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A sonar system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the sonar with a speed of 360kmh^(-1). What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450ms^(-1)

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Solution :
`v = 360 xx ( 5)/( 18) = 100ms^(-1)` `v =1450 ms^(-1)` `f = 40.0 kHz` <br> Apparent frequency `f' = f ((v+v_(s))/(v-v_(s)))` <br> `(v-v_(s))` is related velocity of sound w.r.t. the source <br> `( v + vs )` is relative velocity of sound w.r.t. the observer. <br> i.e. `f' = 40 xx 10^(3) xx((1450 + 100)/(1450-100))` <br> i.e. `f' = ( 40 xx 10^(3) xx 1550)/( 1350) Hz = 45.92 xx 10^(3) Hz` <br> `f' = 45.92 kHz` <br> i.e. `f' = 45.9 kHz`

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