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.

Answer 1

the frequency of SONAR received by enemy submarine will be further reflected back to SONAR which it will receive again with a different frequency .

SONAR frequency ( Vs) = 40 kHz
= 40 × 10³ Hz
speed of enemy submarine ( Ve) = 360 km/h = 360 × 5/18 = 100 m/s
speed of sound in water = 1450 m/s
Apparent frequency recieved by the Submarine ( f') = { ( V + Vo)/V}f
= (1450 + 100)/1450 × 40
= 42.76 kHz

Now, the reflected wave have a different frequency .
f'' = { V/(V + Vs) } f'
= { 1450/(1450 - 100) × 42.76
= 45.93 kHz

Answer 2

operating frequency of the SONAR system =40 kHz

speed of the enemy submarine ve=360km/h=100 m/s

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

the source is at rest and the observer (enemy submarine )is moving toward it. hence the Apparent frequency(v^1) received and reflected by the submarine is given by the relation:

v^1=(v+ve/v)v

=(1450+100/1450)×40=42.76 kHz

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

v^n=(v/v+vs)v^1

where vs=100m/s

v^n=(1450/1450-100)×42.76=45.93 kHz.



hope its helps