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Calculating distance using the speed of sound through two different mediums

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data

    David is swimming when he hears a beaver sap the water with its tail, first through the water (v= 1400 meters/second) and then 0.95 seconds later (after the sound reaches him through the water) through the air (v= 340 meters/second).


    2. Relevant equations
    Find the distance from David to the beaver.


    3. The attempt at a solution
    This one stumped me. I thought maybe I 'd have to compare the two velocities and do some cross multiplication but I really don't know.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 1, 2013 #2

    gneill

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    Staff: Mentor

    What equations do you think apply here? If you happened to know the distance and velocity, how would you find the time? Write the expressions for the time of travel for both modes of transport (air, water) using a variable to represent the unknown distance.
     
  4. Jan 2, 2013 #3
    Well, I obviously know that d=t×v. This would mean that in the air the sound travels 323 m (d= 0.95×340). The problem is that isn't the total distance, there remains an additional distance that I can't figure out. If I sub in the time value into an equation with the sound velocity in water (d=0.95×1400) then the sound will have traveled 1330 m. But I am sure that neither of these two values are the actual distance. There remains another step but I am unsure what to do.
     
  5. Jan 2, 2013 #4
    The distance travelled in both mediums is a constant. 0.95 seconds is a relation between the time taken by the sound wave in both mediums to cover this distance.
     
    Last edited: Jan 2, 2013
  6. Jan 2, 2013 #5

    gneill

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    Staff: Mentor

    No, the 0.95 seconds is the difference in time between the arrival of the sound by the two paths.

    There is only one unknown distance, the distance between the beaver and the person. Call that distance d. Write the expressions for the two arrival times of the sound assuming distance is d; so t1 = ?, t2 = ?.
     
  7. Jan 2, 2013 #6
    OK, so t1 (water)= d/1400
    t2 (air)= d/340
     
  8. Jan 2, 2013 #7

    gneill

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    Staff: Mentor

    Okay. Now, what is the relationship between t1 and t2? (This is where the .95 seconds comes into play).
     
  9. Jan 3, 2013 #8
    Then, t2=t1+0.95
    so: d/1400=d/340+0.95
     
  10. Jan 3, 2013 #9
    [STRIKE]Yes. Now write another equation using speed=distance/time.[/STRIKE]
     
    Last edited: Jan 3, 2013
  11. Jan 3, 2013 #10
    no, sorry, I mean: d/340=d/1400+0.95
     
  12. Jan 3, 2013 #11
    You can edit your posts. :rofl:
     
  13. Jan 3, 2013 #12
    You mean like this: 340=d/(d/1400)+0.95
     
  14. Jan 3, 2013 #13
    Using your equation d/340=d/1400+0.95, can you not solve for d?
     
  15. Jan 4, 2013 #14
    So would the answer be 690m?
     
  16. Jan 4, 2013 #15
  17. Jan 4, 2013 #16
    OK, my bad. I am just unsure at how to rearrange d/340=d/1400+0.95 to isolate d. Any ideas?
     
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