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Homework Help: Graphs of y(x,t)=A Sin(kx - wt + φ) are shown below

  1. May 26, 2017 #1
    1. The problem statement, all variables and given/known data
    graphs.png The first graph shows y vs t for a point at x=0 m. The second shows y vs x for the string at a time of 2 s.

    I have calculated the amplitude of the oscillation to be 1.0m

    I have calculated the angular frequency of the oscillation to be 2.90 rad/s

    I have calculated the wavelength of the oscillation to be 6.28m

    However the following questions I am not quite sure how to do based on the graphs I have linked in the question

    1) Calculate the vertical position of the string at x=0m, t=0s.

    2) Calculate the phase constant, φ, of the motion.

    3) Determine the vertical displacement of the string element at x=0m at time t=2s.

    4) What is the speed of the wave?

    2. Relevant equations


    3. The attempt at a solution
    1) if we let x=0m and t=0m then y=1m sin φ but how do I solve for the vertical displacement if there are two unknowns?

    3) similar to number 1, again not quite sure how to solve for the vertical displacement

    2), 4) not quite sure how to do these either
     
  2. jcsd
  3. May 26, 2017 #2

    Ssnow

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    For the ##1)## you must have solved ##2)##. But you know ##\omega## and ##k## and selecting a precise point ##(x_{0},t_{0})## on the graphs you can deduce ##\varphi## ...
     
  4. May 26, 2017 #3
    So for number 2 I can pick any point?
     
  5. May 26, 2017 #4

    Ssnow

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    yes, for example at ##t=2## and ##x=6## we have ##y=-1/2## ...
     
  6. May 26, 2017 #5
    okay so then -1/2=1m sin Sin(6k - 2w + φ), so then how do I solve for φ?
     
  7. May 26, 2017 #6

    Ssnow

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    you know ##\omega## and ##k## ...
     
  8. May 26, 2017 #7
    so then -1/2=1sin(6(2.90)-(6.28)(2)+φ), that seems to be correct
     
  9. May 26, 2017 #8
    I am not quite sure how to solve for φ in that equation, but thanks for your help anyways!
     
  10. May 26, 2017 #9

    Ssnow

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    It is a goniometric equation ...
    Ssnow
     
  11. May 26, 2017 #10
    -1/2=sin(4.84+φ) so then what is φ? I have never solved a equation like this before
     
  12. May 26, 2017 #11
    I tried to solve 6*2.90-6.28*2+a=-pi/6, which gave a=-5.36rad but this was incorrect
     
  13. May 27, 2017 #12
    I also tried 2.53rad but this was incorrect
     
  14. May 27, 2017 #13

    vela

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    Are you sure about your value for the angular frequency? It looks closer to something like 2.1 rad/s to me.
     
  15. May 27, 2017 #14
    2.90rad/s was correct for the angular fequency, y=-1/2 when x=6 but when t=2 it looks like it is y=-.250, as was stated above to use. I have been stuck on this question for quite some time and am trying to finish it off
     
  16. May 27, 2017 #15

    vela

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    Oops, I calculated the period, not the frequency.

    You just said y=-0.25 m. Did you mean y=-0.50 m?
     
  17. May 27, 2017 #16
    Well on the first graph y vs t when t is 2 it doesn't quite look like y=-0.5, however on the second graph y vs x when x is 6 y=-0.5,
     
  18. May 27, 2017 #17

    vela

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    The first graph is of the motion of the point x=0 on the wave. If you use y=-0.25 m for t=2 s, you need to set x=0 m.
     
  19. May 27, 2017 #18
    Oh okay. So would you be able to help me solve the phase constant then?
     
  20. May 27, 2017 #19

    vela

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    The way you tried was right. You're probably just plugging in the wrong numbers or messing up on the calculator.
     
  21. May 27, 2017 #20
    I did 7pi/6= (6×2.90−6.28×2+a) solved for a and got 2 numbers both in degrees, 145, -35 I changed 145 degrees to radians and got 2.53. What did I do wrong?
     
  22. May 27, 2017 #21

    vela

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    It looks like you're multiplying ##\omega## by ##x##, which doesn't make sense. Also your value for ##k## is wrong.
     
  23. May 27, 2017 #22
    Isn't w=6.28 and k=2.90?
     
  24. May 27, 2017 #23

    vela

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    Nope. A good place to start is to make sure you know what the symbols represent.
     
  25. May 27, 2017 #24
    Yes you are right, so k=2π/λ, then k= 2π/6.28, k = 1. Therefore the equation I should be solving is 7pi/6=1*2-6*6.28+a which gives a=39.35 degrees which is 0.687 rad for the phase constant? Please let that be right lol
     
  26. May 27, 2017 #25
    Apparently 0.687 rad is incorrect also
     
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