How Is the Distance to the Seabed Calculated Using Sonar?

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Homework Help Overview

The discussion revolves around calculating the distance to the seabed using sonar technology, specifically focusing on the time it takes for a sound pulse to travel to the seabed and back. The subject area includes concepts from physics related to sound propagation in water and basic kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the equation d = vt, questioning whether the total distance calculated represents the actual depth to the seabed. There is an exploration of the difference between the total distance traveled by the sound pulse and the depth measurement.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the relationship between the total distance traveled by the sonar pulse and the actual depth to the seabed. Some guidance has been provided regarding the need to consider the time for the sound to reach the seabed only, rather than the round trip.

Contextual Notes

Participants are working under the assumption that the seabed is flat and that the ship is stationary. There is a focus on understanding the implications of the time measurement provided in the problem statement.

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Homework Statement


A depth sounder on a ship measures the distance between the bottom of the ship and the seabed using a high-frequency pulse of sound (sonar stands for SOund Navigation And Ranging). Sound travels through water at 1.48 km/s.

Calculate the distance to the seabed if the reflected pulse is received 0.461 s after the sound is transmitted. Assume that the ship is not moving and that the transmitter and receiver are located at the same position on the boat. Also assume that the seabed is flat, so that the first reflection is received from the surface directly beneath the boat.

Homework Equations


d = vt

v = 1480 m/s
t = 0.461 s

The Attempt at a Solution


d = v t
= 1480 x 0.461
= 682.28 m

Is this the correct equation to use for this question, if not what equation should I use?
 
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Jess_18033152 said:
d = v t
= 1480 x 0.461
= 682.28 m
This is the total distance traveled by the pulse. Is this the same thing as the depth?
 
Orodruin said:
This is the total distance traveled by the pulse. Is this the same thing as the depth?

The answer was using: 1480 x (1/2 x 0.461) = 341 m
 
That does not really answer the question if the total distance traveled is the same thing as the depth. The real question is not if you know what the answer is. The real question is whether or not you understand why the answer is what it is. What is the difference between your attempt and the answer? Why is it different?
 
Orodruin said:
That does not really answer the question if the total distance traveled is the same thing as the depth. The real question is not if you know what the answer is. The real question is whether or not you understand why the answer is what it is. What is the difference between your attempt and the answer? Why is it different?

Yes, it's half the time because we are only want the time to get to the sea bed, whereas we were given the time to reach the seabed and come back to the boat.
 

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