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Doppler Effect of a destroyer

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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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  • #2
SammyS
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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.
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.
 
  • #3
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Thank you, I found out what I was doing wrong.
 

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