Answer: 101 db. db = 10 * log( 1.16 x 10-2 W/m 2 / 1 x W/m 2 ) = 101 db

Transcription

1 54. A machine produces a sound with an intensity of 2.9 x 10-3 W/m 2. What would be the decibel rating if four of these machines occupy the same room? Answer: 101 db Four of these machines would be four times as intense as one machine - that would be an intensity of 1.16 x 10-2 W/m 2. The decibel rating is db = 10 * log( 1.16 x 10-2 W/m 2 / 1 x W/m 2 ) = 101 db 55. The sound in the United Center during a Chicago Bulls basketball game in 1998 was seven times as intense as it is today. If the decibel rating today is 89 db, then what was the intensity rating in 1998? Answer: 5.56 x 10-3 W/m 2 This problem involves finding the intensity level in the stadium today and then multiplying it by seven. The conversion from db to intensity in W/m 2 is done using the usual equation. db = 10 * log( I / 1 x W/m 2 ) The decibel level is substituted into the equation and then each side is divided by 10. Now to solve for I, take the invlog of each side of the equation. Now multiply both sides of the equation by 1 x W/m 2 to solve for I. Now multiply by 7 since the intensity was seven times greater. 89 db = 10 * log( I / 1 x W/m 2 ) 8.9 = log( I / 1 x W/m 2 ) invlog(8.9) = invlog( log( I / 1 x W/m 2 ) ) 7.94 x 10 8 = I / 1 x W/m x 10-4 W/m 2 = I I = 5.56 x 10-3 W/m A sound has an intensity of 8.0 x10-3 W/m 2 at a distance of 2.0 m from its source. What is the intensity at a distance of... a m from the source? b m from the source? c m from the source? d m from the source? e m from the source? See below. This problem tests your understanding of the inverse square law: the intensity of a sound varies inversely with the square of the distance from the source of the sound. I = k (1/R 2 ) a. As the distance is doubled (4 m is two times 2 m), the intensity level is reduced by a factor of 4. The new intensity is 8.0 x10-3 W/m 2 / 4 = 2.0 x10-3 W/m 2. b. As the distance is tripled (6 m is three times 2 m), the intensity level is reduced by a factor of 9. The new intensity is 8.0 x10-3 W/m 2 / 9 = 8.9 x10-4 W/m 2. c. As the distance is quadrupled (8 m is four times 2 m), the intensity level is reduced by a factor of 16. The new intensity is 8.0 x10-3 W/m 2 / 16 = 5.0 x10-4 W/m 2. d. As the distance is increased by a factor of 12 (24 m is 12 times 2 m), the intensity level is reduced by a factor of 144. The new intensity is 8.0 x10-3 W/m 2 / 144 = 1.4 x10-5 W/m 2. e. As the distance is increased by a factor of (46.1 m is times 2 m), the intensity level is reduced by a factor of 531. The new intensity is 8.0 x10-3 W/m 2 / 531 = 1.5 x10-5 W/m Ben Stupid is sitting 2.0 m in front of the speakers on the stage at the Twisted Brother concert. The decibel rating of the sound heard there is 110 db. What would be the decibel rating at a location of... a m from the speaker? b m from the speaker? c m from the speaker? Answer: See answers below. This problem is similar to the last problem in that it tests your understanding of the inverse square law. There is however a slight twist. The decibel rating is given and since the decibel scale is logarithmic, we must first find the intensity level before we do the division. A decibel rating of 110 db is equivalent to an intensity level of times the threshold of hearing (1 x W/m 2 ). So the intensity level at 2.0 m from the speaker is 0.1 W/m 2. Now the inverse square law is applied; the intensity of a sound varies inversely with the square of the distance from the source of the sound. I = k (1/R 2 ) a. As the distance is doubled (4 m is two times 2 m), the intensity level is reduced by a factor of 4. The new intensity is 0.1 W/m 2 / 4 = W/m 2 or 2.5 x10-2 W/m 2. This converts to a decibel rating of 104 db. b. As the distance is tripled (6 m is three times 2 m), the intensity level is reduced by a factor of 9. The new intensity is 0.1 W/m 2 / 9 = W/m 2 or 1.1 x10-2 W/m 2. This converts to a decibel rating of 101 db (100.5 db). c. As the distance is increased by a factor of 10 (20 m is 10 times 2 m), the intensity level is reduced by a factor of 100. The new intensity is 0.1 W/m 2 / 100 = W/m 2 or 1.0 x10-3 W/m 2. This converts to a decibel rating of 90 db. 58. Use the Doppler equation for a moving source to calculate the observed frequency for a 250.-Hz source of sound if it is moving with a speed of. (Assume that the speed of sound in air is 340. m/s.) a. 30. m/s towards the observer. b. 30. m/s away from the observer. c m/s towards the observer.

