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Simple Harmonic Motion - Determine quadrant for phase angle.

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    When completing this problem I am able to find a value for the phase angle but am unsure of how to find the quadrant for the phase angle therefore unable to get the correct phase angle.
    4ddb60e96fad8f621abdd91c42e149c9.png

    2. Relevant equations
    Provided in the question: x = xmcos(wt+ phi)

    3. The attempt at a solution
    After doing a series of calculations I determined sin(phi) = -0.8

    at t =0

    v(0) = -xmwsin(w(0) + phi)
    v(0) = -vmsin(phi)
    4.0cm/s = (-5.0 cm/s)sin(phi)
    sin(phi) = -4/5 = -0.80

    Therefore, one answer is -0.93 rad for the phase angle, phi. However other phase angles would also make sin(phi) = -0.8 such as -2.21. How do I determine which angle in radians is correct?
     
    Last edited: Jan 17, 2016
  2. jcsd
  3. Jan 17, 2016 #2

    TSny

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    One way you might decide is to consider the sign of the acceleration at t = 0.
     
  4. Jan 17, 2016 #3
    Yeah I notice that since when I did my calculations I used t =0 that the acceleration at that point was positive do to the slope of the velocity graph being positive. However how would I determine the angle range? I think it may be important to also know the velocity is positive at t = 0 perhaps.
     
  5. Jan 17, 2016 #4

    TSny

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    To make sure you are on the right track, would you mind showing how you got sin(φ) = -.80? You should not have needed to know anything about the acceleration to get this.

    From the graph, is the acceleration positive or negative at t = 0. This should give you some more information about which quadrant φ is in.
     
  6. Jan 17, 2016 #5
    Thank you for the response here are my calculations:

    at t =0

    v(0) = -xmwsin(w(0) + phi)
    v(0) = -vmsin(phi)
    4.0cm/s = (-5.0 cm/s)sin(phi)
    sin(phi) = -4/5 = -0.80

    EDIT: The acceleration is positive at t = 0 and so is the velocity. How do these quantities apply to the quadrant?
     
  7. Jan 17, 2016 #6

    TSny

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    Great!

    How do you determine the acceleration from a velocity vs time graph?
     
  8. Jan 17, 2016 #7
    acceleration was determined by the slope of the velocity vs time graph. At t = 0 I had observed a negative slope therefore acceleration is actually negative at t=0. Sorry I made an error. How does this negative acceleration relate to phase angle?
     
  9. Jan 17, 2016 #8

    TSny

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    What is the mathematical expression for the acceleration as a function of time? What does it reduce to at t = 0?
     
  10. Jan 17, 2016 #9
    a(t) = -xmw^2 cos(wt + phi)
    a(0) = -xm(w^2)cos(phi)
    a(0) = -xw^2
     
  11. Jan 17, 2016 #10

    TSny

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    OK to here.
    What happened to cos(phi)? [EDIT: Oh, I see. You dropped the subscript m on the position. OK]
     
  12. Jan 17, 2016 #11
    Well I substituted the displacement function x(0)=xmcos(phi) into the velocity equation. I just called this equation x instead of x(0)
     
  13. Jan 17, 2016 #12

    TSny

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    OK. From a(0) = -xm(w^2)cos(phi) can you deduce anything about the sign of cos(phi)?
     
  14. Jan 17, 2016 #13
    since xm and w^2 are always positive the cos(phi) is also positive. Therefore, according to the CAST rule since cos(phi) is positive the quadrant must be either 1 or 4 but sin(phi) is negative so therefore phi is in quadrant for so phi= -0.92 radians (to 2 significant figures) is the answer?
     
  15. Jan 17, 2016 #14

    TSny

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    Yes, 4th quadrant. -0.92 rad looks correct.
     
  16. Jan 17, 2016 #15
    Actually did I make an error, should cos(phi) actually be negative do to the negative in front so the answer should be quadrant 3? since sin and cos are both negative in quadrant 3?
     
  17. Jan 17, 2016 #16

    TSny

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    Sorry, I'm not following how you are getting that cos(phi) should be negative. What is the sign of a(0)?
     
  18. Jan 17, 2016 #17
    Not a problem let me try to explain.

    xm is a positive quantity. Angular frequency, w , a positive quantity
    a(0) = -xm(w^2)cos(phi)
    a(0) = -(positive)(positive)^2(cos(phi))
    negative * positive * positive = negative
    that is,
    a(0) = -cos(phi)
     
  19. Jan 17, 2016 #18

    TSny

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    Exactly.
    So, what is the sign of cos(phi)?
     
  20. Jan 17, 2016 #19
    Is it not negative? Or am I perhaps mislead by the negative sign in front of cos(phi)?
     
  21. Jan 17, 2016 #20

    TSny

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    If cos(phi) were negative, what would be the sign of -cos(phi)?
     
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