Chapter 18 Solutions

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Dominick Lloyd
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1 Chapter 18 Solutions 18.1 he resultant wave function has the form y A 0 cos φ sin kx ω t + φ (a) A A 0 cos φ (5.00) cos (π /4) 9.4 m f ω π 100π π 600 Hz *18. We write the second wave function as hen y A sin(kx ω t + φ) y ( m) sin [π (0.100x 80.0t) + φ] y 1 + y ( m) sin [π (0.100x 80.0t)] + ( m) sin [π (0.100x 80.0t) + φ] ( m) cos φ sin π (0.100x 80.0t) + φ We require ( m) cos φ cos φ 3 φ 60.0 π 3 hen the second wave function is y ( m) sin π 0.100x 80.0t y ( m) sin [π(0.100x 80.0t )]

2 Chapter 18 Solutions 18.3 Suppose the waves are sinusoidal. he sum is (4.00 cm) sin (kx ω t) + (4.00 cm) sin (kx ω t ) (4.00 cm) sin (kx ω t ) cos 45.0 So the amplitude is (8.00 cm) cos cm 18.4 A 0 cos φ A 0, so φ 1 cos π 3 hus, the phase difference is φ 10 π 3 his phase difference results if the time delay is 3 1 3f λ 3v ime delay 3.00 m 3(.00 m/s) s 18.5 Waves reflecting from the near end travel 8.0 m (14.0 m down and 14.0 m back), while waves reflecting from the far end travel 66.0 m. he path difference for the two waves is: r 66.0 m 8.0 m 38.0 m Since λ v f, then r λ ( r)f v (38.0 m)(46 Hz) 343 m/s 7.54 or, r 7.54λ he phase difference between the two reflected waves is then φ 0.54(1 cycle) 0.54(π rad) (a) First we calculate the wavelength: λ v f 344 m/s 1.5 Hz 16.0 m hen we note that the path difference equals 9.00 m 1.00 m 1 λ herefore, the receiver will record a minimum in sound intensity.

3 Chapter 18 Solutions 3 If the receiver is located at point (x, y), then we must solve: (x ) + y (x 5.00) + y 1 λ hen, (x ) + y (x 5.00) + y + 1 λ Square both sides and simplify to get: 0.0x λ 4 λ (x 5.00) + y Upon squaring again, this reduces to: 400x 10.0λ x + λ λ (x 5.00) + λ y Substituting λ 16.0 m, and reducing, we have: 9.00x 16.0y 144 or x 16.0 y (When plotted this yields a curve called a hyperbola.) 18.7 We suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance L + d L. He hears a minimum when this is (n 1)λ / with n 1,, 3,... hen, L + d L (n 1/)v/f L + d (n 1/)v/f + L L + d (n 1/) v /f + L + (n 1/)vL/f L d (n 1/) v /f (n 1/)v/f n 1,, 3,... his will give us the answer to. he path difference starts from nearly zero when the man is very far away and increases to d when L 0. he number of minima he hears is the greatest integer solution to d (n 1/) v/f n greatest integer df/v + 1/ (a) df/v + 1 (4.00 m)(00/s)/330 m/s He hears two minima.

4 4 Chapter 18 Solutions With n 1, L d (1/) v /f (1/)v/f (4.00 m) (330 m/s) /4(00/s) (330 m/s)/00/s L 9.8 m with n L d (3/) v /f (3/)v/f 1.99 m 18.8 Suppose the man's ears are at the same level as the lower speaker. Sound from the upper speaker is delayed by traveling the extra distance r L + d L. He hears a minimum when r (n 1) λ with n 1,, 3,... hen, L + d L (n 1/)(v/f) L + d (n 1/)(v/f) + L L + d (n 1/) (v/f) + (n 1/)(v/f)L + L (1) Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. he path difference r starts from nearly zero when the man is very far away and increases to d when L 0. (a) he number of minima he hears is the greatest integer value for which L 0. his is the same as the greatest integer solution to d (n 1/)(v/f), or number of minima heard n max greatest integer d(f/v) + 1/ From Equation 1, the distances at which minima occur are given by L n d (n 1/) (v/f) (n 1/)(v/f) where n 1,,..., n max 18.9 y (1.50 m) sin (0.400x) cos (00t) A 0 sin kx cos ωt herefore, k π λ rad m π λ rad/m 15.7 m and ω πf, so f ω π 00 rad/s π rad 31.8 Hz