2 d m/s away from the observer. e m/s towards the observer. f. 335 m/s towards the observer. See below. The Doppler equation for determining the observed frequency for a moving source is: f observed = v sound / (v sound ± v source ) f source The + sign is used if the source moves away from the observer The - sign is used if the source moves towards the observer. If applied to this situation, v sound is 340 m/s and f source is 250 Hz. a. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(310. m/s) (250. Hz) = 274 Hz b. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(370. m/s) (250. Hz) = 230. Hz c. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(40. m/s) (250. Hz) = 2130 Hz (2125 Hz) d. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(640. m/s) (250. Hz) = 133 Hz e. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(20. m/s) (250. Hz) = 4250 Hz f. f observed = (340. m/s)/(340. m/s m/s) (250 Hz) = (340. m/s)/(5 m/s) (250. Hz) = 1.70 x 10 4 Hz 59. Shirley Yackin is holding the phone cord in her hand. It is stretched to a length of 2.4 m and has a mass of 1.8 kg. If the tension in the phone cord is 2.5 N, then determine the speed of vibrations within the cord. Answer: 1.8 m/s The telephone cord acts as a guitar string. As such the speed of waves in the cord is given by the equation where F tens is the tension in the cord in Newtons and mu is the linear density in kg/m. The tension is given and the mu value can be calculated: mu = mass/length = (1.8 kg) / (2.4 m) = 0.75 kg/m Now substitute mu and F tens into the equation and solve for speed: v = SQRT[(2.5 N) / (0.75 kg/m)] = SQRT(3.33) = 1.8 m/s ( m/s) 60. (Referring to problem #59.) With what frequency must Shirley vibrate the cord up and down in order to produce the second harmonic within the cord? Answer: 0.76 Hz If Shirley is producing the second harmonic, then there will be one full wavelength in the length of the cord. Since the length of the cord is 2.4 m, the wavelength is 2.4 m as well (see diagram at right). With the wavelength and speed (problem #59) known, the frequency can be calculated. (Note that the unrounded number from problem #59 is used i this calculation.) f = v / lambda = ( m/s) / (2.4 m) = 0.76 Hz ( Hz) 61. (Referring to problem #60.) If Shirley maintains this same frequency and wishes to produce the fourth harmonic, then she will have to alter the speed of the wave by changing the tension. Assuming the same mass density as in #59, and the same frequency as in #60, to what tension must the cord be pulled to produce the fourth harmonic? Answer: 0.63 N In this case, Shirley is changing the medium. So she is producing the fourth harmonic in a different medium (different properties) using the same frequency as in #60. For the fourth harmonic, there are two full wavelengths inside the length of the cord (see diagram at right); so the length of the wave (wavelength) is one-half the length of the cord m. With frequency and wavelength known, the speed can be computed: v = f lambda = ( Hz) (1.2 m) = m/s. Now with speed and mu known, the tension can be calculated from the equation: First, perform algebra to manipulate the equation into a form with F tens by itself. This is done by squaring both sides of the equation and multiplying by mu. v 2 = F tens / mu Now substitute and solve: mu v 2 = F tens F tens = (0.75 kg/m) ( m/s) 2 = 0.63 N ( N) 62. A guitar string has a mass of 32.4 g and a length of 1.12 m. The string is pulled to a tension of 621 N. Determine the speed at which vibrations move within the string. Answer: 147 m/s This problem is very similar to question #59. The speed of waves in the guitar string is given by the equation where F tens is the tension in the string in Newtons and mu is the linear density in kg/m. The tension is given and the mu value can be calculated (one will

3 have to be careful of units): Now substitute mu and F tens into the equation and solve for speed: mu = mass/length = ( kg) / (1.12 m) = kg/m v = SQRT [ (621 N)/( kg/m) ] = SQRT(21467) = 147 m/s ( m/s) 63. (Referring to problem #62.) Stan Dingwaives is playing this guitar. If Stan leaves the string "open" and uses its full length to produce the first harmonic, then what frequency will Stan be playing? Answer: 65.4 Hz For the first harmonic, there is only one-half of a wave inside of the length of the string. So the length of the wave is two times longer than the length of the string (see diagram at right). Thus, the wavelength is 2.24 m. Now with speed and wavelength known, the frequency can be calculated. f = v /lambda = ( m/s) / (2.24 m) = 65.4 Hz ( Hz) 64. (Referring to problem #62 and# 63.) If Stan wishes to increase the frequency by a factor of , then how far (in cm) from the end of the string must he "close" the string (i.e., where must he press his finger down to change the length and produce the desired