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Submarine echo physics problems

  • Thread starter akd28
  • Start date
5
0
1. Homework Statement
A submarine on the ocean surface gets a sonar echo indicating an underwater object. The echo comes back at an angle of 20* above the horizontal and the echo took 2.32s to get back to the submarine. What is the object's depth?

2. Homework Equations
delta of theta = (2pi/lambda) * (delta of L)

I = (P)/(4piR^2) and (I2/I1) = (R1/R2)^2

3. The Attempt at a Solution
I really have no idea where to begin this problem. I can't really visualize it...if the submarine is on the ocean's surface and emits a sonar signal, shouldn't the echo be coming up from under the submarine vertically? Why is it 20* above the horizontal? I tried drawing a diagram but I don't know what it should look like with the information given. Also, I have no idea what equation would be best to use either. (I have more equations, I just thought that these would be the most helpful for the problem.)



1. Homework Statement
A person has a hearing loss of 30dB for a particular frequency. What is the sound intensity that is heard at this frequency that has an intensity of the threshold of pain?

2. Homework Equations
same I equations as above and B = 10log(I/Io)

Ip = 1.0W/m^2

3. The Attempt at a Solution
Here is what I did:

log (I / 1.0W/m^2) = 3B
I = 10^3 W/m^2

Am I totally off on what the question is asking me to do?

THANKS!
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

72
0
Ok, for the first one...

I'm not quite sure if what I did was correct, but I'll give you my two cents.

http://img184.imageshack.us/img184/51/subto4.gif [Broken]

So from that drawing, you can see kind of how I pictured it. If that's the case, you know v = 343 m/s (speed of sound in air...you'll probably need to speed of sound in water for this problem which I don't know). You also know t = 2.32 s, and that the angle is 20 degrees.

I figured that the distance down was d, and the hypotenuse was d/sin(20) by trig. Adding these two numbers, you find the total distance traveled by the sound to be d + d/sin(20), or what I thought was easier, ( d (sin(20) + 1 ) )/sin(20). Since you can assume that the acceleration is zero for sound, you can simply use d = vt.

Your equation should look something similar to:

d = (v * t * sin(20)) / (sin(20) + 1)

I got 202.8 m using the speed of sound in air. Using the correct value for the speed of sound in water should change your answer....But again, I have no idea if I thought of the question correctly either.
 
Last edited by a moderator:
5
0
Your figure makes sense but the problem states that the echo comes back at an "angle 20* ABOVE the horizontal". That's the part that confuses me.
 

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