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 $$1450 m s^{–1}$$ .

Asked by Pragya Singh | 1 year ago |  229

#### 1 Answer

##### Solution :-

Frequency of the SONAR system, f=40 kHz = 40 x 103 Hz

Speed of sound in water, v =1450m/s
Speed of the enemy submarine, v0=360km/h

= 360 x ($$\dfrac{ 5}{18}$$) = 100m/s

The SONAR is at rest and the energy submarine moves towards it. Therefore, the apparent frequency is given by the relation
f’ =$$\dfrac{(v+v_0)}{vf}$$

=  [$$\dfrac{ (1450 + 100)}{1450}$$] x 40 x 103

= 42.75 x 103
This frequency (f’) is reflected by the enemy submarine and it is observed by the SONAR.

Therefore, vs = 360 km/s  = 100m/s

f” = $$\dfrac{v}{v-v_s}$$x f’

= $$\dfrac{ (1450 - 100)}{1450}$$ x 42.75 x 103

= ($$\dfrac{ 1450}{1350}$$) x 42.75 x 103

= 45.91 x 103

Answered by Abhisek | 1 year ago

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