UNIT 3 REVIEW (Page 36) Understanding Concepts total time. (a) T number of cycles 60 s T 4 swings.5 s 60 s T 0 swings 3.0 s (b) T > T Thus the student on the second swing is on the longer swing (the longer the period, the longer the pendulum. (c) delay t 3.0.5 0.5 With eery fifth swing of the second swing the students will be swinging together. The lowest frequency is 0 swings/min. The students will be swinging together 4 swings during that one minute.. If the stroboscope is flashing at 5 Hz, the time between flashes is 0. s. Since the white pendulum is stopped eery half cycle, its period is (0.) 0.40 s. Thus, it takes 0.4 s for the pendulum to make a full cycle. (This is the most likely answer for a pendulum, but theoretically the pendulum could be 5Hz and 7.5 Hz, and still produce the same effect.) 3. f λ 344 m / s 0.38 m f 905 Hz 4. λ f λ f 5. 3.0 08 m/s 8.8 0 7 Hz λ 3.4 m 3.0 0 8 m/s.08 0 8 Hz λ.8 m 6. Copyright 00 Nelson Thomson Learning Chapter 8 Music, Musical Instruments, and Acoustics 05
7. 8. λ f 4.0 0 m/s 86 Hz λ.4 m The distance between successie nodes is λ or 0.70 m apart. 06 Unit 3 Waes and Sound Copyright 00 Nelson Thomson Learning
9. 0. Copyright 00 Nelson Thomson Learning Chapter 8 Music, Musical Instruments, and Acoustics 07
. Circular waes decrease in amplitude as they moe away from the source because the energy is spread oer a larger area. The wae front increases as the circumference of a circle increases.. 33 m / s + 0.59 m/s 33 m / s + 0.59 m/s ( C) 33 m / s + 0.59 m/s ( C) 39 m / s 345 m / s 3. 33 m / s + 0.59 m/s λ f 33 m / s + 0.59 m/s 344 m /s (0 C) 440 Hz 344 m / s λ 0.78 m Since the instrument determines the fundamental frequency, the waelength is constant (fixed by the instrument). At 0ºC, the speed in air is 33 m/s. f λ 33 m /s 0.78 m f 46 Hz 4. fλ (0 Hz)(50 m) 500 m /s or 5. 0 3 m/s 5. d t d t d d t t t t 0.50 s d t ( 3.4 0 m/s)5.36 0 s ( ).8 0 m or d ( 5.0 0 3 m/s)3.6 0 s ( ).8 0 m 6. s 33 m / s + 0.59 m/s 33 m / s + 0.59 m/s (0 C) s 344 m / s 7. d (80 m) d 360 m t d.0 0 3 m 5. 0 3 m/s t 0.39 s t ( t 0.50 s) ( ) ( 3.4 0 m/s) t 5.0 0 3 m/s t 0.50 s ( 3.4 0 m/s) t 5.0 0 3 m/s( t ).5 0 3 s ( 4.66 0 3 m/s) t.5 0 3 s t 5.36 0 s d t t 3.6 0 s o ( 344 m / s) ( 0.6) o 55.0 m / sor 98 km / h 360 m s 360 m / sor 3.6 0 m/s 08 Unit 3 Waes and Sound Copyright 00 Nelson Thomson Learning
8. d (75 m) d 550 m d t 550 m.6 s 344 m / sor 3.4 0 m/s 9. The amplitude of the sound wae would increase as the band member blows hard into the flute. The intensity and loudness of the sound would also increase. The frequency and waelength would stay constant for a gien note. If the band member blows too hard, howeer, the sound will be distorted. 0. The speed of sound is much faster in metal than in air (see Table, Chapter 7, section 7.3, page 46). The frequency would remain constant. Thus the waelength would decrease in air (λ α ).. beat frequency f a f b, where f a > f b 0.0 Hz 56 Hz f b f b 46 Hz. 33 m / s + 0.59 m/s 33 m / s + 0.59 m/s (0 C) 344 m / s f λ, where λ λ 344 m /s (440 Hz)λ λ 0.78 m 33 m /s f λ 33 m /s 0.78 m f 45.6 Hz beat frequency f f 440 Hz 46 Hz beat frequency 4 Hz 3. (a) lower frequency, longer waelength. The lower frequency responds to.33 m waelength. (b) 3 beats beat frequency f λ f λ 5.0 s f (.9 m) f (.33 m) beat frequency.6 Hz.33 m f f.9 m f f beat frequency f f (.0 m) f f.6 Substitute for f :.33 m f f.9 m.6.6 f.33 m.9 m f λ (49 Hz)(.33 m) 347 m /s or 3.5 0 m/s f 49 Hz 4. Two situations are possible: the string is fingered at a quarter of its length from the right or a quarter of its length from the left. From the right From the left f L, where L f L 3 4 L f L, where L f L 4 L L L f 96 Hz f 96 Hz 3 L L 4 4 f 6 Hz f 784 Hz 5. first harmonic (f 0 ) 56 Hz fourth harmonic (4f 0 ) 4(56 Hz) 04 Hz 6. An unskilled player will not hae good fingering and bowing. Instead of two strings being in harmony, they will produce dissonance. Instead of being consonant with the fundamental, the notes will hae dissonance. Copyright 00 Nelson Thomson Learning Chapter 8 Music, Musical Instruments, and Acoustics 09
