, L n, and the requency is Thus, n T n L μ T n n + and also + L μ Thus, n Tn 5. kg g n T 6. kg g + ( ) ( ) n+ Thereore, n + 5n, or n 5. m 5. kg 9. 8 m s Then, ( ) ( )( ). kg m 35 Hz The largest mass will correspond to a standing wae o loop m( 98. ms) (n ) so 35 Hz (. m ). kg m yielding m kg P8. For the whole string ibrating, d NN 6. m ; 8. m The speed o a pulse on the string is m m s s (a) When the string is stopped at the ret, d NN 3 6. m ;. 853 m ms 95 Hz. 853 m FIG. P8.(a) The light touch at a point one third o the way along the string damps out ibration in the two lowest ibration states o the string as a whole. The whole string ibrates in its third resonance possibility: 3d NN. 6 m 3. 7 m P8. d NN. 7 m ms. 7 m 99 Hz FIG. P8.. m 38 ms T (. 3 ) (. 7) (a) T 63 N With one-third the distance between nodes, the requency is 3 3 Hz 66 Hz FIG. P8. 379_8_ch8_p73-96.indd 8 /3/7 8:5:3 PM

11 Superposition and Standing Waes 83 P8. G (. 35 m) ; L G G G LG L LG LG LG (. 35 m ) m Thus, L L. 38 m. 35 m. 38 m. 3 m, or the inger should be placed G 3. cm rom the bridge. T L μ ; dl dt T ; dl dt μ L T dt dl. 6 cm 38. % T L ( ) cm P8.3 In the undamental mode, the string aboe the rod has only two nodes, at and B, with an anti-node halway between and B. Thus, B L or L cosθ cosθ Since the undamental requency is, the wae speed in this segment o string is L cosθ L q B lso, M T T μ mb TL m cosθ T where T is the tension in this part o the string. Thus, L TL or cosθ m cosθ L TL cos θ m cosθ q F and the mass o string aboe the rod is: m T cosθ L [] Mg FIG. P8.3 Now, consider the tension in the string. The light rod would rotate about point P i the string exerted any ertical orce on it. Thereore, recalling Newton s third law, the rod must exert only a horizontal orce on the string. Consider a ree-body diagram o the string segment in contact with the end o the rod. Mg Fy T sinθ Mg T sinθ Then, rom Equation [], the mass o string aboe the rod is Mg Mg m cosθ sin θ L L tanθ 379_8_ch8_p73-96.indd 83 /3/7 8:5: PM

12 8 Chapter 8 P8. Let m ρv represent the mass o the copper cylinder. The original tension in the wire is T mg ρvg. The water exerts a buoyant orce ρ V water g on the cylinder, to reduce the tension to T V ρvg ρ g ρ ρ water water Vg The speed o a wae on the string changes rom T μ to T. The requency changes rom μ T μ to T μ where we assume L is constant. Then T T 8. 3 Hz 89. ρ ρwater ρ Hz P8.5 Comparing y (. m) sin ((π rad m)x) cos (( π rad s)t) with y sinkxcosωt π we ind k π m,. m, and ω π π s : 5. Hz (a) Then the distance between adjacent nodes is d NN. m L 3. m and on the string are 3 loops. m d NN 5 s m m s For the speed we hae ( ) b In the simplest standing wae ibration, dnn 3. m, b 6. m and a b b ms 6. m 6. 7 Hz (c) In T, i the tension increases to Tc T μ 9 and the string does not stretch, the speed increases to c 9T T 3 3 3( ms) 3 μ μ ms Then c 3 ms c 6. m d NN 5 s c a 3. m and one loop its onto the string. 379_8_ch8_p73-96.indd 8 /3/7 8:5:5 PM

