496_PartUniversity Physics Solution

496_PartUniversity - 16-4 Chapter 16(c area =(5.00 mm)2 = 2.5 105 m 2 Part(a(1 1012 W/m 2(2.5 105 m 2 = 2.5 1017 J/s 16.15 Part(b(3.2 103 W/m 2(2.5

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16-4 Chapter 16 (c) 25 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): 32 5 2 8 (3.2 10 W/m )(2.5 10 m ) 8.0 10 J/s. = × EVALUATE: For faint sounds the displacement and pressure variation amplitudes are very small. Intensities for audible sounds vary over a very wide range. 16.15. IDENTIFY: Apply Eq.(16.12) and solve for A . / , vf λ = with /. vB ρ = SET UP: 2. f ω π = For air, 5 1.42 10 Pa. B EXECUTE: (a) The amplitude is 62 11 2 39 2 22 ( 3 . 0 0 1 0 W m ) 9.44 10 m. (1000 kg m )(2.18 10 Pa)(2 (3400 Hz)) Ι A ρΒω π × == = × × The wavelength is 93 (2.18 10 Pa) (1000 kg m ) 0.434 m. 3400 Hz B ρ v ff × = = (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 = EVALUATE: (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. IDENTIFY and SET UP: Use Eq.(16.7) to eliminate either v or B in 2 max . 2 vp I B = EXECUTE: From Eq. (19.21), 2 . = Using Eq.(16.7) to eliminate ( ) max max ,2 2 . vI B ρ pB p ρ B Using Eq. (16.7) to eliminate B , 222 max max 2( ) 2 . Iv p v ρ p ρ v EVALUATE: The equation in this form shows the dependence of I on the density of the material in which the wave propagates. 16.17. IDENTIFY and SET UP: Apply Eqs.(16.5), (16.11) and (16.15). EXECUTE: (a) 2 (2 rad)(150 Hz) 942.5 rad/s f ππ = 2 2 942.5 rad/s 2.74 rad/m 344 m/s f k vv ππ ω = = = 5 1.42 10 Pa B (Example 16.1) Then 56 max (1.42 10 Pa)(2.74 rad/m)(5.00 10 m) 1.95 Pa. k A × × = (b) Eq.(16.11): 2 1 2 IB k A = 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( / ), II β = with 12 2 0 11 0 W /m. I ( ) 1 (10 dB)log (4.58 10 W/m )/(1 10 W/m ) 96.6 dB. × = EVALUATE: 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. IDENTIFY: Apply 0 (10 dB)log( / ). = In part (b), use Eq.(16.14) to calculate I from the information that is given. SET UP: 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 . EXECUTE: (a) 2 12 2 0.500 W/m (10 dB) log 57 dB. 10 W/m β μ ⎛⎞ ⎜⎟ ⎝⎠ (b) 2 52 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), 12 2 2.73 10 W m (10 dB) log 74.4 dB. 10 W m β × EVALUATE: As expected, the sound intensity is larger for the jack hammer. 16.19. IDENTIFY: Use Eq.(16.13) to relate I and max . p 0 (10 dB)log( / ). = Eq.(16.4) says the pressure amplitude and displacement amplitude are related by max 2 . f p BkA B A v SET UP: At 20 C ° the bulk modulus for air is 5 × and 344 m/s. v = 12 2 0 I

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Sound and Hearing 16-5 EXECUTE: (a) 25 2 12 2 max 5 (344 m/s)(6.0 10 Pa) 4.4 10 W/m 22 ( 1 . 4 2 1 0 P a ) vp I B × == = × × (b) 12 2 12 2 4.4 10 W/m (10 dB)log 6.4 dB 11 0 W /m β ⎛⎞ × ⎜⎟ × ⎝⎠ (c) 5 11 max 5 5.8 10 m 2 2 (400 Hz)(1.42 10 Pa) vp A fB ππ × = × × EVALUATE: 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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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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496_PartUniversity - 16-4 Chapter 16(c area =(5.00 mm)2 = 2.5 105 m 2 Part(a(1 1012 W/m 2(2.5 105 m 2 = 2.5 1017 J/s 16.15 Part(b(3.2 103 W/m 2(2.5

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