IMPORTANT NOTE ABOUT WEBASSIGN:


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1 Week 13 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions o these problems, various details have been changed, so that the answers will come out dierently. The method to ind the solution is the same, but you will need to repeat part o the calculation to ind out what your answer should have been. WebAssign Problem 1: The drawing shows a loudspeaker A and point C, where a listener is positioned. A second loudspeaker B is located somewhere to the right o A. Both speakers vibrate in phase and are playing a 68.6Hz tone. The speed o sound is 343 m/s. What is the closest to speaker A that speaker B can be located, so that the listener hears no sound? REASONING The geometry o the positions o the loudspeakers and the listener is shown in the ollowing drawing. C d = m 1 d y A B x 1 x
2 The listener at C will hear either a loud sound or no sound, depending upon whether the intererence occurring at C is constructive or destructive. I the listener hears no sound, destructive intererence occurs, so nλ d d1 = n = 1, 3, 5, K (1) SOLUTION Since v = λ, according to Equation 16.1, the wavelength o the tone is v 343 m/s λ = = = 5.00 m 68.6 Hz Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so d nλ 5.00 m = + d = m = 3.50 m 1 From the igure above we have that, x1 = (1.00 m) cos 60.0 = m Then y = (1.00 m) sin 60.0 = m x + y = d = (3.50 m) or x = (3.50 m) (0.866 m) = 3.39 m Thereore, the closest that speaker A can be to speaker B so that the listener hears no sound is x1 + x = m m = 3.89 m. WebAssign Problem : Speakers A and B are vibrating in phase. They are directly acing each other, are 7.80 m apart, and are each playing a 73.0Hz tone. The speed o sound is 343 m/s. On the line between the speakers there are three points where constructive intererence occurs. What are the distances o these three points rom speaker A? REASONING AND SOLUTION Since v = λ, the wavelength o the tone is λ v 343 m/s = = = 4.70 m 73.0 Hz The igure below shows the line between the two speakers and the distances in question.
3 A x L  x P L B Constructive intererence will occur when the dierence in the distances traveled by the two sound waves in reaching point P is an integer number o wavelengths. That is, when (L x) x = nλ where n is an integer (or zero). Solving or x gives x = L nλ (1) When n = 0, x = L/ = (7.80 m)/ = m. This corresponds to the point halway between the two speakers. Clearly in this case, each wave has traveled the same distance and thereore, they will arrive in phase. When n = 1, (7.80 m) (4.70 m) x = = 1.55 m Thus, there is a point o constructive intererence m r o m s p e a k e r A. The points o constructive intererence will occur symmetrically about the center point at L/, so there is also a point o constructive intererence 1.55 m rom speaker B, that is at the point 7.80 m 1.55 m = 6. 5 m r o m s p e a k e r A. When n > 1, the values o x obtained rom Equation (1) will be negative. These values correspond to positions o constructive intererence that lie to the let o A or to the right o C. They do not lie on the line between the speakers. WebAssign Problem 3: A row o seats is parallel to a stage at a distance o 8.7 m rom it. At the center and ront o the stage is a diraction horn loudspeaker. This speaker sends out its sound through an opening that is like a small doorway with a width D o 7.5 cm. The speaker is playing a tone that has a requency o Hz. The speed o sound is 343 m/s. What is the distance between two seats, located near the center o the row, at which the tone cannot be heard?
4 REASONING AND SOLUTION The igure at the right shows the geometry o the situation. D i r a c t i o n h o r n S t a g e The tone will not be heard at seats located at the irst diraction minimum. This occurs when λ sin θ = = D v D That is, the angle θ is given by 8. 7 m θ C θ x F i r s t r o w o s e a t s θ = sin = sin = 7. D ( Hz)(0.075 m) 1 v m/s 4 From the igure at the right, we see that x tan 7. = x = (8.7 m)(tan 7. ) = 4.47 m 8. 7 m θ 8.7 m x Thus, seats at which the tone cannot be heard are a distance x on either side o the center seat C. Thus, the distance between the two seats is x = (4.47 m) = 8. 9 m WebAssign Problem 4: Two pure tones are sounded together. The drawing shows the pressure variations o the two sound waves, measured with respect to atmospheric pressure. What is the beat requency? REASONING The beat requency o two sound waves is the dierence between the two sound requencies. From the graphs, we see that the period o the wave in the 1 upper igure is 0.00 s, so its requency is = 1/ T = 1/(0.00 s)= Hz. The requency o the wave in the lower igure is = 1/(0.04 s)=4. 10 Hz.