13 Superposition and Standing Waes 85 Section 8. Resonance P8.6 The wae speed is gd ( 98. ms)( 36. m) 88. ms The bay has one end open and one closed. Its simplest resonance is with a node o horizontal elocity, which is also an antinode o ertical displacement, at the head o the bay and an antinode o elocity, which is a node o displacement, at the mouth. The ibration o the water in the bay is like that in one hal o the pond shown in Figure P8.7. Then, d N m and 8 3 m Thereore, the period is m T 7. s h min 8.8 m s The natural requency o the water sloshing in the bay agrees precisely with that o lunar excitation, so we identiy the extra-high tides as ampliied by resonance. 95. m P8.7 (a) The wae speed is 366. ms.5 s From the igure, there are antinodes at both ends o the pond, so the distance between adjacent antinodes is d 95. m and the waelength is 8. 3 m 366. ms The requency is then 8. 3 m. Hz We hae assumed the wae speed is the same or all waelengths. P8.8 The distance between adjacent nodes is one-quarter o the circumerence. so and d NN. cm d 5. cm. cm 9 ms 9 Hz 9. khz. m The singer must match this requency quite precisely or some interal o time to eed enough energy into the glass to crack it. 379_8_ch8_p73-96.indd 85 /3/7 8:5:6 PM

14 86 Chapter 8 Section 8.5 Standing Waes in ir Columns P8.9 (a) For the undamental mode in a closed pipe, L, as in the diagram. But, thereore L So, 33 ms s L ( ). 357 m For an open pipe, L, as in the diagram. l/ L N l/ N So, L 33 ms s ( ). 75 m FIG. P8.9 P8.3 d. 3 m;. 6 m (a) 53 Hz. 85 m; d. 5 mm 33 ms P8.3 The waelength is 3. m 6.6 s FIG. P8.3 so the length o the open pipe ibrating in its simplest (-N-) mode is d to. 656 m closed pipe has (N-) or its simplest resonance, (N--N-) or the second, and (N--N--N-) or the third. 5 5 Here, the pipe length is 5 d Nto ( 3. m) 6. m P8.3 The air in the auditory canal, about 3 cm long, can ibrate with a node at the closed end and antinode at the open end, with d Nto 3 cm so. m and 33 ms. m 3 khz small-amplitude external excitation at this requency can, oer time, eed energy into a larger-amplitude resonance ibration o the air in the canal, making it audible. 379_8_ch8_p73-96.indd 86 /3/7 8:5:7 PM

15 Superposition and Standing Waes 87 P8.33 For a closed box, the resonant requencies will hae nodes at both sides, so the permitted waelengths will be L n, ( n 3,,, ). i.e., n n L and n L Thereore, with L.86 m and L. m, the resonant requencies are n n( 6 Hz ) or L.86 m or each n rom to 9 and n n( 8. 5 Hz ) or L. m or each n rom to 3. P8.3 The waelength o sound is The distance between water leels at resonance is and d r t π R πr Rt πr d P8.35 For both open and closed pipes, resonant requencies are equally spaced as numbers. The set o resonant requencies then must be 65 Hz, 55 Hz, 5 Hz, 35 Hz, 5 Hz, 5 Hz, 5 Hz. These are odd-integer multipliers o the undamental requency o 5. Hz. Then the pipe length is d N 3 ms 5 ( ).7 m. s *P8.36 (a) The open ends o the tunnel are antinodes, so d m n with n,, 3,. Then d m n. nd (33 m s) ( m n).858n Hz,withn,,3, It is a good rule. ny car horn would produce seeral or many o the closely-spaced resonance requencies o the air in the tunnel, so it would be greatly ampliied. Other driers might be rightened directly into dangerous behaior, or might blow their horns also. P8.37 For resonance in a narrow tube open at one end, n L (a) ssuming n and n 3, ( n 35,,, ) 38 3 (. 8 ) and 38 (. 683 ) In either case, 35 m s. 38 Hz warm air.8 cm 68.3 cm For the next resonance n 5, and L 5 ( 35 ms ) 5 ( ) s 38. m FIG. P _8_ch8_p73-96.indd 87 /3/7 8:5:8 PM

