#### Question

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 360 km h^{–1}. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^{–1}.

#### Solution 1

Her frequency of Sonar (source) = `40.0 kHz = 40xx10^3 "Hz"`

Speed of sound waves, `v= 1450 ms^(-1)`

Speed of observers,` v_0 = 360 "km/h" = 360 xx 5/18 = 100 ms^(-1)`

Since the source is at rest and obsever moves toward the source (SONAR)

`:. v' = (v+v_0)/v.v = (1450+100)/1450 xx 40xx 10^3` = `4.276 xx 10^(4) Hz`

This frequency (v') is reflected by the enemy ship and is observed by the SONAR (which now act as observer). Therefore, in this case `v_s = 360` km/h = `100 ms^(-1)`

:. Apparent frequency, `v" = v/(v - v_s) v' = 1450/(1450 - 100) xx 4.276 xx 10^4`

`= 4.59 xx 10^4 Hz = 45.9 kHz`

#### Solution 2

Operating frequency of the SONAR system, ν = 40 kHz

Speed of the enemy submarine, *v*_{e} = 360 km/h = 100 m/s

Speed of sound in water, *v *= 1450 m/s

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency (V') received and reflected by the submarine is given by the relation:

`v' = ((v+v_e)/v) v`

= `((1450+100)/1450) xx 40 = 42.76 kHz`

The frequency (v") received by the enemy submarine is given by the relation:

`v" = (v/(v+v_s))v'`

where `v_s = 100 "m/s"`

`:. v" = (1450/(1450 - 100))xx 42.76 = 45.93` kHz