5 SOLUTION The beat requency o the two sound waves is 1 1 beat = 1 = Hz Hz = 8 Hz WebAssign Problem 5: A string has a linear density o kg/m and is under a tension o 80 N. The string is 1.8 m long, is ixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) requency o the traveling waves that make up the standing wave. REASONING A standing wave is composed o two oppositely traveling waves. The F speed v o these waves is given by v = (Equation 16.), where F is the m / L tension in the string and m/l is its linear density (mass per unit length). Both F and m/l are given in the statement o the problem. The wavelength λ o the waves can be obtained by visually inspecting the standing wave pattern. The requency o the waves is related to the speed o the waves and their wavelength by = v/λ (Equation 16.1). SOLUTION a. The speed o the waves is F 80 N v = = = m / L kg/m 180 m/s b. Two loops o any standing wave comprise one wavelength. Since the string is 1.8 m long and consists o three loops (see the drawing), the wavelength is 1.8 m 3 ( ) λ = 1.8 m = 1. m c. The requency o the waves is v 180 m/s = = = λ 1. m 150 Hz λ
6 WebAssign Problem 6: Divers working in underwater chambers at great depths must deal with the danger o nitrogen narcosis (the bends ), in which nitrogen dissolves into the blood at toxic levels. One way to avoid this danger is or divers to breathe a mixture containing only helium and oxygen. Helium, however, has the eect o giving the voice a highpitched quality, like that o Donald Duck s voice. To see why this occurs, assume or simplicity that the voice is generated by the vocal cords vibrating above a gasilled cylindrical tube that is open only at one end. The quality o the voice depends on the harmonic requencies generated by the tube; larger requencies lead to higherpitched voices. Consider two such tubes at 0 C. One is illed with air, in which the speed o sound is 343 m/s. The other is illed with helium, in which the speed o sound is m/s. To see the eect o helium on voice quality, calculate the ratio o the n th natural requency o the heliumilled tube to the n th natural requency o the airilled tube. REASONING For a tube open at only one end, the series o natural requencies is given by n = nv / ( 4L ) (Equation 17.5), where n has the values 1, 3, 5, etc., v is the speed o sound, and L is the tube length. We will apply this expression to both the airilled and the heliumilled tube in order to determine the desired ratio. SOLUTION According to Equation 17.5, we have nv nv = = 4L 4L air n, air and n, helium helium Dividing the expression or helium by the expression or air, we ind that nvhelium 3 n, helium 4L vhelium m/s = = = =.9 nv n, air air vair 343 m/s 4L WebAssign Problem 7: Two ultrasonic sound waves combine and orm a beat requency that is in the range o human hearing. The requency o one o the ultrasonic waves is 70 khz. What is (a) the smallest possible and (b) the largest possible value or the requency o the other ultrasonic wave? REASONING AND SOLUTION Two ultrasonic sound waves combine and orm a beat requency that is in the range o human hearing. We know that the requency range o human hearing is rom 0 Hz to 0 khz. The requency o one o the ultrasonic waves is 70 khz. The beat requency is the dierence between the two sound requencies. The smallest possible value or the ultrasonic requency can be ound by subtracting the upper limit o human hearing rom the value o 70 khz. The largest possible value or the ultrasonic requency can be determined by adding the upper limit o human hearing to the value o 70 khz.
