A sonar system fixed in a submariene operatres at a frequency 40KHz. An enemey submarine moves towards the sonar with a speed of `360kmh^(-1)`. What is the frequency of sound reflected by the submarine? Speed of sound in water `1450ms^(-1)?`

Answer

`46 KHz`

Explanation

Apparent frequency
`f' = ((v + v_(0))/(v))f`
`= ((1450 + 100)/(1450)) xx 40 xx 10^(3)`
This frequency is reflected by the enemy submarine (now source) and is observed by `SONAR`(now observer)
`:.` Apparent frequency ,
`f'' = ((v)/(v - v_(S)))f' = ((1450)/(1450 - 100))4.276 xx 10^(4) = 45.9 xx 10^(3) Hz cong 46 KHz`