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Finding Phase Constant

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    An air-track glider attached to a spring oscillates with a period of 2.3 s. At t=0 s the glider is 3.09 cm left of the equilibrium position and moving to the right at 35.07 cm/s.
    What is the phase constant? (in degrees)



    2. Relevant equations

    v0x= -wAsin(ro), x=Acos(ro)

    3. The attempt at a solution
    I divided v0x/x=wtan(ro). I solved for angular speed which was 2.73 rad/s. Now I solved for (ro) by taking tan inverse of (v/-wx). I received 73 as an answer and that's wrong. Would you please help me? Thank you.
     
  2. jcsd
  3. Jun 16, 2009 #2
    Since you are to the left, you have to subtract 180. The anwser you came up with is to the right of the equilibrium point.
     
  4. Jun 16, 2009 #3
    I minus that by 180 so it's 107 but that's still wrong. What do you think is wrong?
     
  5. Jun 16, 2009 #4

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    Did you get

    [tex] r_{0} = \tan^{-1} \frac{35.07}{3.09 \cdot 2.73}? [/tex]
     
  6. Jun 16, 2009 #5
    yeah, I tried that as well which was 76.4751. Then I minus it from 180 so it's 103.525 but that's wrong.
     
  7. Jun 16, 2009 #6

    dx

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    Both 76.47 and 76.47 - 180 have the same tangent. You can see by drawing a pic that the one we want is (76.47 - 180).

    -103.5 degrees = -1.807 radians.
     
  8. Jun 16, 2009 #7
    is it because it's negative and the distance is negative?
     
  9. Jun 16, 2009 #8
    Also, in this case, is angular speed the same as velocity? If I wanted to find the phase at t=0 s would I just need to use wt+ro, w being the velocity?
     
  10. Jun 16, 2009 #9

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    Yes, the cosine of 76.4 is positive, but the initial distance is negative.
     
  11. Jun 16, 2009 #10
    Oh sorry w=2.73. Sorry about that.
     
  12. Jun 16, 2009 #11

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    No, angular speed and velocity are different. The angular speed is constant, while the velocity is changing.

    v = -Aωsin(ωt + a0)
     
  13. Jun 16, 2009 #12
    ok, so for t=.5s I get -102.115 but that's wrong. All I did is wt+constant(-103.48). But that's wrong for some reason. Does the constant change at that time?
     
  14. Jun 16, 2009 #13

    dx

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    Write -103.48 in radians.
     
  15. Jun 16, 2009 #14
    -1.80607 rad
     
  16. Jun 16, 2009 #15

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    Do you get the right answer now? (I'm not sure what you're trying to calculate BTW, did you get the correct answer for the phase constant?)
     
  17. Jun 16, 2009 #16
    I got the correct answer for the phase constant but I'm trying the phase for t=.5s. Sorry for the confusion.
     
  18. Jun 16, 2009 #17

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    Ok, so at t = 0.5, you have ωt + a0 = 2.73(0.5) - 1.806. This is in radians. If you want to convert it to degrees, multiply by 360/2pi.
     
  19. Jun 16, 2009 #18
    The question has several parts and it's asking me for the phase at t=.5s. I used the same formula but it's wrong. Could you please let me know what's wrong? thank you.
     
  20. Jun 16, 2009 #19
    Ok thank you very much for all your help. I greatly appreciate it.
     
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