16 88 Chapter 8 P8.38 The length corresponding to the undamental satisies L : L 3 ( m. ) Since L >. cm, the next two modes will be obsered, corresponding to or L 3 3 and L 5 L 3. 5 m and L m *P8.39 Call L the depth o the well and the speed o sound. Then or some integer n ( n )( 33 ms) L ( n ) ( n ) 55 (. s ) and or the next resonance n + 33 ms L [ ( n+ ) ] ( n+ ) 6. s Thus, ( n )( 33 ms) ( n+ )( 33 ms) 55 (. s ) 6 (. s ) and we require an integer solution to n+ n The equation gies n 656., so the best itting integer is n 7. 7 Then the results [ ( 7) ]( 33 ms) L. 6 m 55 (. s ) and [ ( 7) + ]( 33 ms) L. m 6 (. s ) suggest that we can say ( )( ) ( ) the depth o the well is (.5 ±.) m. The data suggest.6-hz uncertainty in the requency measurements, which is only a little more than %. P8. (a) For the undamental mode o an open tube, L 33 ms ( ) 88 s ms ) 38 ms 73 We ignore the thermal expansion o the metal. 38 ms L (. 95 m ) The lute is lat by a semitone.. 95 m 8 Hz 379_8_ch8_p73-96.indd 88 /3/7 8:5:9 PM

17 Superposition and Standing Waes 89 Section 8.6 Standing Waes in Rod and Membranes 5 P8. (a) 59. khz L ( )(. 6 ) (c) Since it is held in the center, there must be a node in the center as well as antinodes at the ends. The een harmonics hae an antinode at the center so only the odd harmonics are present khz L ( )(. 6 ) P8. When the rod is clamped at one-quarter o its length, the ibration pattern reads NN and the rod length is L d. Thereore, L 5 ms Hz 6. m Section 8.7 Beats: Intererence in Time P8.3 T new 5. Hz 6 Δ s. s 5. 6 beats s P8. (a) The string could be tuned to either 5 Hz or 55 Hz rom this eidence. (c) Tightening the string raises the wae speed and requency. I the requency were originally 5 Hz, the beats would slow down. Instead, the requency must hae started at 55 Hz to become 56 Hz. T μ T From L L μ T 53 and T T T 989T T Hz. 56 Hz The ractional change that should be made in the tension is then T T ractional change % lower T The tension should be reduced by.%. ( + s ) P8.5 For an echo the beat requency is b. ( s ) ( s ) Soling or b, gies b when approaching wall. ( ) (a) b ( 56 ( ) ) 33. ( ) s 99. Hz beat requency When he is moing away rom the wall, s changes sign. Soling or s gies s b b ( 5 )( 33 ) ( )( 56) m s 379_8_ch8_p73-96.indd 89 /3/7 8:5: PM

18 9 Chapter 8 P8.6 Using the and -oot pipes produces actual requencies o 3 Hz and 96 Hz and a combination tone at ( 96 3) Hz 65. Hz, so this pair supplies the so-called missing undamental. 3 The and -oot pipes produce a combination tone ( 6 3) Hz 3 Hz, so this does not work. The 3 and -oot pipes produce a combination tone at ( 6 96) Hz 65. Hz, so this works. lso,,, and -oot pipes all playing together produce the 65.-Hz combination tone. 3 Section 8.8 Nonsinusoidal Wae Patterns P8.7 We list the requencies o the harmonics o each note in Hz: P8.8 We ealuate Harmonic Note C# E The second harmonic o E is close the the third harmonic o, and the ourth harmonicoc#is close to the ith harmonic o. s sinθ + 57sin θ sin3θ + 5sinθ sin 5θ sin 6θ sin 7θ where s represents particle displacement in nanometers and θ represents the phase o the wae in radians. s θ adances by π, time adances by ( 53) s. Here is the result: FIG. P8.8 dditional Problems *P8.9 (a) (c) The yo-yo s downward speed is dl dt + (.8 m s )(. s).96 m s. The instantaneous waelength o the undamental string wae is gien by d NN L so L and d dt dl dt (.96 m s).9 m s. For the second harmonic, the waelength is equal to the length o the string. Then the rate o change o waelength is equal to dl dt.96 m/s, hal as much as or the irst harmonic. yo-yo o dierent mass will hold the string under dierent tension to make each string wae ibrate with a dierent requency, but the geometrical argument gien in parts (a) and still applies to the waelength. The answers are unchanged : d dt.9 m s and d dt.96 m s. 379_8_ch8_p73-96.indd 9 /3/7 8:5: PM