7. (a) 33 m / s + 0.59 m/s 33 m / s + 0.59 m/s (5 C) 34 m / s f λ 34 m /s.80 m f Hz f λ 34 m /s 3.0 m f 0 Hz beat frequency f f Hz 0 Hz beat frequency Hz or beats / s (b) The regions of loudest sound occur at the nodes λ. λ of beats f 34 m/s 4 m Hz λ λ of beats 8 m 8. The engine of the car transfers ibrations to other parts of the car with the frequency corresponding to the rotation of the crankshaft. If any part is free to rotate (a piece of metal), it will hae a natural frequency determined by its physical characteristics (such as length). When the frequency of the crankshaft is equal to the natural frequency of the loose piece of metal, it rattles. Another source of ibration could be a wheel, especially one that requires balancing. In either case, the rattle only occurs at a specific speed where the frequencies are equal. Other examples of this phenomenon include unbalanced wheels, particularly in the front, which will produce resonance in the whole front end of the car. A brake drum or disk may be out-of-round producing a shimmy in the wheel that could cause other parts of the car to ibrate. Car windows may also rattle in resonance if they are loosely mounted. 9. (a) Assume fundamental resonance. 33 m / s + 0.59 m/s λ f 344.4 m /s 33 m / s + 0.59 m/s ( C) 5 Hz λ 0.67 m 344.4 m / s closed column λ 7 cm 4 (b) The other resonances will occur at 3 4 λ 0.50 m (or 50 cm) and 5 λ 0.84 m (or 84 cm). 4 30. Assume adjacent resonance. λ.09 m 0.647 m 0.443 m λ 0.886 m 3. λ f 344 m /s 440 Hz λ 0.78 m 3. (a) open organ pipe (0.50 m) f 0 λ 0.50 m λ.00 m st oertone (f 0 ) (340 Hz) 680 Hz fλ (384 Hz)(0.886 m) 340 m/s 4 λ 0.0 m 3 4 λ 0.59 m f λ 340m /s.00 m f 340Hz 0 Unit 3 Waes and Sound Copyright 00 Nelson Thomson Learning
(b) closed organ pipe f 0 4 λ 0.50 m λ.00 m st oertone 3(f 0 ) 3(70 Hz) 50 Hz 33. (a) first oertone is 3 4 λ f 340m / s.00 m 70Hz st oertone 3 (0.50 m) 4 0.375 m or 0.38 m (b) 33 m / s + 0.59 m/s 33 m / s + 0.59 m/s (0.0 C) 344 m / s 4 λ 0.38 m λ.5 m f λ 344 m / s.5 m f 6 Hz 34. T λ λ T.5 cm 0.0 cm / s d t T 0.075 s 5.0 mm 0.075 s 67 mm /sor 6.7 cm /s 35. The high-pitched sounds of a person who inhales helium results from the higher speed of sound in helium compared to air (see Chapter 7, Table, page 46). The ocal cords ibrate at the same frequencies, but the resonant frequency of the column formed by the larynx, mouth, and nasal caities is higher. 36. (a) without wind 33 m / s + 0.59 m/s d t (344 m /s)(4.0 s) 33 m / s + 0.59 m/s d 376 m or.4 0 3 m (0 C) 344 m / s (b) with wind at 48 km/h or 3.3 m/s s a w 344 m / s 3.3 m / s s 330.7 m / s t d 376 m 330.7 m / s t 4.6 s The sound would take 0.6 s longer (4.6 s 4.0 s) with the wind blowing. 37. (a) T f T 5.0 0 Hz T.0 0 3 s T 5.0 0 4 Hz T.0 0 5 s Copyright 00 Nelson Thomson Learning Chapter 8 Music, Musical Instruments, and Acoustics