5 Chapter 18 Solutions 5 he speed of waves in the medium is v λf λ π πf ω k 00 rad/s 500 m/s rad/m y m cos x cos (40t) (a) nodes occur where y 0: x (n + 1) π so x (n + 1)π π, 3π, 5π,... y max m cos m he facing speakers produce a standing wave in the space between them, with the spacing between nodes being d NN λ v f 343 m/s (800 s 1 ) 0.14 m If the speakers vibrate in phase, the point halfway between them is an antinode, at 1.5 m 0.65 m from either speaker. hen there is a node at 0.65 m 0.14 m m, a node at m 0.14 m m, a node at m 0.14 m m, a node at m m 0.73 m, a node at 0.73 m m m, and a node at m m 1.16 m from either speaker

6 6 Chapter 18 Solutions *18.1 (a) he resultant wave is y A sin kx + φ cos ω t φ he nodes are located at kx + φ nπ so x nπ k φ k which means that each node is shifted φ k to the left. he separation of nodes is x (n + 1) π k φ k nπ k φ k x π k λ he nodes are still separated by half a wavelength y sin [π(x t)] cm y 3.00 sin [π(x 0.600t)] cm y y 1 + y [3.00 sin (πx) cos (0.600 πt) sin (πx) cos (0.600πt)] cm (6.00 cm) sin (πx) cos (0.600πt) (a) We can take cos(0.600πt) 1 to get the maximum y. At x 0.50 cm, y max (6.00 cm) sin (0.50π) 4.4 cm At x cm, y max (6.00 cm) sin (0.500π) 6.00 cm (c) Now take cos (0.600πt) 1 to get y max : At x 1.50 cm, y max (6.00 cm) sin (1.50π)( 1) 6.00 cm (d) he antinodes occur when x nλ/4 (n 1, 3, 5,... ). But k π/λ π, so λ.00 cm, and x 1 λ/ cm as in x 3λ/ cm as in (c) x 3 5λ/4.50 cm

7 Chapter 18 Solutions (a) Using the given parameters, the wave function is y π sin πx cos (10πt) We need to find values of x for which sin πx 1 his condition requires that πx π n + 1 For n 0, x 1.00 cm and for n 1, x 3.00 cm ; n 0, 1,,... herefore, the distance between antinodes, x.00 cm A π sin πx ; when x 0.50 cm, A.40 cm y A 0 sin kx cos ωt y x A 0k sin kx cos ωt y t A 0ω sin kx cos ωt Substitution into the wave equation gives A 0 k sin kx cos ωt 1 v ( A 0ω sin kx cos ωt) his is satisfied, provided that v ω k kg µ.00 m kg/m v µ (0.0 kg m/s ) kg/m 0.0 m/s For the simplest vibration possibility, NAN, d NN.00 m λ λ 4.00 m f v λ (0.0 m/s) 4.00 m 5.00 Hz

8 8 Chapter 18 Solutions For the second state NANAN, d NN 1.00 m f (0.0 m/s).00 m λ.00 m 10.0 Hz For the third resonance, NANANAN, d NN.00 m 3 λ 1.33 m f 15.0 Hz he mode mentioned in the problem has d NN m λ m f 5.0 Hz It is the fifth allowed state L 30.0 m µ kg/m 0.0 N f 1 v L where v µ so f / 47.1 m/s Hz f f Hz f 3 3f 1.36 Hz f 4 4f Hz Goal Solution G: he string described in the problem is very long, loose, and somewhat massive, so it should have a very low fundamental frequency, maybe only a few vibrations per second. O: he tension and linear density of the string can be used to find the wave speed, which can then be used along with the required wavelength to find the fundamental frequency. A: he wave speed is v F µ 0 N m/s kg/m For a vibrating string of length L fixed at both ends, the wavelength of the fundamental frequency is λ L 60.0 m; and the frequency is f 1 v λ v L 47.1 m/s 60 m Hz