19 Superposition and Standing Waes 9 *P8.5 (a) Use the Doppler ormula ( ± ) ( s ) With requency o the speaker in ront o student and requency o the speaker behind the student. ( 33 ms+. 5 ms) ( 56 Hz) 58 Hz ( 33 ms ) ( 33 ms. 5 ms) ( 56 Hz) 5 Hz ( 33 ms+ ) Thereore, b 399. Hz 33 ms The waes broadcast by both speakers hae. 75 m. The standing 56 s wae between them has d. 376 m. The student walks rom one maximum to the. 376 m next in time Δt. 5 s, so the requency at which she hears maxima is 5. ms T 399. Hz (c) The answers are identical. The models are equally alid. We may think o the intererence o the two waes as intererence in space or in time, linked to space by the steady motion o the student. P Hz speed o sound in air: a 3 m s (a) b ( 87. s )(. m) b 3. 8 m s a L a a L a 3 ms 87. s ( ). 977 m FIG. P _8_ch8_p73-96.indd 9 /3/7 8:5:3 PM

20 9 Chapter 8 P8.5 N I N N II N N N III d N d N d N d N d N FIG. P kg (a) μ m 3 kg m T 3. kg m s 3 μ 6. kg m. 3 ms In state I, d N. 86 m (c) 3. m. 3 ms 3. m. Hz In state II, d N ( m) m. 3 ms (. 87 m). 5 m 5. m In state III, d N ( m) m. 3 ms (. 7 m ). 7 Hz. Hz P8.53 I the train is moing away rom station, its requency is depressed: Hz: 78 8 Soling or gies (. )( 33) 78 Thereore, 385. m s away rom station I it is moing toward the station, the requency is enhanced: Hz: Soling or gies (. )( ) 8 Thereore, 377. m s toward the station P8.5 ( 8. )(. ) ms 3 8. d NN a. m;. m; a 33 ms m. Hz 7. 7 Hz ( ) 379_8_ch8_p73-96.indd 9 /3/7 8:5: PM

21 Superposition and Standing Waes 93 P8.55 (a) Since the irst node is at the weld, the waelength in the thin wire is L or 8. cm. The requency and tension are the same in both sections, so T 6. L ( ) 3 μ Hz s the thick wire is twice the diameter, the linear density is times that o the thin wire. μ 8. g m T so L μ L ( )( 599. ) cm hal the length o the thin wire. *P8.56 The waelength stays constant at.96 m while the waespeed rises according to (T μ) / [(5 +.86t).6] / [ t] / so the requency rises as [ t] /.96 [ t] / The number o cycles is dt in each incremental bit o time, or altogether 35. ( ) / 35. t dt ( t) [ t] / 938 dt 3 / / 3 / ( ) ( ) 96 7 cycles n T P8.57 (a) L μ so L L L L The requency should be haled to get the same number o antinodes or twice the length. n T n T so T n T n The tension must be T n n + n T n + (c) nl nl T T so T n L T nl T 3 T T T 9 6 to get twice as many antinodes. 379_8_ch8_p73-96.indd 93 /5/7 6:56:33 PM

22 9 Chapter 8 P8.58 (a) For the block: Fx T Mgsin 3. so T Mgsin 3. Mg The length o the section o string parallel to the incline is h h. The total length o the string is then 3h. sin 3. FIG. P8.58 (c) The mass per unit length o the string is μ m 3h T (d) The speed o waes in the string is Mg 3h m 3Mgh μ m (e) In the undamental mode, the segment o length h ibrates as one loop. The distance between adjacent nodes is then d NN h, so the waelength is h. 3Mgh 3Mg The requency is h m 8mh (g) When the ertical segment o string ibrates with loops (i.e., 3 nodes), then h and the waelength is h ( ) The period o the standing wae o 3 nodes (or two loops) is m T h 3Mgh mh 3Mg ( ) ( ) (h) b... 3Mg 8mh P8.59 We look or a solution o the orm 5. sin (. x. t) +. cos (. x. t) ( ) sin. x. t + φ sin (. x. t) cosφ + cos (. x. t) sinφ This will be true i both 5. cosφ and. sinφ, requiring ( 5. ) + (. ). and φ 63. The resultant wae. sin (. x. t ) is sinusoidal. 379_8_ch8_p73-96.indd 9 /3/7 8:5:6 PM

