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Unformatted text preview: Version PREVIEW – Homework 2 – Savrasov – (39825) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Displacement on a Wave 001 10.0 points Calculate the displacement at position 17 . 5 m and time 1 . 48 s for a sound wave in air described by cos( x, t, φ ) , where φ = 0 when x = 0 and t = 0 . The wavelength is 0 . 0639 m, the speed of sound is 343 m / s, the density of the air is 1 . 35 kg / m 3 , and the pressure ampli tude is 0 . 262 Pa. Correct answer: − 7 . 84692 × 10 − 9 m. Explanation: Let : x = 17 . 5 m , t = 1 . 48 s , λ = 0 . 0639 m , v = 343 m / s , ρ = 1 . 35 kg / m 3 , and Δ P max = 0 . 262 Pa . The wave number is k = 2 π λ , its angular frequency is ω = 2 π f = 2 π v λ , the maximum amplitude is s m = Δ P max ρ v ω = Δ P max λ ρ v · 2 π v = (0 . 262 Pa) (0 . 0639 m) (1 . 35 kg / m 3 ) (343 m / s) · 2 π (343 m / s) = 1 . 67765 × 10 − 8 m , and the angle is θ = k x − ω t = 2 π x λ − 2 π v t λ = 2 π λ ( x − v t ) = 2 π . 0639 m (17 . 5 m − 343 m / s 1 . 48 s) = − 48194 . 7 rad , so the displacment is s = s m sin( k x − ω t ) = (1 . 67765 × 10 − 8 m) × sin( − 48194 . 7 rad) = − 7 . 84692 × 10 − 9 m . Explosion in the Atmosphere 002 10.0 points Sound waves in air are absorbed at a rate of approximately 9 . 52 dB / km. An explosive charge is detonated at a height of several kilo meters in the atmosphere. At a distance of 564 m from the explosion the acoustic pres sure reaches a maximum of 7 . 16 N / m 2 . What will be the sound level 6 . 28 km from the explosion? Consider the atmosphere to be homogeneous over the distances considered. The density of the air is 1 . 24 kg / m 3 and the speed of sound is 343 m / s....
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.
 Fall '09
 RichardScalettar
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