9 Chapter 18 Solutions 9 he next three harmonics are f f Hz f 3 3f 1.36 Hz f 4 4f Hz L: he fundamental frequency is even lower than expected, less than 1 Hz. In fact, all 4 of the lowest resonant frequencies are below the normal human hearing range (0 to Hz), so these harmonics are not even audible L 10 cm f 10 Hz ( a ) For four segments, L λ or λ 60.0 cm m v λ f 7.0 m/s d NN m f 1 v L 7.0 (1.0) λ 1.40 m 30.0 Hz f λ v 308 m/s ( )/(0.700) (a) 163 N f Hz

10 10 Chapter 18 Solutions Goal Solution G: he tension should be less than 100 lbs. (~500 N) since excessive force on the 4 cello strings would break the neck of the instrument. If the string vibrates in three segments, there will be three antinodes (instead of one for the fundamental mode), so the frequency should be three times greater than the fundamental. O: he length of the string can be used to find the wavelength, which can be used with the fundamental frequency to find the wave speed. he tension can then be found from the wave speed and linear mass density of the string. A: When the string vibrates in the lowest frequency mode, the length of string forms a standing wave where L λ/ (see Figure 18.b), so the fundamental harmonic wavelength is λ L (0.700 m) 1.40 m and the velocity is v fλ (0 s 1 )(1.40 m) 308 m/s From the tension equation v µ m/l we get (a) v m L (308 m/s) ( kg) m 163 N For the third harmonic, the tension, linear density, and speed are the same. However, the string vibrates in three segments so that the wavelength is one third as long as in the fundamental (see Figure 18.d). λ 3 λ/3 From the equation λf v, we find that the frequency is three times as high: f 3 v 3 v 3f 660 Hz λ 3 λ L: he tension seems reasonable, and the third harmonic is three times the fundamental frequency as expected. Related to part, some stringed instrument players use a technique to double the frequency of a note by cutting a vibrating string in half. When the string is suddenly held at its midpoint to form a node, the second harmonic is formed, and the resulting note is one octave higher (twice the original fundamental frequency).

11 Chapter 18 Solutions 11 *18.0 f 1 v L, where v µ 1/ (a) If L is doubled, then f 1 L 1 will be reduced by a factor 1. If µ is doubled, then f 1 µ 1/ will be reduced by a factor 1. (c) If is doubled, then f 1 will increase by a factor of L 60.0 cm m 50.0 N µ g/cm kg/m where f n nv L v µ 1/ 70.7 m/s f n n n 0,000 Hz Largest n 339 f khz 18. f v λ µ since µ M L 1 λ ρv L 4 ρπ d L ρ AL L f new 4 old 4 ρ old π (d old ) L old / old 4 ρ old π d old L old f old 800 Hz 18.3 λ G (0.350 m) v f G λ A L A v f A L G L A L G f G f A L G L G Error! ) (0.350 m) Error! m

12 1 Chapter 18 Solutions hus, L A L G m m m 0.31 m, or the finger should be placed 31. cm from the bridge. L A v 1 f A f A µ d dl A 4 f A µ dl A 1 d L A d dl A cm L A ( ) cm 3.84% 18.4 In the fundamental mode, the string above the rod has only two nodes, at A and B, with an anti-node halfway between A and B. hus, A λ AB L cos θ or λ L cos θ Since the fundamental frequency is f, the wave speed in this segment of string is L θ B v λf Lf cos θ M Also, v µ L m cos θ m/ab where is the tension in this part of the string. hus, θ F Lf cos θ L m cos θ or 4L f cos θ L m cos θ and the mass of string above the rod is: Mg m cos θ [Equation 1] 4Lf Now, consider the tension in the string. he light rod would rotate about point P if the string exerted any vertical force on it. herefore, recalling Newton s third law, the rod must exert only a horizontal force on the string. Consider a free-body diagram of the string segment in contact with the end of the rod. F y sin θ Mg 0 Mg sin θ