23 Superposition and Standing Waes 95 P8.6 d m is the distance between antinodes. Then. 3 m and ms 5 6. Hz 3. m The crystal can be tuned to ibrate at 8 Hz, so that binary counters can derie rom it a signal at precisely Hz. FIG. P8.6 P8.6 (a) Let θ represent the angle each slanted rope makes with the ertical. In the diagram, obsere that:. m sinθ.5 m 3 Considering the mass, or θ. 8 F y : T cosθ mg. kg 9. 8 m s cos. 8 or T ( )( ) N FIG. P8.6 T N The speed o transerse waes in the string is 8 ms μ. kg m For the standing wae pattern shown (3 loops), d 3 or (. m ) 33. m 3 8 ms Thus, the required requency is Hz 33. m NSWERS TO EVEN PROBLEMS P8. see the solution P cm P8.6 (a) 3.33 rad 83 Hz P8.8 (a) The number is the greatest integer d + n,,, nmax ( ) ( ) ( )( ) d n L n n where P8. (a) Point is one hal waelength arther rom one speaker than rom the other. The waes it receies interere destructiely. long the hyperbola 9x 6y. (c) Yes; along the straight line through the origin with slope.75 or the straight line through the origin with slope.75. P8. see the solution P8. (a) see the solution m is the distance between crests. (c) 5 Hz. The oscillation at any point starts to repeat ater a period o ms, and T. (d) m. By comparison with equation 8.3, k π, and π k. (e) 5 Hz. By comparison with equation 8.3, ω π π. 379_8_ch8_p73-96.indd 95 /3/7 8:5:7 PM

24 96 Chapter 8 P8.6 (a) Yes. The resultant wae contains points o no motion. and (c) The nodes are still separated by. They are all shited by the distance φ k to the let. P Hz P8. (a) 95 Hz 99 Hz P8. 3. cm rom the bridge; 3.8% P8. 9 Hz P8.6 The natural requency o the water sloshing in the bay agrees precisely with that o lunar excitation, so we identiy the extra-high tides as ampliied by resonance. P khz P8.3 (a) 53 Hz.5 mm P8.3 3 khz; a small-amplitude external excitation at this requency can, oer times, eed energy into a larger-amplitude resonance ibration o the air in the canal, making it audible. r P8.3 Δt π R P8.36 (a).858 n Hz, with n,, 3, It is a good rule. ny car horn would produce seeral or many o the closely-spaced resonance requencies o the air in the tunnel, so it would be greatly ampliied. Other driers might be rightened directly into dangerous behaior, or might blow their horns also. P m;.837 m P8. (a).95 m 8 m P8..6 m P8. (a) 5 Hz or 55 Hz 56 Hz (c) reduce by.% P8.6 -oot and 3 -oot; 3 and -oot; and all three together P8.8 see the solution P8.5 (a) 3.99 beats s 3.99 beats s (c) The answers are identical. The models are equally alid. We may think o the intererence o the two waes as intererence in space or in time, linked to space by the steady motion o the student. P8.5 (a).3 m s 86. cm, 8.7 cm, 7. cm (c). Hz,. Hz,.7 Hz P m P cycles P8.58 (a) Mg 3h (c) m 3h (d) 3Mgh m (e) 3Mg 8mh ( ) mh 3Mg ( ) (g) h (h). 3Mg 8mh P8.6 6 khz 379_8_ch8_p73-96.indd 96 /3/7 8:5:8 PM