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518_Physics ProblemsTechnical Physics

# 518_Physics ProblemsTechnical Physics - 520 P17.68 Sound...

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520 Sound Waves P17.68 The total output sound energy is eE t =℘∆ , where is the power radiated. Thus, t eE eE IA eE r I eE d I = = = = 4 4 2 2 π π e j . But, β = F H G I K J 10 0 log I I . Therefore, I I = 0 10 10 β e j and t eE d I = 4 10 2 0 10 π β . P17.69 (a) If the source and the observer are moving away from each other, we have: θ θ S = ° 0 180 , and since cos180 1 °= − , we get Equation 17.12 with negative values for both v O and v S . (b) If v O = 0 m s then ′ = f v v v f S S cos θ Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection, cos θ S = 4 5 so ′ = f 343 343 0 800 25 0 500 m s m s m s Hz . . b g a f or ′ = f 531 Hz . Note that as the train approaches, passes, and departs from the intersection, θ S varies from 0° to 180° and the frequency heard by the observer varies from: = ° = = = ° = + = f v v v f f v v v f S S max min cos . cos . 0 343 343 25 0 500 539 180 343 343 25 0 500 466 m s m s m s Hz Hz m s m s m s Hz Hz a f a f P17.70 Let T represent the period of the source vibration, and E be the energy put into each wavefront.
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