520 Sound WavesP17.68The total output sound energy is eEt=℘∆, where ℘is the power radiated.Thus, ∆teEeEIAeErIeEdI=℘===4422πej.But, β=FHGIKJ100logII. Therefore, II=01010and ∆teE=4102010.P17.69(a)If the source and the observer are moving away from each other, we have: θθS−= °0180 ,and since cos1801°=−, we get Equation 17.12 with negative values for both vOand vS.(b)If vO=0 m s then ′=−fvvvfSScosθAlso, when the train is 40.0 m from the intersection, and the car is 30.0 m from theintersection,cosS=45so ′ =−f3433430 800 25 0500msHz..bgafor f531 Hz .Note that as the train approaches, passes, and departs from the intersection, Svaries from0° to 180° and the frequency heard by the observer varies from:−°=−==+=fvffvfSSmaxmincos.cos.034334325 050053918034334325 0500466HzHzHzHzafP17.70Let Trepresent the period of the source vibration, and Ebe the energy put into each wavefront.Then ℘=avET. 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 .