(b) λ T λ (.5 0 3 m /s)(.0 0 3 s) λ 3.0 m λ (.5 0 3 m /s)(.0 0 5 s) λ.0 0 m 38. (a) T f T 3.0 0 4 Hz T 3.3 0 5 s (b) 33 m / s + 0.59 m/s λ f 33 m / s + 0.59 m/s 349.7 m /s (30.0 C) 3.0 0 4 Hz 349.7 m / s λ.7 0 m or. 0 m (c) t 4.6 0 s d t (349.7 m / s)(.3 0 s) t.3 0 s d 8.0 m 39. The second, fourth, sixth, etc., harmonics hae a node exactly in the middle. 40. When the temperature decreases, the speed decreases. Because the waelength is constant and f, the frequency, and λ thus the pitch, is proportional to the speed. 4. There must be reflection from behind the listener, or else the sound would not be reflected back to the end wall again. 33 m / s + 0.59 m/s time for one reflection d t 33 m / s + 0.59 m/s ( C) ( 60.0 m) 344 m / s 344 m/s t 0.93 s 6.0 s number of reflections 0.93 s 6.5 times Reflections can occur only as whole numbers, so the actual number of reflections is. The decimal portion of the answer represents the partial distance the sound traels before it dies out. 4. The Indian earthquake at 7.9 on the Richter scale is logs greater than an earthquake at 5.9. This corresponds to an amplitude increase of 0 or 00 times. Compared to an earthquake of 6.9, the amplitude increases 0 or ten times larger. 43 When an elastic band is wrapped around one prong of a tuning fork, the frequency of the fork decreases because of the slightly greater mass. According to Newton s first law, increasing the mass increases the inertia. As a result, the frequency decreases as the inertia of the tuning fork prong increases. According to Newton s second law, if the mass increases and force remains constant (the restoratie force of the prong), then the acceleration will decrease aα and the prong will m ibrate at a lower frequency. Of course, the answers are the same since the first law is a special case of the second law. Unit 3 Waes and Sound Copyright 00 Nelson Thomson Learning
44. (a) (b) diameter 6 cm circumference π d 8.8 cm λ 8.8 cm λ 9.4 cm 45. As the gas tank is filled, the resonant length of the closed air column decreases and the resonant frequency of the gas tank increases. Since pitch is related to frequency, the pitch also increases. The source of the original sound is the gas hitting the surface of the gas in the tank. 46. (a) Longer waelength results in a perceied decrease in frequency because the speed of light is constant. This is called the red shift. (b) According to the Doppler effect, f f ± ; if the frequency decreases (f <f ), the factor s ± <. s Therefore the stars are moing away from us. 47. A megaphone directs sound in one direction because of reflection inside the megaphone. This makes the sound more intense because the sound energy does not spread out until it reaches the end of the megaphone. 48. The V rockets traelled faster than the speed of sound. The warhead could hit the ground and explode before the sound of the rocket could be heard, since the sound traelled slower than the rocket. 49. length of bathtub λ fλ (0.7 Hz)(.0 m) 0.85m/s or 85 cm/s λ (60.0 cm) 0 cm or.0 m Applying Inquiry Skills 50. If the speed of sound aried with frequency, we would lie in a world where high frequency sounds would trael at a different speed than low frequency sounds. In this were the case, when a baritone and a soprano are singing a harmonious duet, the music would reach our ears at different times producing dissonance. This is obiously not the case. 5. Lower frequency sound waes diffract more than higher frequency sound waes because they hae longer waelengths. FM radio and teleision signals use high frequency sound waes that do not diffract around hills. Thus, to receie FM and TV signals oer the air, there should be a line of sight between the transmitting tower and the antenna receiing the signal. AM radio signals hae a lower frequency (longer waelength) and thus, diffract better around hills and oer the horizon. At night, when there are fewer disturbances, AM radio waes can trael ery long distances (>000 km). 5. A guitar string would hae to be ery long to produce the low frequencies required. Fine wire is wrapped around cord in a guitar string to increase its density and its diameter. Both factors will lower the frequency of the string without increasing the length. (see Table, page 8). Copyright 00 Nelson Thomson Learning Chapter 8 Music, Musical Instruments, and Acoustics 3
