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496_PartUniversity Physics Solution

496_PartUniversity Physics Solution - 16-4 Chapter 16(c...

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16-4 Chapter 16 (c) 2 5 2 area = (5.00 mm) 2.5 10 m . = × Part (a): 12 2 5 2 17 (1 10 W/m )(2.5 10 m ) 2.5 10 J/s. × × = × Part (b): 3 2 5 2 8 (3.2 10 W/m )(2.5 10 m ) 8.0 10 J/s. × × = × E VALUATE : For faint sounds the displacement and pressure variation amplitudes are very small. Intensities for audible sounds vary over a very wide range. 16.15. I DENTIFY : Apply Eq.(16.12) and solve for A . / , v f λ = with / . v B ρ = S ET U P : 2 . f ω π = For air, 5 1.42 10 Pa. B = × E XECUTE : (a) The amplitude is 6 2 11 2 3 9 2 2 2(3.00 10 W m ) 9.44 10 m. (1000 kg m )(2.18 10 Pa)(2 (3400 Hz)) Ι A ρΒω π × = = = × × The wavelength is 9 3 (2.18 10 Pa) (1000 kg m ) 0.434 m. 3400 Hz B ρ v f f λ × = = = = (b) Repeating the above with 5 1.42 10 Pa B = × and the density of air gives 9 5.66 10 m and 0.100 m. A λ = × = E VALUATE : (c) The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense air molecules must have a larger amplitude to transfer the same amount of energy. 16.16. I DENTIFY and S ET U P : Use Eq.(16.7) to eliminate either v or B in 2 max . 2 vp I B = E XECUTE : From Eq. (19.21), 2 . v B ρ = Using Eq.(16.7) to eliminate ( ) 2 2 max max , 2 2 . v I B ρ p B p ρ B = = Using Eq. (16.7) to eliminate B , 2 2 2 max max 2( ) 2 . I vp v ρ p ρ v = = E VALUATE : The equation in this form shows the dependence of I on the density of the material in which the wave propagates. 16.17. I DENTIFY and S ET U P : Apply Eqs.(16.5), (16.11) and (16.15). E XECUTE : (a) 2 (2 rad)(150 Hz) 942.5 rad/s f ω π π = = = 2 2 942.5 rad/s 2.74 rad/m 344 m/s f k v v π π ω λ = = = = = 5 1.42 10 Pa B = × (Example 16.1) Then 5 6 max (1.42 10 Pa)(2.74 rad/m)(5.00 10 m) 1.95 Pa. p BkA = = × × = (b) Eq.(16.11): 2 1 2 I BkA ω = 5 6 2 3 2 1 2 (942.5 rad/s)(1.42 10 Pa)(27.4 rad/m)(5.00 10 m) 4.58 10 W/m . I = × × = × (c) Eq.(16.15): 0 (10 dB)log( / ), I I β = with 12 2 0 1 10 W/m . I = × ( ) 3 2 12 2 (10 dB)log (4.58 10 W/m )/(1 10 W/m ) 96.6 dB. β = × × = E VALUATE : Even though the displacement amplitude is very small, this is a very intense sound. Compare the sound intensity level to the values in Table 16.2. 16.18. I DENTIFY : Apply 0 (10 dB)log( / ). I I β = In part (b), use Eq.(16.14) to calculate I from the information that is given. S ET U P : 12 2 0 10 W/m . I = From Table 16.1 the speed of sound in air at 20.0 C ° is 344 m/s. The density of air at that temperature is 3 1.20kg m . E XECUTE : (a) 2 12 2 0.500 W/m (10 dB) log 57 dB. 10 W/m β μ = = (b) 2 2 2 5 2 max 3 (0.150 N m ) 2.73 10 W m . 2 2(1.20 kg m )(344 m s) p I ρ v = = = × Using this in Equation (16.15), 5 2 12 2 2.73 10 W m (10 dB) log 74.4 dB. 10 W m β × = = E VALUATE : As expected, the sound intensity is larger for the jack hammer. 16.19. I DENTIFY : Use Eq.(16.13) to relate I and max . p 0 (10 dB)log( / ). I I β = Eq.(16.4) says the pressure amplitude and displacement amplitude are related by max 2 . f p BkA B A v π = = S ET U P : At 20 C ° the bulk modulus for air is 5 1.42 10 Pa × and 344 m/s. v = 12 2 0 1 10 W/m . I = ×
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Sound and Hearing 16-5 E XECUTE : (a) 2 5 2 12 2 max 5 (344 m/s)(6.0 10 Pa) 4.4 10 W/m 2 2(1.42 10 Pa) vp I B × = = = × × (b) 12 2 12 2 4.4 10 W/m (10 dB)log 6.4 dB 1 10 W/m β × = = × (c) 5 11 max 5 (344 m/s)(6.0 10 Pa) 5.8 10 m 2 2 (400 Hz)(1.42 10 Pa) vp A fB π π × = = = × × E VALUATE : This is a very faint sound and the displacement and pressure amplitudes are very small. Note that the displacement amplitude depends on the frequency but the pressure amplitude does not.
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