2.  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. Topic 11.1 Extended A – Sound intensity  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.

3.  We will define the intensity of sound I to be the average rate per unit area at which energy is transmitted by the wave. Topic 11.1 Extended A – Sound intensity  Since an energy rate is the same thing as power, we can write Intensity = Power Area  Since the area of a sphere is A sphere = 4  r 2, we can write I = P4r2P4r2 Intensity of a Spherical Wavefront 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 / m 2 ).

4. 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. Topic 11.1 Extended A – Sound intensity What happens to the intensity of sound if you double your distance from the source? Since I  1 / r 2 if distance doubles, intensity is reduced to 1/4 of the original.  In fact we can look at two arbitrary distances r 1 and r 2 and write I1 =I1 = P4r12P4r12 and I2 =I2 = P4r22P4r22 so that I1I2I1I2 = P4r12P4r12 P4r22P4r22 = P4r12P4r12 · 4r22P4r22P = r2r1r2r1 2  Thus I1I2I1I2 = r2r1r2r1 2 Inverse Square Relationship between Intensity and Distance

5. Topic 11.1 Extended A – Sound intensity  To solidify this concept consider the point source shown, and three expanding wavefronts of radii R, 2R and 3R as shown:  If we spread out our wavefronts to look at them separately, we can focus on 9 "squares" on the wavefront as it expands: 0 R 2R 3R3R FYI: The energy in the three pictured regions is the same. It is just spreading out over subsequently greater areas. 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:

6. Topic 11.1 Extended A – Sound intensity  The human ear is sensitive to a wide range of intensities, from about I 0 = 10 -12 W/m 2 (known as the threshold of hearing), to levels higher than I pain = 1 W/m 2 (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:  = 10 log II0II0 Sound Level (dB) I 0 = 10 -12 W/m 2  Why, you may ask, do we use the logarithm function?  What is log 10? It is 1.  What is log 100? It is 2.  What is log 10 12 ? It is 12.  log extracts the power of 10 from the number you are taking the log of.

7. Topic 11.1 Extended A – Sound intensity  The human ear is sensitive to a wide range of intensities, from about I 0 = 10 -12 W/m 2 (known as the threshold of hearing), to levels higher than I pain = 1 W/m 2 (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:  = 10 log II0II0 Sound Level (dB) I 0 = 10 -12 W/m 2  Suppose we take the log of the ratio of the threshold of pain to the threshold of hearing: log II0II0 = 1 10 -12 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. 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. FYI: The 10 is placed in front of the log for historic reasons.

8. Topic 11.1 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  (dB) I/I 0 010 0 = 1 1010 1 = 10 2010 2 = 100 3010 3 = 1000 4010 4 = 10000 5010 5 = 100000 : 12010 12 Table 2 Some Sound Levels (dB) 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

9. 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 1000 = 3 log 10000 = 4 log f =  3.3 log f = 3.3 10 log f = 10 3.3 f = 1995 Hz Here

10. 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.

11. Topic 11.1 Extended A – Sound intensity  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? FYI: The INTENSITIES add, not the decibel levels. First, find the intensity of the 90 dB jackhammer sound waves. Then add them, then convert back to dB:  = 10 log II0II0 90 = 10 log II0II0 9 = log II0II0 II0II0 = 10 9 I = 10 9 I 0 I total = I + I I total =10 9 I 0 + I total = 2  10 9 I 0 II0II0 = 2  10 9  = 10 log II0II0  = 10 log 2  10 9  = 93 dB FYI: In other words, the difference in sound levels between one and two jackhammers is almost indistinguishable by the human ear.