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Finding Phase Constant

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Part (a) of the figure below is a partial graph of the position function x(t) for a simple harmonic oscillator with an angular frequency of 1.35 rad/s; Part (b) of the figure is a partial graph of the corresponding velocity function v(t). The vertical axis scales are set by xs = 6.5 cm and vs = 7.0 cm/s. What is the phase constant of the SHM if the position function x(t) is given by the form x = xmcos(ωt + ϕ)?

    15-20.gif

    2. Relevant equations
    x = xmcos(ωt + ϕ)
    v = -ωxmsin(ωt + ϕ)

    3. The attempt at a solution

    For the X vs. t graph the line crosses t=0 when x = 2.6. For the V vs. t graph the line crosses t=0 when v=-5.6.

    I thought then I could just plug all the number in and find out when they are equal

    2.6 = 6.5cos(1.35*0+ϕ)
    -5.6 = -8.775sin(1.35*0+ϕ)

    I subtracted 2.6 from both sides for the first equation and added 5.6 to both sides for the second. I then set them equal. I used my calculator to attempt to solve them.

    Looking at them separately it looks like it should be 1.15 radians for X and .69 radians for V (Roughly).

    I can't figure out what I'm doing wrong. Perhaps I shouldn't be reading the graph like I am. or I am simply reading it wrong.
     
  2. jcsd
  3. Nov 12, 2009 #2
    Found an answer. I don't understand why this is correct, but dividing velocity by position gives the following

    v/x = tan-1([tex]\frac{Velocity @ t=0}{Position @ t=0 TIMES Angular Frequency}[/tex])

    So that leaves


    tan-1([tex]\frac{-5.6}{2.6 * 1.35}[/tex]) = -1.01

    I think then since it is shifted I changed it to positive 1.01.

    That answer was taken as correct.
     
  4. Nov 12, 2009 #3
    One equation represents position and the other represents velocity. Since the units are different, you can never "set them equal."
     
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