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Speed of Sound Waves: Echoes in a Canyon

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot and hears it echoes. The second echo arrives 1.92s after the first and 1.47s before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. Take the speed of sound as 340 m/s. What is the distance between the cliffs? If he could hear a fourth echo, how long after the third echo does it arrive?


    2. Relevant equations
    I was thinking that I could just try multiplying 340m/s by 1.92s, but this isn't correct. I thought it has something to do with the fact that he isn't standing in the center of the two cliffs. Since this didn't work I am now stumped how to get started.


    3. The attempt at a solution
     
  2. jcsd
  3. Jun 24, 2010 #2

    Doc Al

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    Attack it step by step. Name all the variables: Distance to closer cliff = d1; further cliff = d2. Write expressions for the time it takes to hear all three echoes. Then apply the given info.
     
  4. Jun 24, 2010 #3
    OK I have been trying to come up with an expression for time, but I'm having trouble. Would it just be: t1= [tex]\frac{340}{d1}[/tex] t2= [tex]\frac{340}{d2}[/tex]?
     
  5. Jun 24, 2010 #4

    Doc Al

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    What's the total distance the sound has to travel?

    (FYI: Don't use 'sub' tags within Latex. Use d_1 and d_2 instead.)
     
  6. Jun 24, 2010 #5
    OK so it's d1+d2= (340)t1+(340)t2?
     
  7. Jun 24, 2010 #6

    Doc Al

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    No. Just consider the first echo, which bounces off the near cliff. What total distance does that sound travel? Use that distance to calculate the time for that echo's arrival.

    Then do the same for the second and third echoes.
     
  8. Jun 24, 2010 #7
    I think what I'm having trouble seeing is whether or not all of these echoes are bouncing off of the closer wall. I see that it should be 2d_1= 340t_1 but then would it be 2d_1= 340t_2?
     
  9. Jun 24, 2010 #8

    Doc Al

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    They don't all bounce off the closer wall. The first one does, but the second one bounces off the far wall.
    Right.
    No, the second echo has nothing to do with the near wall.

    Draw a diagram and trace the path of each echo.

    (What about that third echo?)
     
  10. Jun 24, 2010 #9
    OK so the first echo which has no time attached to it will be off the close wall and the second echo which is 1.92s is off the far wall. The third echo which is 1.47s should be off the close wall again.
     
  11. Jun 24, 2010 #10

    Doc Al

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    Correct for the first two echoes. Note that those times are really differences in times, not what we're calling t1, t2, t3.

    What's the complete path of that third echo? How far does that sound travel?
     
  12. Jun 24, 2010 #11
    Alright I think I've got it Doc. Would the second one be 2d2= 340(1.92+t1 ? I'm still thinking about the third one a little, but would it have bounced off both walls?
     
  13. Jun 24, 2010 #12

    Doc Al

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    Sounds good to me.
    Exactly.

    Tip: Don't think you have to combine these times in your head. I would first write expressions for all three times, and then worry about the time differences. For example:
    t1 = 2d1/v
    t2 = 2d2/v

    Then you can add that t2 - t1 = 1.92, and so on.
     
  14. Jun 24, 2010 #13
    Excellent!! Thanks so much for all your help again Doc. You're my hero!!
     
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