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!

Phase Difference

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A tuning fork generates sound waves with a frequency of 246Hz. The waves travel in
    opposite directions along hallway, are reflected by end walls, and return. The hallway
    is 47.0m long and the tuning fork is located 14.0m from one end. What is the phase
    difference between the reflected waves when they meet at the tuning fork? The speed of
    sound in air is 343 m/s.

    2. Relevant equations
    delta r = Phi/2pie x Lambda
    f = v/Lambda

    3. The attempt at a solution
    The first thing I did was delta r = r2 - r1 = (33.0m - 14.0m) = 19.0m
    then f = v/Lambda or Lambda = v/f = 343m/s / 246Hz = 1.39 units?
    using delta r = Phi/2pie x Lambda
    19.0m = Phi/2pie x 1.39
    I get like 86.5 degrees and it should be 91.3 degrees
    Any help would be appriciated.
    Kevin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 30, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that the sound reflects from the walls and thus makes a round trip.
    OK.
    I don't understand your equation. Each wavelength of path difference gives you 360 degrees (or 2 pi radians) of phase difference.
     
  4. Nov 30, 2008 #3
    Doc,
    Thanks but now I don't either

    delta r = (66.0m - 28.0m) = 38.0m

    Is the phase difference = kd
    where k = 2pie/delta r

    Thanks,
    Kevin
     
  5. Nov 30, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    How many wavelengths does that correspond to?
     
  6. Nov 30, 2008 #5
    Not sure but trying to figure out
     
  7. Nov 30, 2008 #6

    Doc Al

    User Avatar

    Staff: Mentor

    You already found the wavelength. Just divide that wavelength into the path difference.
     
  8. Nov 30, 2008 #7
    I am confused as I get the wavelength to be 1.39m
    from Lambda = v/f

    and path difference is 38.0m
    Are these the figures that I use
    Kevin
     
  9. Nov 30, 2008 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, those are the numbers that you need.

    It's the same as this example: Say you have 423 pennies. Since 1 dollar equals 100 pennies, how many dollars do you have? (I suspect you have no problem solving that one.) Now do this: You have a path difference of 38 m. Since 1 wavelength equals 1.39 m, how many wavelengths do you have?
     
  10. Nov 30, 2008 #9
    27.3 wavelengths. I got that but not sure where to put that to find
    phase difference.
    Thanks for the help. My brain isn't working this early
    Kevin
     
  11. Nov 30, 2008 #10

    Doc Al

    User Avatar

    Staff: Mentor

    OK, but don't round off until the last step.

    Whole numbers of wavelengths don't lead to a phase difference. Only the fraction left over after getting rid of integral numbers contributes to the phase difference.
     
  12. Nov 30, 2008 #11
    Thanks but I don't get this. I'm not sure what to do with this number
     
  13. Nov 30, 2008 #12
    Doc,
    I still don't understand can you give me a clue what to do with these numbers
    Thanks,
    Kevin
     
  14. Nov 30, 2008 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Answer these questions. If two waves differ by exactly one wavelength, what's their phase difference? What if they differed by exactly 12 wavelength? What if they differed by 1/3 of a wavelength? What about 2.33 wavelengths?
     
  15. Nov 30, 2008 #14
    I think that is where my problem understand is. I'm looking some more but
    I believe that if they differ one wavelength they are in the same phase as well
    as 12. I'm not sure though
    Kevin
     
  16. Nov 30, 2008 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Exactly right! If the path difference is an integral number of wavelengths, then there is no phase difference.

    What about when there's a fraction of a wavelength difference?
     
  17. Nov 30, 2008 #16
    I think that when it is a phase shift of 90 degrees it is pie/2
    and 180 is pie
    270 is 3pie/2
    But I'm not sure how to use those to find any other fraction
    Kevin
     
  18. Nov 30, 2008 #17

    Doc Al

    User Avatar

    Staff: Mentor

    You are comparing degrees with radians. 90 degrees = pi/2 radians.

    In terms of wavelengths, one wavelength = 360 degrees (or 2 pi radians).

    So, for example, a 1/4 wavelength path length difference corresponds to 1/4 * 360 = 90 degrees phase difference.

    When figuring out the phase difference based on the path length difference, all that matters is the fractional part of the path length difference. If the path difference is 2.4 wavelengths, then all that counts is the 0.4 wavelengths. And 0.4 wavelengths corresponds to a phase difference of 0.4*360 = 144 degrees.
     
  19. Nov 30, 2008 #18
    It's starting to make more sense so going back we have 38.0m / 1.39m
    =27.338129496 number of waves. So they are out of phase by
    .338129496 or 121.7 degrees?
    Doesn't match the answer but I just want to understand the concept
    Kevin
     
  20. Nov 30, 2008 #19
    When I carry the numbers correctly it works out
    to be .25364432 or 91.3 degrees
    Thanks,
    Kevin
     
  21. Nov 30, 2008 #20

    Doc Al

    User Avatar

    Staff: Mentor

    Sweet! (You made it.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Phase Difference
  1. Phase difference (Replies: 1)

  2. Phase difference! (Replies: 1)

  3. Phase Difference (Replies: 3)

  4. Phase difference. (Replies: 2)

  5. Phase difference (Replies: 5)

Loading...