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A source and an observer are approaching one another with the relative velocity 40 ms^(-1). If the true source frequency is 1200 Hz, deduce the observed frequency under the following conditions: (i) All velocity is in the source alone. (ii) All velocity is in the observer alone. The source moves in air at 100 ms^(-1) towards the observer, but the observer also moves with the velocity v_(0) in the same direction. |
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Answer» SOLUTION :Here V=1200Hz, `v=340ms^(-1)` RELATIVE velocity `=40ms^(-1)` (i) Here source move TOWARDS the stationary observer, `v_(s)=+40ms^(-1),v_(0)=0` ![]() `"Train"overset(S""+ve""v_(s))tooverset(*)"Observer "v_(0)=0` `v'=((v-v_(0)))/((v-v_(0)))xxv=(340-0)/(340-40)xx1200` `=(340xx1200)/(300)=1360Hz`. (ii)Here observer moves towards the stationary source, `v_(0)=-40ms^(-1),v_(s)=0` `S*v_(s)=0""overset(-ve""v_(0)""O)to` `v'=(v-v_(0))/(v-v_(s))xxv=(340+40)/(340-0)xx1200` `=(380)/(340)xx1200=1341Hz`. (iii)Here observer and source move in the same direction, `v_(s)=100ms^(-1),v_(0)=100-40=60ms^(-1)` `overset(S""+ve""v_(s))to""overset(S""+ve""v_(s))to` `v'=(v-v_(0))/(v-v_(s))xxv=(340-60)/(340-100)xx1200` `=(280)/(240)xx1200=1400Hz`.
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