Topic 4.4 Extended A – Sound intensity Consider a sound point source and the sound pulse emitted by it: The disturbance leaves the source at the speed of sound, and in the form of a spherical wavefront. Whatever energy is contained in the first wavefront , is contained in the subsequent wavefronts. FYI: It should be clear, then, that the farther you are from the source, the more spread out the energy of the wavefront. © 2006 By Timothy K. Lund

Topic 4.4 Extended A – Sound intensity We will define the intensity of sound I to be the average rate per unit area at which energy is transmitted by the wave. Since an energy rate is the same thing as power, we can write Power Area Intensity = , we can write Since the area of a sphere is Asphere = 4r2 Intensity of a Spherical Wavefront P 4r2 © 2006 By Timothy K. Lund I = r = distance from source FYI: The intensity of a sound wave is thus proportional to the power, and inversely proportional to the square of the distance. FYI: The units of intensity are watts per square meter (W / m2).

Topic 4.4 Extended A – Sound intensity Suppose the power output of a sound source doubles. What happens to the intensity of the sound? Since I  P if power doubles, intensity doubles. What happens to the intensity of sound if you double your distance from the source? Since I  1 / r2 if distance doubles, intensity is reduced to 1/4 of the original. In fact we can look at two arbitrary distances r1 and r2 and write © 2006 By Timothy K. Lund P 4r12 P 4r22 I1 = and I2 = so that P 4r12 I1 I2 P 4r12 4r22 P r2 r1 2 = = · = P 4r22 Thus Inverse Square Relationship between Intensity and Distance I1 I2 r2 r1 2 =

Topic 4.4 Extended A – Sound intensity FYI: The energy in the three pictured regions is the same. It is just spreading out over subsequently greater areas. Topic 4.4 Extended A – Sound intensity FYI: At 2R, the energy is spread out over FOUR times the area at R: FYI: At 3R, the energy is spread out over NINE times the area at R: To solidify this concept consider the point source shown, and three expanding wavefronts of radii R, 2R and 3R as shown: R 3R 2R © 2006 By Timothy K. Lund If we spread out our wavefronts to look at them separately, we can focus on 9 "squares" on the wavefront as it expands:

Topic 4.4 Extended A – Sound intensity The human ear is sensitive to a wide range of intensities, from about I0 = 10-12 W/m2 (known as the threshold of hearing), to levels higher than Ipain = 1 W/m2 (known as the threshold of pain). Since sound intensity has such a wide range of values we will define a sound level , measured in the decibel (dB), using logs: Sound Level (dB) I I0  = 10 log I0 = 10-12 W/m2 Why, you may ask, do we use the logarithm function? © 2006 By Timothy K. Lund What is log 10? It is 1. What is log 100? It is 2. What is log 1012? It is 12. log extracts the power of 10 from the number you are taking the log of.

Topic 4.4 Extended A – Sound intensity FYI: It turns out that the ear can distinguish intensities differing by factors of ten, so the formula below is more useful than looking at intensities. Topic 4.4 Extended A – Sound intensity FYI: The 10 is placed in front of the log for historic reasons. The human ear is sensitive to a wide range of intensities, from about I0 = 10-12 W/m2 (known as the threshold of hearing), to levels higher than Ipain = 1 W/m2 (known as the threshold of pain). Since sound intensity has such a wide range of values we will define a sound level , measured in the decibel (dB), using logs: Sound Level (dB) I I0  = 10 log I0 = 10-12 W/m2 Suppose we take the log of the ratio of the threshold of pain to the threshold of hearing: © 2006 By Timothy K. Lund I I0 1 10-12 log = log = 12 which tells us that the intensity at the threshold of pain is "12 factors of ten" more than the intensity at the threshold of hearing.

Topic 4.4 Extended A – Sound intensity The following tables illustrate use of the sound level formula, and gives some common decibel levels: Table 1 Values from the Sound Level equation Table 2 Some Sound Levels (dB) (dB) I/I0 0 100= 1 10 101= 10 20 102= 100 30 103= 1000 40 104= 10000 50 105= 100000 : 120 1012 Threshold of hearing 0 Rustle of leaves 10 Whisper (1 m) 20 City street no traffic 30 Office or classroom 50 Normal conversation (1 m) 60 Jackhammer (1 m) 90 Rock group 110 Threshold of pain 120 Jet engine (50 m) 130 Saturn V rocket (50 m) 200 © 2006 By Timothy K. Lund

FYI: At what frequency does the ear appear to be most sensitive? FYI: This graph shows the relation between sound level and frequency. Since we hear only within the audible range, it should not come as a surprise that the thresholds are also related to the frequencies of the sound. FYI: At what frequency does the ear appear to be most sensitive? log f = 3.3 10 log f = 103.3 f = 1995 Hz © 2006 By Timothy K. Lund log f = 3.3 log 1000 = 3 log 10000 = 4 Here

FYI: If the sound level is high enough, and the frequency is correct, resonance can be fed in a glass to the point that the standing waves in the glass exceed the maximum amplitude that the glass can carry. © 2006 By Timothy K. Lund

Topic 4.4 Extended A – Sound intensity FYI: The INTENSITIES add, not the decibel levels. FYI: In other words, the difference in sound levels between one and two jackhammers is almost indistinguishable by the human ear. The following problem illustrates a subtle difference between intensity and sound level. Suppose you are watching road construction and you happen to be located 50 m from each of two jackhammers. If each jackhammer has a sound level of 90 dB, what is the sound level of the two together? First, find the intensity of the 90 dB jackhammer sound waves. Then add them, then convert back to dB: I I0 Itotal = I + I  = 10 log Itotal = 109I0 + 109I0 I I0 © 2006 By Timothy K. Lund 90 = 10 log Itotal = 2109I0 I I0 = 2109 I I0 9 = log I I0 I I0  = 10 log = 109  = 10 log 2109 I = 109I0  = 93 dB