How Does the Doppler Effect Calculate Submarine Depth and Velocity?

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SUMMARY

The discussion focuses on calculating submarine depth and vertical velocity using sonar technology. A stationary destroyer emits sound pulses at 30.000 MHz, receiving reflected pulses at a frequency of 29.958 MHz with a time delay of 60 ms. The correct depth of the submarine is determined to be 93 meters, accounting for the round trip of the sound waves. The vertical velocity of the submarine is calculated using the equation v = λ * f, resulting in a velocity of 1,552.17 m/s.

PREREQUISITES
  • Understanding of sonar technology and its applications
  • Knowledge of the Doppler Effect and its implications in wave frequency
  • Familiarity with basic physics concepts such as velocity, wavelength, and frequency
  • Ability to perform calculations involving sound speed in seawater
NEXT STEPS
  • Study the principles of the Doppler Effect in depth
  • Learn about sonar signal processing techniques
  • Explore the physics of sound propagation in different mediums
  • Investigate advanced sonar applications in underwater navigation
USEFUL FOR

Marine engineers, physicists, and anyone involved in underwater acoustics or submarine technology will benefit from this discussion.

krtica
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A stationary destroyer is equipped with sonar that sends out pulses of sound at 30.000 MHz. Reflected pulses are received from a submarine directly below with a time delay of 60 ms at a frequency of 29.958 MHz.
(a) If the speed of sound in seawater is 1.55 km/s, find the depth of the submarine.(b) Find its vertical velocity. (Take upward to be the positive direction.)

Well, for (a) I used the definition of velocity and multiplied 1,550 m/s by the time delay, .06s. My answer is 93m, which is wrong.

For (b), I utilized the equation v=lamba*f, where lambda is the wavelength and f is the frequency. Lambda is equivalent to (v+-u_s)/f_s, where u_s is the velocity of source relative to the medium, v is the velocity of waves in motion, and f_s is the frequency of the source. So v=((v+-u_s)/f_s)*f, where I assigned v=1550 m/s, u_s= 0 m/s, f_s = 29.958E6 Hz, and f=30E6 Hz. My answer is 1,552.17 m/s.

Please help!
 
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krtica said:
A stationary destroyer is equipped with sonar that sends out pulses of sound at 30.000 MHz. Reflected pulses are received from a submarine directly below with a time delay of 60 ms at a frequency of 29.958 MHz.
(a) If the speed of sound in seawater is 1.55 km/s, find the depth of the submarine.

(b) Find its vertical velocity. (Take upward to be the positive direction.)

Well, for (a) I used the definition of velocity and multiplied 1,550 m/s by the time delay, .06s. My answer is 93m, which is wrong.
How is the distance traveled by the sound (down & back - hint, hint) related to the depth of the sub?
For (b), I utilized the equation v=lamba*f, where lambda is the wavelength and f is the frequency. Lambda is equivalent to (v+-u_s)/f_s, where u_s is the velocity of source relative to the medium, v is the velocity of waves in motion, and f_s is the frequency of the source. So v=((v+-u_s)/f_s)*f, where I assigned v=1550 m/s, u_s= 0 m/s, f_s = 29.958E6 Hz, and f=30E6 Hz. My answer is 1,552.17 m/s.

Please help!
For (b): The distance the sound waves travel changes at twice the rate at which the sub changes depth.
 
Thank you, I found out what I was doing wrong.
 

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