7 a. The smallest possible requency o the other ultrasonic wave is = 7 0 k H z 0 k H z = 5 0 k H z which results in a beat requency o 7 0 k H z 5 0 k H z = 0 k H z. b. The largest possible requency or the other wave is = 7 0 k H z + 0 k H z = 9 0 k H z which results in a beat requency o 9 0 k H z 7 0 k H z = 0 k H z. WebAssign Problem 8: The undamental requency o a vibrating system is 400 Hz. For each o the ollowing systems, give the three lowest requencies (excluding the undamental) at which standing waves can occur: (a) a string ixed at both ends, (b) a cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end open. REASONING AND SOLUTION a. For a string ixed at both ends the undamental requency is 1 = v/(l) so n = n 1. = 800 Hz, = 100 Hz, = 1600 Hz 3 4 b. For a pipe with both ends open the undamental requency is 1 = v/(l) so n = n 1. = 800 Hz, = 100 Hz, = 1600 Hz 3 4 c. For a pipe open at one end only the undamental requency is 1 = v/(4l) so n = n 1 with n odd. = 100 Hz, = 000 Hz, = 800 Hz WebAssign Problem 9: The A string on a string bass is tuned to vibrate at a undamental requency o 55.0 Hz. I the tension in the string were increased by a actor o our, what would be the new undamental requency? REASONING For standing waves on a string that is clamped at both ends, Equations 17.3 and 16. indicate that the standing wave requencies are v F n = n where v = L m / L
8 Combining these two expressions, we have, with n = 1 or the undamental requency, 1 1 F = L m / L This expression can be used to ind the ratio o the two undamental requencies. SOLUTION The ratio o the two undamental requencies is Since Fnew = 4F, we have old old new 1 Fold = L m / L = 1 Fnew L m / L F F old new F 4 F = = = 4 = (55.0 Hz) () = Hz new old new old old old Fold Fold
9 Practice conceptual problems: 1. Does the principle o linear superposition imply that two sound waves, passing through the same place at the same time, always create a louder sound than either wave alone? Explain. REASONING AND SOLUTION The principle o linear superposition states that when two or more waves are present simultaneously at the same place, the resultant wave is the sum o the individual waves. This principle does not imply that two sound waves, passing through the same place at the same time, always create a louder sound than either wave alone. The resultant wave pattern depends on the relative phases o the two sound waves when they meet. I two sound waves arrive at the same place at the same time, and they are exactly in phase, then the two waves will interere constructively and create a louder sound than either wave alone. On the other hand, i two waves arrive at the same place at the same time, and they are exactly out o phase, destructive intererence will occur; the net eect is a mutual cancellation o the sound. I the two sound waves have the same amplitude and requency, they will completely cancel each other and no sound will be heard. 6. A tuning ork has a requency o 440 Hz. The string o a violin and this tuning ork, when sounded together, produce a beat requency o 1 Hz. From these two pieces o inormation alone, is it possible to determine the exact requency o the violin string? Explain. REASONING AND SOLUTION A tuning ork has a requency o 440 Hz. The string o a violin and this tuning ork, when sounded together, produce a beat requency o 1 Hz. This beat requency is the dierence between the requency o the tuning ork and the requency o the violin string. A violin string o requency 439 Hz, as well as a violin string o 441 Hz, will produce a beat requency o 1 Hz, when sounded together with the 440 Hz tuning ork. We conclude that, rom these two pieces o inormation alone, it is not possible to distinguish between these two possibilities. Thereore, it is not possible to determine the requency o the violin string. 10. A string is vibrating back and orth as in Figure 17.18a. The tension in the string is decreased by a actor o our, with the requency and the length o the string remaining the same. Draw the new standing wave pattern that develops on the string. Give your reasoning. REASONING AND SOLUTION A string is being vibrated back and orth as in Figure 17.18a. The tension in the string is decreased by a actor o our, with the requency and the length o the string remaining the same. From Equation 16., v = F /( m / L), we see that decreasing the tension F by a actor o our results in decreasing the wave speed v by a actor o 4 or. From the relationship v = λ, we see that, when the requency is ixed, decreasing the wave speed by a actor o results in decreasing the wavelength λ by a actor o. Thereore, when the tension in the string is decreased by a actor o 4, the wavelength o the resulting standing
10 wave pattern will decrease by a actor o. In Figure 17.18a, the wavelength o the standing wave is equal to twice the length o the string. Since decreasing the tension reduces the wavelength by a actor o, the new wavelength is equal to the length o the string. With this new wavelength, the pattern shown below results. 13. Standing waves can ruin the acoustics o a concert hall i there is excessive relection o the sound waves that the perormers generate. For example, suppose a perormer generates a 093Hz tone. I a largeamplitude standing wave is present, it is possible or a listener to move a distance o only 4.1 cm and hear the loudness o the tone change rom loud to aint. Account or this observation in terms o standing waves, pointing out why the distance is 4.1 cm. REASONING AND SOLUTION When a concert perormer generates a 093Hz tone, the wavelength o the sound is λ = v 343 m/s = 093 Hz = m or 16.4 cm I there is excessive relection o the sound that the perormer generates, a large amplitude standing wave can result. The distance between an antinode (maximum loudness) and the next adjacent node (zero loudness) on a standing wave is onequarter o a wavelength. For a standing sound wave with a requency o 093 Hz, the distance between an antinode and the adjacent node is (16.4/4) cm or 4.1 cm. It would be possible, thereore, or a listener to move a distance o only 4.1 cm and hear the loudness o the tone change rom loud to aint.
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