A submarine sonar system sends a burst of sound with a frequency of

AI Thread Summary
A submarine sonar system emits a sound burst at 325Hz, which bounces off an underwater rock face and returns in 8.50 seconds. The wavelength of the sound is 4.71m, allowing the calculation of sound velocity as 1530.75m/s. The total distance traveled by the sound is calculated using the formula d=vt, resulting in 13011.375m for the round trip. Dividing by two gives the distance to the rock face as 6.51km. The discussion highlights the importance of accounting for the round trip of sound in sonar calculations.
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A submarine sonar system sends a burst of sound with a frequency of 325Hz. The sound wave bounces off an underwater rock face and returns to the submarine in 8.50s. If the wavelength of the sound is 4.71m, how far away is the rock face? (Ans: 6.51km )

--
v=fλ

λ=wavelength
f=frequency
v=velocity
--
v=fλ
v=325Hz X 4.71m
=1530.75m/s

v=d/t
d=vt
= 1530.75m/s X 8.50s
=13011.375m
=13011.375m/2
=6505.6875m

=6.51km.

What exactly am I doing wrong here? ..
 
Last edited:
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k0k said:
v=dt
d=v/t
This is not correct. What's the definition of speed? (When in doubt, check the units.)

Also: Don't forget that the sound makes a round trip.
 


Oh, my bad, a silly error. I got it now. Thanks for the tip on sound making a round trip. : )
 

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