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.
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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
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
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