13 Chapter 18 Solutions 13 hen, from Equation 1, the mass of string above the rod is m Mg cos θ sin θ 4Lf Mg 4Lf tan θ 18.5 (a) Let n be the number of nodes in the standing wave resulting from the 5.0-kg mass. hen n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For standing waves, λ L/n, and the frequency is f v/λ. hus, f n L n µ, and also f n + 1 L n + 1 µ hus, n + 1 n n n + 1 (5.0 kg)g (16.0 kg)g 5 4 herefore, 4n + 4 5n, or n 4 hen, f 4 (.00 m) (5.0 kg)(9.80 m/s ) kg/m 350 Hz he largest mass will correspond to a standing wave of 1 loop 1 (n 1), so 350 Hz (.00 m) m(9.80 m/s ) kg/m yielding m 400 kg *18.6 Using the frets does not change the speed of the wave. herefore, if d NN is the distance between adjacent nodes, λ 1 f 1 λ f d NN1 f 1 d NN f or d NN d NN1 f 1 f 1.4 cm 349 Hz 17 Hz.7 cm hus, the distance between frets is d NN d NN1.7 cm 1.4 cm 1.7 cm *18.7 he natural frequency is f 1 1 π g L 1 π 9.80 m/s.00 m 0.35 Hz he big brother must push at this same frequency of 0.35 Hz to produce resonance.

14 14 Chapter 18 Solutions *18.8 he distance between adjacent nodes is one-quarter of the circumference. d NN d AA λ 0.0 cm cm so λ 10.0 cm and f v λ 900 m/s m 9000 Hz 9.00 khz he singer must match this frequency quite precisely for some interval of time to feed enough energy into the glass to crack it. *18.9 (a) he wave speed is v 9.15 m.50 s 3.66 m/s From Figure P18.9, there are antinodes at both ends, so the distance between adjacent antinodes is d AA λ 9.15 m, and the wavelength is λ 18.3 m he frequency is then f v λ 3.66 m/s 18.3 m 0.00 Hz We have assumed the wave speed is the same for all wavelengths. *18.30 he wave speed is v gd (9.80 m/s )(36.1 m) 18.8 m/s he bay has one end open and one end closed, so its simplest resonance is with a node (of velocity, antinode of displacement) at the head of the bay and an antinode (of velocity, node of displacement) at the mouth. hen, d NA m λ 4 and λ m herefore, the period is 1 f λ v m 18.8 m/s s 1 h 4 min his agrees precisely with the period of the lunar excitation, so we identify the extrahigh tides as amplified by resonance (a) For the fundamental mode in a closed pipe, λ 4L. (see Figure 18.3b) But v fλ, therefore L v 4f 343 m/s So, L 4(40/s) m For an open pipe, λ L. (see Figure 18.3a) So, L v f 343 m/s (40/s) m

15 Chapter 18 Solutions 15 *18.3 λ d AA L n or L nλ for n 1,, 3,... Since λ v f, L n v f for n 1,, 3,... With v 343 m/s, and f 680 Hz, L n 343 m/s (680 Hz) n(0.5 m) for n 1,, 3,... Possible lengths for resonance are: L 0.5 m, m, m,..., n(0.5) m d AA 0.30 m λ m (a) f v 531 Hz λ λ m d AA 4.5 mm *18.34 he wavelength is λ v f 343 m/s 61.6/s 1.31 m so the length of the open pipe vibrating in its simplest (A-N-A) mode is d A to A 1 λ m A closed pipe has (N-A) for its simplest resonance, (N-A-N-A) for the second, and (N-A-N-A- N-A) for the third. Here, the pipe length is 5d N to A 5λ 4 5 (1.31 m) 1.64 m 4 *18.35 he air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and antinode at the open end, with d N to A 3 cm λ 4 so λ 0.1 m and f v λ 343 m/s 0.1 m 3 khz A small-amplitude external excitation at this frequency can, over time, feed energy into a larger-amplitude resonance vibration of the air in the canal, making it audible.

