How Does the Doppler Effect Calculate Submarine Depth and Velocity?

AI Thread Summary
The discussion focuses on calculating the depth and vertical velocity of a submarine using sonar data from a stationary destroyer. For depth, the initial calculation of 93 meters was incorrect due to not accounting for the round trip of the sound waves, which travel down and back. The correct depth requires halving the product of the speed of sound in seawater and the time delay. For vertical velocity, the participant used the Doppler effect equations, leading to a calculated velocity of 1,552.17 m/s, which was later clarified as needing adjustments based on the correct depth calculation. The conversation emphasizes the importance of understanding sound wave travel in relation to submarine depth.
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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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