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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