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Waves and optics pertaining to sinusoidal waves

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  1. Sep 29, 2014 #1
    1. The problem statement, all variables and given/known data
    https://us-mg5.mail.yahoo.com/neo/launch?.rand=768lpb97vo87e#4636076903 The picture displayed shows three graphs and in each problem the objective is to find the phase constant. I'm having an issue attempting any form of strategy to solve these problems. Any help would be greatly appreciated thanks.

    2. Relevant equations
    For the first graph D(x=0,t=0)=asin(kx-wt+phase constant)
    1=2sin(0-0+phase constant)
    phase constant = 30 degrees

    For the second graph
    1=2sin(0-0+phase constant)
    phase constant = 30 degrees

    For the third graph
    1=2sin((2pi*4)/4 - (2pi)(T/2)/4+phase constant)

    Don't have an idea how to solve for the third graph


    3. The attempt at a solution
    1. For the first graph I answered 30 degrees
    2. For the second graph I answered 30 degrees
     

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  2. jcsd
  3. Sep 29, 2014 #2

    BvU

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    Your answers would imply that the two pictures both represent ##D = 2 \sin(kx-\omega t + \pi/6)## and are therefore exactly the same.
    Clearly, they are not. Perhaps the equation ##\sin\phi = 1/2## has another solution ?
     
  4. Sep 29, 2014 #3
    I agree there are two different solutions however I'm not sure which problem has the solution pi/6 and why it would be considered an answer over the other problem.
     
  5. Sep 29, 2014 #4

    BvU

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    Did you read the 'Note' ? (I have to ask, since in your 1 and 2 answers you do have the same ##\phi_0##)
     
  6. Sep 29, 2014 #5

    BvU

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    From another angle :): In the first picture, is D increasing or decreasing when you take a small step in time ?
     
  7. Sep 29, 2014 #6
    Yes I did read your note, and I just read your second one now. It seems much more clearer and I have a better understanding. So the first graph has a phase constant of pi/6 since d is increasing while in the second graph d is decreasing.
     
  8. Sep 29, 2014 #7
    So in that case the second graph would indicate the answer would be 5pi/6
     
  9. Sep 29, 2014 #8

    BvU

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    :)I didn't mean my note, but the note under the exercise!

    Then: if you make a little time step in the first picture, does D really increase ?

    Because, if I fill in x = 0, I get ##D = 2 \sin(-\omega t + \pi/6)## and what is the time derivative ##d\;D\over dt## at time zero for that ?
     
  10. Sep 30, 2014 #9
    The derivative of time for that equation would be D=-2wcos(−ωt+π/6) but what does this equation have to do with the problem?

    I suppose time doesn't increase or decrease since it is following a pattern.
     
  11. Sep 30, 2014 #10

    BvU

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    Time is running. The picture is a snapshot at t=0. There is even a big fat arrow showing which way the wave is going.
    So what is this derivative at t=0 ? Is it positive or negative ? This derivative has everything to do with the problem: to determine the phase angle, you have to know if the D is increasing or decreasing.

    Look at it this way: If you are sitting on your surfboard at x=0 on t=0 and this huge wave moves as is indicated by the arrow, are you going up or down ?
     
  12. Sep 30, 2014 #11
    -pi/6 is the derivative at t=0, so d is decreasing. It is going down
     
  13. Oct 1, 2014 #12

    BvU

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    And in the second picture, same procedure ?
     
  14. Oct 1, 2014 #13
    I believe the same method is applicable but your guidance is widely accepted
     
  15. Oct 1, 2014 #14

    BvU

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    :) The idea is that you are now in a good position to handle the second picture by yourself ! And yes, the approach is the same.
    You see that ##D(0,0) = 2 \sin\phi_0 = 1## which has two solutions (*). Which one is the right one, you decide by looking at the time derivative of D at x=0, t=0.

    (*) Actually, it has infinitely many solutions, but only two are in the range ##[0,2\pi]##
     
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