frequency)? Use the same mass density and speed as in problem #62. Answer: 0.23 m For the same speed (same properties of the medium), frequency and wavelength are inversely related. Increasing the frequency by a factor of will be the result of decreasing the wavelength by the same factor. So the new wavelength will be lambda = (2.24 m) / = 1.78 m Since the wavelength is twice the effective length of the string, the string must have an effective length of 0.89 m or 89 cm. So, Stan must place his finger a distance of 0.23 m or 23 cm (1.12 m m) from the end of the string 65. A guitar string has a fundamental frequency of 262 Hz. Determine the frequency of the... a.... second harmonic. b.... third harmonic. c.... fifth harmonic. d.... eighth harmonic. The frequencies of the various harmonics of an instrument are whole number rations of the fundamental frequency. The frequency of the second harmonic is two times the fundamental frequency; the frequency of the third harmonic is three times the fundamental frequency and so on. The frequencies of the harmonics can be found utilizing the formula f n = n * f 1 a. f 2 = 2 * f 1 = 2 * 262 Hz = 524 Hz b. f 3 = 3 * f 1 = 3 * 262 Hz = 786 Hz c. f 5 = 5 * f 1 = 5 * 262 Hz = 1310 Hz d. f 8 =8 * f 1 = 8 * 262 Hz = 2.10x 10 3 Hz 66. Determine the speed of sound through air if the temperature is... a degrees Celsius. b degrees Celsius. c degrees Celsius. d degrees Celsius. There are numerous equations for computing the speed of sound through air based on the temperature of air. A common equation found in books is v = 331 m/s * SQRT (1 + T/273) where T is the Celsius temperature. The following answers were found using this equation. a. v = 331 m/s * SQRT (1 + 0/273) = 331 m/s b. v = 331 m/s * SQRT (1 + 12/273) = 338 m/s c. v = 331 m/s * SQRT (1 + 25/273) = 346 m/s d. v = 331 m/s * SQRT (1 + 40/273) = 354 m/s An alternative equation which is comonly used is v = 331 m/s + (.60 m/s/c)*t where T is the temperature is degrees celsius. Using this equation yields nearly the same same speed values m/s, 338 m/s, 346 m/s, and 355 m/s repsectively. 67. A wind chime is an open-end air column. Determine the fundamental frequencies of a 62.5-cm chime when the temperature is.... a degrees on a cold autumn evening. b degrees on a summer evening. c degrees during a hot summer day.

4 The frequency of the sound produced by a wind chime is related to the speed of air in the wind chime and the wavelength of the standing wave pattern of the resonating air column. The speed of the wave in air depends on the properties of air (temperature); these values were just computed in problem #66. The wavelength (lambda) of the resonating air column can be determined using a good diagram accompanied by the length of the air column. As depicted in the diagram at the right, the wavelength of the wave for the fundamental frequency is two times the length of the open-end air column. So in each of these cases, the wavelength is 2*62.5 cm = 125 cm = 1.25 m. With speed and wavelength known, the frequency values can be computed. a. f 1 = v / lambda = (338 m/s) / (1.25 m) = 271 Hz b. f 1 = v / lambda = (346 m/s) / (1.25 m) = 277 Hz c. f 1 = v / lambda = (354 m/s) / (1.25 m) = 284 Hz 68. An organ pipe has a length of 2.45 m and is open at both ends. Determine the fundamental frequency of the pipe if the temperature in the room is 25 degrees Celsius. Answer: 70.6 Hz For an open-end air column, the wavelength of the fundamental's standing wave pattern is two times the length of the air column; this relationship is depicted in the diagram at the right. So the wavelength of the wave is 4.90 m. The speed of the sound wave in air is dependent upon temperature. This speed was calculated in problem #66; it is 346 m/s. The frequency of the fundamental can now be calculated: f 1 = v / lambda f 1 = (346 m/s) / (4.90 m) = 70.6 Hz 69. (Referring to problem #68.) Determine the fundamental frequency of the pipe if it is closed at one end. Answer: 35.3 Hz The easy way to solve this problem is to recognize that by changing an open end to a closed end has the effect of lengthening the wave by a factor of two. If the wavelength is twice as large, then the frequency is twice as small. Thus, divide the original answer of 70.6 Hz (question #68) by 2. Alternatively, one could repeat the entire process of re-computing the wavelength from a diagram of the standing wave of the fundamental in a closed-end air column. For the fundamental, the wavelength is four times the length of the air column. In this case, the wavelength would be 9.80 m. Now divide the speed of the sound wave (346 m/s) by the wavelength value to obtain the frequency. 