1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sound problems

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data
    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. Relevant 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. The problem statement, all variables and given/known data
    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. Relevant 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. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 30, 2007 #2
    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

    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.
     
  4. Jan 30, 2007 #3
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Sound problems
  1. Sound problem (Replies: 1)

  2. Problems with Sound. (Replies: 11)

  3. A sound problem (Replies: 2)

  4. Sound problem (Replies: 1)

Loading...