16 16 Chapter 18 Solutions λ v f 343 m/s 440/s m d N to A λ m length of resonant air column Water height m m 0.05 m m ρv ρah (1000 kg/m 3 )(0.100 m )(0.05 m) 0.5 kg For a closed box, the resonant frequencies will have nodes at both sides, so the permitted wavelengths will be L nλ, (n 1,, 3,... ). i.e., and L nλ nv f f nv L herefore, with L m and L'.10 m, the resonant frequencies are f n n(06 Hz) for L m for each n from 1 to 9 and f' n n(84.5 Hz) for L'.10 m for each n from to We suppose these are the lowest resonances of the enclosed air columns. For one, λ v f (343 m/s) 56/s 1.34 m length d AA λ m For the other, λ v f 343 m/s 440/s m length m So, original length 1.06 m λ d AA.1 m (a) f (343 m/s).1 m 16 Hz

17 Chapter 18 Solutions he fork radiates sound with λ v f he distance between successive water levels at resonance is 00 Hz d NN v f So Rt π r v f t π r v R f t π ( m) (343 m/s) ( m 3 /s)(00/s) 39 s 18.0 cm 3 /s he wavelength of sound is λ v f he distance between water levels at resonance is d v f Rt π r d π r v f and t π r v R f he length corresponding to the fundamental satisfies f v 4L, giving L 1 v 4f 343 4(51) m Since L > 0.0 cm, the next two modes will be observed, corresponding to f 3v 4L and f 5v 4L 3 or L 3v 4f 0.50 m and L 3 5v 4f m

18 18 Chapter 18 Solutions 18.4 Call L the depth of the well and v the speed of sound. hen for some integer n L (n 1) λ 1 4 (n 1) v 4f 1 (n 1)(343 m/s) 4(51.5 s 1 ) and for the next resonance hus, L [(n + 1) 1] λ 4 (n 1)(343 m/s) 4(51.5 s 1 ) (n + 1) v 4f (n + 1)(343 m/s) 4(60.0 s 1, ) (n + 1)(343 m/s) 4(60.0 s 1 ) and we require an integer solution to n n he equation gives n hen L [(7) 1](343 m/s) 4(51.5 s 1 ) 6.56, so the best fitting integer is n m and L [(7) + 1](343 m/s) 4(60.0 s 1 ) 1.4 m suggest the best value for the depth of the well is 1.5 m For resonance in a tube open at one end, f n v (n 1, 3, 5,...) Equation L ( a ) Assuming n 1 and n 3,.8 cm v 384 4(0.8) 3v and 384 4(0.683) 68.3 cm In either case, v 350 m/s For the next resonance, n 5, and L 5v 4f 5(350 m/s) 4(384 s m )

19 Chapter 18 Solutions (a) f 1 v λ v 4L m/s 4(4.88 m) 17.0 Hz f 1 v λ v L 34.0 Hz (c) For the closed pipe, f v(0.0 C) v(0 C) f f Hz For the open pipe, f f Hz *18.45 (a) For the fundamental mode of an open tube, L λ v f 343 m/s (880 s 1 ) m v 331 m/s 1 + ( 5.00) m/s We ignore the thermal expansion of the metal. f v λ v L 38 m/s 841 Hz (0.195 m) he flute is flat by a semitone When the rod is clamped at one-quarter of its length, the fundamental frequency corresponds to a mode of vibration in which L λ. herefore, L v f 5100 m/s 4400 Hz 1.16 m (a) f v L khz ()(1.60) Since it is held in the center, there must be a node in the center as well as antinodes at the ends. he even harmonics have an antinode at the center so only the odd harmonics are present. (c) f v ' L khz ()(1.60) v 4500 m/s λ 1 4L 40 cm.40 m so, f 1 v λ 1 v 4L khz

20 0 Chapter 18 Solutions f v f new Hz f 5.64 beats/s Goal Solution G: Beat frequencies are usually only a few Hertz, so we should not expect a frequency much greater than this. O: As in previous problems, the two wave speed equations can be used together to find the frequency of vibration that corresponds to a certain tension. he beat frequency is then just the difference in the two resulting frequencies from the two strings with different tensions. A: Combining the velocity equation v f/λ and the tension equation v f µλ µ we find that and since µ and λ are constant, we can divide to get f f 1 With f Hz, N, and 540 N: f (110 Hz) N 600 N Hz he beat frequency is: f b f 1 f 110 Hz Hz 5.64 Hz L: As expected, the beat frequency is only a few cycles per second. his result from the interference of the two sound waves with slightly different frequencies has a tone that varies in amplitude over time, similar to the sound made by saying wa-wa-wa Note: he beat frequency above is written with three significant figures on the assumption that the data and known precisely enough to warrant them. his assumption implies that the original frequency is known more precisely than to the three significant digits quoted in "110 Hz." For example, if the original frequency of the strings were Hz, the beat frequency would be 5.6 Hz. *18.50 (a) he string could be tuned to either 51 Hz or 55 Hz from this evidence. ightening the string raises the wave speed and frequency. If the frequency were originally 51 Hz, the beats would slow down. Instead, the frequency must have started at 55 Hz to become 56 Hz.

