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

Explanation

v = 40 kHz = `40 xx 10^(3)` Hz
V = speed of sound in water = 1450 m/s
`V_(L)` = speed of enemy submarine = 360 km `h^(-1)`
`= (360 xx 1000)/(3600)`
= 100 m/s
As the source is at rest and observer is moving towards the source.
Therefore, apparent frequency received by the enemy submarine `= v' = (V - (-V_(L)))/(V - V_(S)) xx v`
`= (V + V_(L))/(V - V_(S)) xx v`
`= (1450 + 100)/(1450 - 0) xx 40`
= 42.76 kHz
The frequency reflected by enemy submarine can be found as :
Now, `V_(S) = 360` km/hr = 100 m/s
`V_(L) = 0`
Apparent frequency of reflected sound
`= v'' = (V - V_(L))/(V-V_(S)) xx v'`
`= (1450 - 0)/(1450 - 100) xx 42.76`
= 45.93 kHz