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 rI eE dI = == = 4 4 2 2 π ej . But, β = F H G I K J 10 0 log I I . Therefore, II = 0 10 10 and t eE = 41 0 2 0 10 . P17.69 (a) If the source and the observer are moving away from each other, we have: θθ 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 vv f SS cos θ Also, when the train is 40.0 m from the intersection, and the car is 30.0 m from the intersection, cos S = 4 5 so ′ = f 343 343 0 800 25 0 500 ms Hz .. bg af or f 531 Hz . Note that as the train approaches, passes, and departs from the intersection, S varies from 0° to 180° and the frequency heard by the observer varies from: −° = = = + = f v f f v f S S max min cos . cos . 0 343 343 25 0 500 539 180 343 343 25 0 500 466 Hz Hz Hz Hz a f P17.70 Let T represent the period of the source vibration, and E be the energy put into each wavefront. Then ℘= av E T . When the observer is at distance
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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