Speed aircraft carrier based on reflected pulse frequency

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SUMMARY

The discussion focuses on calculating the speed of an aircraft carrier based on the frequency shift of a sonar pulse emitted by a submarine traveling at 12 m/s. The sonar pulse has an initial frequency of 2400 Hz and returns at a frequency of 2310 Hz. Using the Doppler effect equations, the calculated speed of the aircraft carrier is determined to be 40.65 m/s, indicating it is moving away from the submarine. The speed of sound in water is given as 1500 m/s, which is crucial for the calculations.

PREREQUISITES
  • Doppler effect equations for sound waves
  • Understanding of relative motion in physics
  • Basic algebra for solving equations
  • Knowledge of wave frequency and speed relationships
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  • Learn about the applications of sonar technology in marine navigation
  • Explore advanced topics in wave mechanics and frequency analysis
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Physics students, marine engineers, and anyone interested in sonar technology and wave mechanics will benefit from this discussion.

euphtone06
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Homework Statement


Submarine is traveling at 12 m/s toward an aircraft carrier emits a 2400 Hz sonar pulse. The reflected pulse returns with a frequency 2310 Hz.
What is the speed of the aircraft carrier?
(Positive/negative means the carrier is moving toward/away from the submarine.) The speed of sound in water is 1500 m/s.


Homework Equations


-C-Vair/-C-Vsub*Freq1ofSub = Freq1ofAircraft
C - Vsub/C - Vair*Freq2ofair = Freq2ofSub


The Attempt at a Solution


Im not sure how to solve the problem using those equations :frown:
 
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First find the expression for frequency n1 of the pulse reaching the aircraft carrier with the source and the detector moving. Now consider the pulse of frequency n1 (reflected) is emitted by the aircraft carrier (moving source). Find expression of the frequency n2 received by moving detector (submarine). This frequency n2 is equal to 2310 Hz. solve the equations.
 
I tried working it out again but I am not sure if its right.

n ' = n [v - vo] / [v - vs]
c>>>carrier, b>>submarine
all v (sound), v0(observer), vs (source) are moving in + x direction

when submarine is source: carrier observer (reflecting)
n ' = 2400 [1500 - vc] / [1500 - vb]
n ' = [2400/1488] [1500 - vc] ----(1)


only difference being that v is now opposite to convention
v changed to (- v)

n'' = n ' [- v - vo] / [- v - vs]
n'' = n ' [- 1500 - 12] / [- 1500 - vc]
n'' = n ' [1512] / [1500 + vc]
2310 = n ' [1512] / [1500 + vc]
2310 [1500 + vc] = 1512 n' >>>use (1) put n'
2310*1488 [1500 + vc] = 1512*2400 [1500 - vc]
7161 [1500 + vc] = 7560 [1500 - vc]
7161 [1500 + vc] = 7560 [1500 - vc]
14721 vc = 1500 [7560 - 7161]
vc = 40.65 m/s moving away from submarine in the same direction.

dont think its right can someone help me?
 

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