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Slinky physics problem

  1. Nov 6, 2005 #1
    I know this seems very basic, but I have a question about standing waves.
    For example, A slinky of length 3.15 m is tied with a light string on both ends such that both ends are free to move. When a person holding on to one of the light strings moves her arm at just the right frequency, a standing wave with 5 nodes is formed.
    (a)How far from one free end of the slinky is the node closest to that free end formed?

    This seems VERY simple, but wouldn't you just divide the length of the slinky by the number of antinodes to find the distance between nodes and use that as your distance from the free end to the node?

    If this is so, wouldn't it come out to 0.525? If so, this is the answer that I provided for the problem, which was then marked incorrect. Is there something I am doing wrong or is there a different way to approach it?

    Please help, I am quite confused as to why my answer is incorrect
     
  2. jcsd
  3. Nov 6, 2005 #2

    Doc Al

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    Hint: Are the ends of the slinky nodes or anti-nodes?
     
  4. Nov 6, 2005 #3
    Re: nodes vs. antinodes

    Shouldn't they be antinodes since they are not fixed? Does this matter because the distance to the nearest node should be the same regardless (?). UNLESS: isn't the antinode the peak of the wave so you would have to add the amplitude to the distance to the node to get the complete distance from the free end to the node???? Am I on the right track, or completely off, because it seemed to make sense a second ago, but now I think it may not work. ???
     
    Last edited: Nov 6, 2005
  5. Nov 6, 2005 #4

    Doc Al

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    Right.
    Well... if the ends are antinodes, then the question amounts to the distance between an antinode and an adjacent node. How does that distance relate to the length of the slinky? (Draw a picture.)

    (Amplitude has nothing to do with it.)
     
  6. Nov 6, 2005 #5
    Re:

    It seems that the distance in this case (slinky length = 3.15 m) and there are 5 nodes (6 antinodes), that the distance between the free end and the node is the same as 1/6th of the whole length of the slinky. So, 1/6 of 3.15 m = 0.525 m (my proposed answer before).

    I am sorry that this whole thing keeps dragging on and I STILL haven't come to the right answer, thanks for your patience with me...

    Am I even remotely on the right track???
     
    Last edited: Nov 6, 2005
  7. Nov 6, 2005 #6

    Doc Al

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    Right.
    Wrong!

    Try this:
    (1) Find the spacing between adjacent antinodes
    (2) Use that to find the spacing between an end (antinode) and an adjacent node
     
  8. Nov 6, 2005 #7
    I think I've got it!

    Is the distance between two nodes or two antinodes expressed in the formula lambda/2? If this is so, would the correct answer be 1.575????
     
  9. Nov 6, 2005 #8
    Does that seem right? I saw a graphic illustrating this, showing that the value of lambda is the length of the whole slinky, which wouldn't make sense outside of the second harmonic. The follow up question to the one I am asking is the length between the free end and the second closest node, which, if you follow the formula, will end up giving you the length of the entire slinky.

    This is a fairly new concept for me, so I am not too familiar with this material-- is there a set formula to find the distance between nodes??
     
  10. Nov 6, 2005 #9

    Doc Al

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    Yes.
    No. (Why did you divide 3.15 by 2?)

    Draw a picture of the slinky.
     
  11. Nov 6, 2005 #10
    By drawing out the wave, it looks like there are 3 complete waves for the 6 antinodes and the wavelength should be 3.15/3= 1.05 and that divided by 2 comes out to be 0.525.
    I feel like I am making no progress.:cry:
     
  12. Nov 6, 2005 #11

    Doc Al

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    There aren't 3 complete waves in my drawing. (Note: The distance between adjacent antinodes is one-half of a wave.) So: How many half-waves does the slinky contain?

    Another hint: If the distance between adjacent antinodes is one-half wave, what is the distance between an anti-node and an adjacent node?
     
  13. Nov 6, 2005 #12
    1/4 a wave!
     
  14. Nov 6, 2005 #13

    Doc Al

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    Right! Now use this to get the answer.
     
  15. Nov 6, 2005 #14
    I drew out the wave again and calculated that the distance from one antinode to the next is 1.26 because 3.15/2.5= 1.26. I then divided 1.26 by two to find the value of 1/4 a wave which was 0.63 and added them together to get 1.89 meters. Is this correct?
     
  16. Nov 6, 2005 #15
    WAIT WAIT WAIT-- I see what I just did wrong!
     
  17. Nov 6, 2005 #16
    Should it be 0.945?
     
  18. Nov 6, 2005 #17

    Doc Al

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    No. Remember there are 6 antinodes along the slinky (and thus 5 spaces between them).
     
  19. Nov 6, 2005 #18
    thanks i've actually got it now! Thanks so much!
     
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