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Speed of sound problem

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data
    The human beings are standing at equal distances from a big and high wall. Distance between them is 150 m. When one fires the gun the other hears two shots in an interval of 2 seconds, using the speed of sound to be 340.298 m/s calculate their distance to the wall.

    2. Relevant equations

    3. The attempt at a solution
    I dont know what two shots mean, i thought it means the sound is reflected from the wall, so i calculated time to reach girl 2 to girl 1 to be 0.44s subtractes from 2s and got 1.55s. Taking the time from guy1 to wall and wall to guy2 to be 1.55/2 i calculated distance to be 260m but thats not even close. What is wrong?
  2. jcsd
  3. Aug 3, 2015 #2
    The second distance from the guy to the wall is calculated via pythagoras:

    [tex] 340.298=\frac{ \sqrt{ 150^2+x^2} }{(2-0.44)} [/tex]
  4. Aug 3, 2015 #3
    Draw a diagram of the wall and two people and really think about how the times and distances relate given the sound echoing off the wall.
  5. Aug 3, 2015 #4
    I got closer to the answer by writing that V*1.56=2x-V*0.44 but i get 340m which is less than true answer. My logic is this. It takes 0.44 s to get to guy 2 and by that time sound has traveled distance V*0.44 along x side of triangle whose base is 150 m. And third side is x...
  6. Aug 3, 2015 #5
    Correction to my first post. I think this should do it: [tex]340.298=\frac{ x+ \sqrt{ 150^2+x^2} }{(2-0.44)}[/tex]
  7. Aug 3, 2015 #6
    So, how much longer does the sound travel after the 0.44s toward the wall before the other guy hears the echo?
  8. Aug 3, 2015 #7
    [tex]340.298=\frac{ x+ \sqrt{ 150^2+x^2} -150}{(2-0.44)}[/tex]
  9. Aug 3, 2015 #8
    It travels for another 1.56 second the distance x + (x-V*0.44) right?
  10. Aug 3, 2015 #9
    No. Re-read the problem. What is the time between the two sounds the other guy hears?
  11. Aug 4, 2015 #10
    For sure the question is a posed vaguely. "In an interval of 2 seconds" can different interpretations.

    Look at the dashed lines:

    Code (Text):

    O    <--150m-->   O
    |             /
    |         /
    |    /
    |/______________ wall
    Last edited: Aug 4, 2015
  12. Aug 4, 2015 #11
    To me, this clearly means that 2 seconds elapse between the two sounds. But, given the possible confusion, I'll just give my answer: 408.5m, and we'll see if this is what the author of the problem intended.
  13. Aug 4, 2015 #12
    I disagree. It would've been clearer it the statement was like this:

    When one fires the gun the other hears two shots in an interval of 2 seconds between the shots.

    I get 401.76 m by the way.
    Last edited: Aug 4, 2015
  14. Aug 4, 2015 #13
    I have the same equation as you do but with a denominator being 2 rather than (2-0.44). I took it as 2 seconds between the shots. The answer book gives is 406 and i get 410.
  15. Aug 4, 2015 #14
    I can't tell if your sketch accurately indicates that the sound will echo from the mid-point on the wall between the men.
  16. Aug 4, 2015 #15
    Im sure the equation V=((x-150)+(150^2+x^2)^1:2)/2 accuretly describes the problem if its ment that 2 sec is the time between shots and 1.56 if its total time from begg of shot 1 to end of echo.
  17. Aug 4, 2015 #16
    Then solve for x (again) and see what you get.
  18. Aug 4, 2015 #17
    I get the answer 412, which is correct.
  19. Aug 4, 2015 #18
    When I put 412 into your equation:

    V = [(x-150)+(150^2+x^2)^1/2]/2

    I get V = 350, not 340????
  20. Aug 4, 2015 #19
    Its 402 not 412 sorry, i rounded up wrong but the book tells its 412, so i guess they got it wrong.
  21. Aug 4, 2015 #20
    Well, at least you've learned to always check your work by plugging your answer back into the equation.

    You don't appear to be using the fact that the sound will echo off the wall at the mid-point between the men.
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