70. The auditory canal of the outer ear acts as a closed end resonator which has a natural frequency of around 3500 Hz. This canal serves to amplify sounds with frequencies around this value, thus making us more sensitive to such frequencies. If the speed of waves inside the canal is 350 m/s, then what is the estimated length of the canal? Answer: 2.5 cm For closed end resonators, the standing wave patterns of the natural frequencies are characterized by a node at the closed end and an antinode at the open end. For the first harmonic (and it must be assumed that the 3500 Hz corresponds to the first harmonic), the wavelength is four times the length of the air column. The strategy in this problem involves determining the wavelength (lambda) from the speed and the frequency and then determining the length from the wavelength-length relationship. lambda = v / f = (350 m/s) / (3500 Hz) = 0.1 m L = 0.25 * lambda = 0.25 * (0.1 m) = m = 2.5 cm 71. Determine the frequency of the lowest three harmonics at which a closed-end air column would sound out at 25 degrees Celsius if its length is 135 cm. Answer: 64.0 Hz, 192 Hz, 320 Hz The fundamental frequency of a closed-end air column is characterized by a standing wave pattern in which there is a node at the closed end and an antinode at the open end. There is one-fourth of a wavelength present within the length of the tube; thus, the wavelength is four times the length. In the case of this air column, the wavelength of the first harmonic is lambda 1 = 4*135 cm = 540 cm = 5.40 m The speed of sound waves in air at 25 degree Celsius was computed in problem #66 to be 346 m/s. The frequency of the first harmonic can now be calculated from the speed and the wavelength. f 1 = v / lambda 1 = (346 m/s) / (5.40 m) = 64.0 Hz The next two lowest frequencies are the third and the fifth harmonics. (Recall the closed-end air columns do not have even-numbered harmonics.) The frequencies of these harmonics are multiples of the first harmonic frequency. f 3 = 3 * f 1 = 3 * 64.0 Hz = 192 Hz f 5 = 5 * f 1 = 5 * 64.0 Hz = 320 Hz

5 72. Suppose that a sound is produced in a helium-filled air column rather than an air-filled air column. By what factor will this change in medium alter the frequency of the sound. (GIVEN: v air = 331 m/s; v He = 970 m/s) Answer: 2.93 The frequency of a resonating air column is dependent upon the speed of sound in that air column; the relationship is a direct relationship. If the air is replaced by another gas that transmits sound waves with a different speed, then the frequency will be increased by the same factor by which the speed is increased. In this problem, the speed of sound is increased by a factor of Thus, the frequency is increased by a factor of Factor = (970 m/s) / (331 m/s) = An organ pipe is used to produce the lowest note audible to the human ear Hz. If the temperature is 25 C, then how long is the organ pipe? (First decide whether it will produce this low note as a closed- or as an open-end air column.) Answer: 4.3 m Harmonics produced by closed-end air columns are characterized by longer wavelengths than the harmonics of open-end air columns of the same length. Thus, a closed-end air column produces lower frequencies. For this reason, we'll assume that the organ pipe in this problem is a closed-end air column. (Note, your answer will be two times larger if you assumed this to be an open-end air column.) At a temperature of 25 C, sound waves travel at 346 m/s (see problem #66). The frequency and speed can be used to determine the wavelength (lambda): lambda = v / f = (346 m/s) / (20 Hz) = 17.3 m The wavelength of the first harmonic of a closed-end air column is four times the length of the column; so the length of the column is one-fourth the wavelength. L = 0.25 * 17.3 m = 4.3 m 74. Determine the length of an open-end air column which would produce a 262 Hz frequency on a balmy day when the temperature is 12 degrees. Answer: 64.5 cm The speed of sound waves at 12 C is 338 m/s (see problem #66). The wavelength (lambda) of a 262 Hz sound at this temperature is lambda = v / f = (338 m/s) / (262 Hz) = 1.29 m For an open-end air column, the wavelength of the first harmonic is two times the length of the air column. So the length is one-half the wavelength: L = 0.5 * 1.29 m = m = 64.5 cm 75. A 440.-Hz tuning fork is held above the open end of a water-filled pop bottle and resonance is heard. The length of the pop bottle (bottom to top) is 28.2 cm. If the speed of sound is 345 m/s, then to what height is the pop bottle filled with water? Answer: 19.6 cm The wavelength (lambda) of a 440 Hz sound can be determine from the wave equation (v = f*lambda) lambda = v / f = (345 m/s) / (440. Hz) = m For a closed-end air column, the wavelength of the first harmonic is four times the length of the air column. So the length is one-fourth the wavelength: L = 0.25 * m = m = 19.6 cm If the pop bottle is 28.2 cm tall, then it should be filled with 8.6 cm of water in order for there to remain 19.6 cm of air.