Speed aircraft carrier based on reflected pulse frequency

  • Thread starter euphtone06
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  • #1
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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:
 

Answers and Replies

  • #2
mukundpa
Homework Helper
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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.
 
  • #3
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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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