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Finding location of nodes and antinodes.

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Two transverse waves traveling on a string combine to a standing wave. The displacements for the traveling waves are Y1(x,t) = 0.0200 m sin(2.00 m−1 x − 2.90 s−1 t + 0.40) and Y2(x,t) = 0.0200 m sin(2.00 m−1 x + 2.90 s−1 t + 0.80), respectively, where x is position along the string and t is time.
    Find the location of the first antinode for the standing wave at x > 0.
    Find the first t > 0 instance when the displacement for the standing wave vanishes everywhere.

    2. Relevant equations



    3. The attempt at a solution
    I got the first part. The correct answer was .485 m. Now I wasnt sure if you should take the location where x is a maximum(anitnode) or where x is a minimum(node). Any help would be greatly appreciated. Thank you.
     
  2. jcsd
  3. Nov 17, 2008 #2

    Redbelly98

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    At a node, the displacement Y is zero at all times, so that won't help with solving the problem. Try using the antinode location you found.
     
  4. Nov 17, 2008 #3
    negative. the antinode I found does not work for when the standing wave vanishes completely. thanks for the thought.
     
  5. Nov 18, 2008 #4

    Redbelly98

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    That's weird, I was able to find t that makes Y1+Y2=0 at the antinode x=0.485.

    Oh well, good luck.
     
  6. Nov 18, 2008 #5
    Yeah i found a t of 2.23s at that x. Finding that t value is to find where sin would go to 0 right? am i understanding that correctly? that is where the wave will vanish right?
     
  7. Nov 18, 2008 #6
    bump. anyone?
     
  8. Nov 18, 2008 #7

    Redbelly98

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    You're thinking is correct, for x at an antinode. However, for your value of t we get:

    sin(2*0.4854 - 2.9*2.23 + 0.4)
    = sin(-5.096 radians)
    = 0.927, not zero.

    Maybe you made an algebra error somewhere?
     
  9. Nov 18, 2008 #8
    that might be it. but let me make sure I comprehend what you have just stated. That .927 is the position where the wave vanishes? not the time? Or is that the time where the wave would vanish? This one has got me wound up hah
     
  10. Nov 18, 2008 #9

    Redbelly98

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    0.927 is the value of the sine function, for x=0.4854m and t=2.23s.

    We want to find the values of t that will make that sine = 0, and then choose the smallest positive t for which that is true.
     
  11. Nov 18, 2008 #10
    alright. so the sin(2pi)=0 so set the equation equal to 2pi. solve for t, and thats the time? What about wavelength, how would I know if thats the smallest t? I tried graphing it on excel but im not sure what to keep constant when solving for time.
     
  12. Nov 18, 2008 #11
    also, do I use the combined equation?
     
  13. Nov 18, 2008 #12
    err. sin(pi)=0. Im using that one
     
  14. Nov 18, 2008 #13
    I solved for t=.611s. so when i plug in;

    sin(2*.485 + 2.9*.611 + .4) => about equal to sin(3.1419) => about equal to sin(pi)=0

    is there where the wave would vanish, pending no mathematical error/ is this the right equation to use?
     
  15. Nov 19, 2008 #14

    Redbelly98

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    This is not one of the original expressions.

    There are also other numbers besides 2pi that give sin(number)=0
     
  16. Nov 19, 2008 #15
    pi and 0? pi would be the one to choose. so set either equation? equal to pi and solve for t. thats is where I am getting tripped up at. which equation to use. does it make a difference because they have different phases?
     
  17. Nov 19, 2008 #16

    Redbelly98

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    Those, yes, and also infinitely many others including -pi, -2pi, etc.
    Examine the values of t you get for each, before deciding which to choose.

    It shouldn't matter which equation, I happened to use the first one and then checked that answer in the 2nd equation to be sure.

    Also, did you catch the error I pointed out at the beginning of post #14?
     
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