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Phase of an oscillation

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x = 5sin(pi*t + pi/3). The phase (in rad) of the motion at t = 2s is

    a) (7pi)/3 b) pi/3 c) pi d) (5pi)/3 e) 2pi

    2. Relevant equations


    3. The attempt at a solution
    I plugged in the value for t=2 but I really did not know how to past this point. The answer is supposed to be a)
     
  2. jcsd
  3. Feb 18, 2015 #2

    berkeman

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    Staff: Mentor

    Please show your work. What do you get for x(2)?
     
  4. Feb 18, 2015 #3
    x(2) = 4.33. But how do I determine the phase? Is the phase simply what is in the brackets? So the phase is equal to 2pi+(pi/3)?
     
  5. Feb 18, 2015 #4

    berkeman

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    Staff: Mentor

    Yes. What is that in radians?
     
  6. Feb 18, 2015 #5
    Its 7pi/3. So for any equation of a particle in SHM, the phase is the part inside the brackets?
     
  7. Feb 18, 2015 #6

    berkeman

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    Sort of. For a sinusoidal function (sin or cos), you can picture the value in 2-dimensions on a circle. The amplitude is the radius of the circle, and the point that rotates around the circle with time has some phase angle θ with the positive horizontal axis. If you have a sin() function like you do in this problem, then yes, the value in the () is the phase angle θ with the horizontal axis.

    http://images.tutorcircle.com/cms/images/106/unit-circle-example.png
     

    Attached Files:

  8. Feb 18, 2015 #7
    Now I understand, thank you so much!
     
  9. Feb 18, 2015 #8

    berkeman

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    It may be true in general, but sometimes we define phase with respect to something, so I'm not sure it is a general statement. Others can correct that if appropriate. :smile:
     
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