21 Chapter 18 Solutions 1 (c) From f v λ /µ L 1 L µ, f f 1 f f 1 and 1 53 Hz Hz 1 he fractional change that should be made in the tension is then fractional change % lower he tension should be reduced by 1.14% For an echo f ' f (v + v s) (v v s ) the beat frequency is f b f ' f Solving for f b gives f b f (v s ) (v v s ) when approaching wall. (a) f b (56) ()(1.33) 1.99 Hz beat frequency ( ) When moving away from wall, v s changes sign. Solving for v s gives f b v (5)(343) v s 3.38 m/s f f b ()(56) 5

22 Chapter 18 Solutions *18.5 We evaluate s 100 sin θ sin θ sin 3θ sin 4θ sin 5θ sin 6θ sin 7θ where s represents particle displacement in nanometers and θ represents the phase of the wave in radians. As θ advances by π, time advances by (1/53) s. Here is the result: Flute Waveform s (nm) Phase (rad) *18.53 We list the frequencies of the harmonics of each note in Hz: Harmonic Note A C# E he second harmonic of E is close to the third harmonic of A, and the fourth harmonic of C# is close to the fifth harmonic of A.

23 Chapter 18 Solutions (a) For the block: F x Mg sin so Mg sin Mg h he length of the section of string parallel to the incline is h/sin 30.0 h. he total length of the θ M string is then 3h. (c) he mass per unit length of the string is µ m/3h. (d) he speed of waves in the string is v µ Mg 3h m 3Mgh m (e) In the fundamental mode, the segment of length h vibrates as one loop. he distance between adjacent nodes is then d NN λ/ h, so the wavelength is λ h. he frequency is f v λ 1 h 3Mgh m 3Mg 8mh (g) (f) When the vertical segment of string vibrates with loops (i.e., 3 nodes), then h λ and the wavelength is λ h. he period of the standing wave of 3 nodes (or two loops) is 1 f λ v h m 3Mgh mh 3Mg (h) f b 1.0f f ( )f ( ) 3Mg 8mh (a) x ( ) m he wavelength is λ v f 343 m/s 300 Hz 1.14 m hus, x λ of a wave, or φ π (0.530) 3.33 rad

24 4 Chapter 18 Solutions

25 Chapter 18 Solutions 5 For destructive interference, we want x λ f x v where x is a constant in this set up. f v x 343 ()(0.606) 83 Hz f 87.0 Hz speed of sound in air: v a 340 m/s l m (a) λ b l v f λ b (87.0 s 1 )(0.400 m) v 34.8 m/s L λ a 4L v a λ a f L v a 4f 340 m/s 4(87.0 s 1 ) m Moving away from station, frequency is depressed: f ' Hz (343) (343 + v) Solving for v gives herefore, v (.00)(343) 178 v 3.85 m/s away from station Moving towards the station, the frequency is enhanced: f ' Hz (343) (343 v)

26 6 Chapter 18 Solutions Solving for v gives herefore, v (.00)(343) 18 v 3.77 m/s towards the station *18.58 Use the Doppler formula f ' f (v ± v 0) (v + v s ) With f ' 1 frequency of the speaker in front of student and f ' frequency of the speaker behind the student. (343 m/s m/s) f 1 ' (456 Hz) (343 m/s 0) (343 m/s 1.50 m/s) f ' (456 Hz) (343 m/s + 0) 458 Hz 454 Hz herefore, f b f ' 1 f ' Hz From the leading train she hears f' 1 f v + 0 v + v s f 343 m/s 343 m/s m/s From the still-approaching train, f' f hen, 4.00 Hz f ' f 1 ' 1.039f 0.977f f 4.00 Hz Hz v (48.0)(.00) m/s d NN 1.00 m λ.00 m f v 70.7 Hz λ λ a v a f 343 m/s 70.7 Hz 4.85 m