Making Connections 53. Auditorium B would be better acoustically, een though auditorium A may look better architecturally. In auditorium B, the sound from the stage is reflected uniformly to all parts of the auditorium, whereas in auditorium A the sound is concentrated on the main floor. Also, the straight walls at the back will tend to increase echoing and prolong the reerberation time. 54. The reerberation time of music will prolong the sound enhancing the quality of the sound. Howeer, a longer reerberation time with speech will result in the reflected sound interfering with the projected sound making difficult (if not impossible) to understand what the speaker is saying. 55. (a) P waes, or compression waes, are longitudinal waes S waes, or shear waes, are transerse waes. Rayleigh waes are transerse waes. Loe waes are longitudinal waes. (b) (i) Loe waes are most likely to cause damage to buildings because they are surface waes that moe the earth from side to side. (c) P 9.0 km/s dp ds S 5.0 km/s t S > t P and t S t P +.0 0 P tp S ts s (9.0 km/s) tp (5.0 km/s)( tp +.0 0 s) (9.0 km/s) tp (5.0 km/s) tp + 6.0 0 s 6.0 0 s (9.0 km/s) tp (5.0 km/s) tp tp.5 0 s d P P t P (9.0 km/s)(.5 0 s) d P.35 0 3 km or.4 0 3 km (d) Three stations are needed to locate the epicentre by triangulation. One seismic station could only tell you how far away the epicentre was, not the direction. 59. (a) Factors that affect the pitch are the factors that affect the frequency. In humans the tension in the ocal chord is a primary factor, as well as the resonance created in the mouth, nasal passages, and throat. (b) In men the resonant passages (mouth, nasal passages, and throat) are larger and, therefore can create lower resonant frequencies than women. (c) When boys reach puberty there are significant changes in the body including the ocal chords and resonant passages, all leading to a oice with a lower pitch. 60. When you listen to your recorded oice, you hear what others hear all the time. When you hear your own oice, the sound traels from the air to your ear, as well as internally through the bones in your head. The result is not what others hear. 6. Deer whistles are high frequency whistles mounted on the front of the car. When the car is at a certain speed, air pressure actiates the whistle and it emits an ultrasonic sound that can be heard by the deer. 6. Students should whistle a few notes of arying frequencies. Air forced through the lips sets up a resonant frequency. The pitch can be increased or decreased by curling the tongue, changing the size of the air resonant column in the mouth 63. (a) Oscillating air in the tubing produces the musical note. (b) Air flows from the middle of the tubing to the whirling end where the rapid motion of the whirling end reduces the air pressure there. As the air flows through the tube and oer the corrugations, it begins to oscillate. The spacing of the corrugations and the speed of the airflow determine the frequency of these oscillations. From the small range of oscillations produced with a gien whirling speed, a resonant frequency is reached, which is the tone you hear from the tube. A faster twirling speed changes the air oscillations and higher resonant frequencies are heard. 4 Unit 3 Waes and Sound Copyright 00 Nelson Thomson Learning