27 Chapter 18 Solutions 7 *18.61 he second standing wave mode of the air in the pipe reads ANAN, with d NA λ m 3 so λ.33 m and f v λ 343 m/s.33 m 147 Hz For the string, λ and v are different but f is the same. λ d NN m so λ m v λf (0.400 m)(147 Hz) 58.8 m/s /µ µv ( kg/m)(58.8 m/s) 31.1 N 18.6 (a) L v 4 f so L' L f f' Letting the longest L be 1, the ratio is 1 : 4 5 : 3 : 1 or in integers 30 : 4 : 0 : 15 L cm (4)(56) his is the longest pipe, so using the ratios the lengths are: 33.5, 6.8,.3, 16.7 cm (c) he frequencies are using the ratio 56, 30, 384, and 51 Hz. hese represent notes C, E, G, and C' on the physical pitch scale (a) Since the first node is at the weld, the wavelength in the thin wire is L or 80.0 cm. he frequency and tension are the same in both sections, so f 1 L µ 1 (0.400) Hz

28 8 Chapter 18 Solutions As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire. µ' 8.00 g/m so L' 1 f µ L' 1 ()(59.9) (4.60) ( ) 0.0 cm half the length of the thin wire f B f A λ B 1 3 λ A v B 1 3 v A v B 1 9 v A v µ B A v B v A (a) f n L µ so f' f L L' L L 1 he frequency should be halved to get the same number of antinodes for twice the length. n' n ' so ' n n' n (n + 1) he tension must be ' n (n + 1)

29 Chapter 18 Solutions 9 (c) f ' f n' L n L' ' so ' n f 'L ' n' f L ' 3 ' 9 16 to get twice as many antinodes For the wire, kg µ.00 m kg/m v µ (00 kg m/s ) kg/m v 00 m/s If it vibrates in its simplest state, d NN.00 m λ f v λ (00 m/s) 4.00 m 50.0 Hz (a) he tuning fork can have frequencies 45.0 Hz or 55.0 Hz If f 45.0 Hz, hen, v f λ (45.0/s) 4.00 m 180 m/s v µ (180 m/s) ( kg/m) 16 N or if f 55.0 Hz v µ f λ µ (55.0/s) (4.00 m) ( kg/m) 4 N

30 30 Chapter 18 Solutions he odd-numbered harmonics of the organ-pipe vibration are: 650 Hz, 550 Hz, 450 Hz, 350 Hz, 50 Hz, 150 Hz, 50.0 Hz Closed f Hz λ 6.80 m L 1.70 m We look for a solution of the form 5.00 sin (.00x 10.0t) cos(.00x 10.0t) A sin (.00x 10.0t + φ) A sin (.00x 10.0t)cos φ + A cos (.00x 10.0t) sin φ his will be true if both 5.00 A cos φ and 10.0 A sin φ, requiring (5.00) + (10.0) A A 11. and φ 63.4 he resultant wave 11. sin(.00x 10.0t ) is sinusoidal. *18.69 (a) With k Error! and ω π f Error! y(x, t) A sin kx cos ω t A sin π x λ cos π vt λ For the fundamental vibration, λ 1 L so y 1 (x, t) A sin π x L cos π vt L (c) For the second harmonic λ L and y (x, t) A sin π x cos π vt L L (d) In general, λ n L n and y n (x, t) A sin nπ x cos nπ vt L L

31 Chapter 18 Solutions (a) In the diagram, observe that: sin θ 1.00 m 1.50 m 3 Considering the mass, or θ 41.8 F y 0 gives cos θ mg d g d or (1.0 kg)(9.80 m/s ) cos N he speed of transverse waves in the string is m (a) m v µ 78.9 N 81 m/s kg/m For the standing wave pattern shown (3 loops), d 3 λ, or λ (.00 m) m hus, the required frequency is f v λ 81 m/s 1.33 m 11 Hz

Waves Part 3A: Standing Waves

Waves Part 3A: Standing Waves Last modified: 24/01/2018 Contents Links Contents Superposition Standing Waves Definition Nodes Anti-Nodes Standing Waves Summary Standing Waves on a String Standing Waves