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How to determine phase constant?

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A 225g mass attached to a horizontal spring oscillates at a frequency of 4.00 Hz. At t=0s, the mass is at x = 5.00 cm and has v_x = -37.0cm/s . Determine the phase constant.

    2. Relevant equations

    velocity = v(t)
    v(t) = Aω·cos( ωt + ϕ )

    3. The attempt at a solution

    -37 cm/s = ( 5.2122*10^-2 m )( 25.133 rad/sec )cos( 0 + ϕ )
    -0.282 = cos( ϕ )
    ϕ = cos⁻¹( -0.282 )
    ϕ = 106.40º = 1.857 radians ◄---

    However that answer is wrong, and I am not sure why. Can anyone help?
     
    Last edited: Apr 20, 2012
  2. jcsd
  3. Apr 20, 2012 #2

    ehild

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    The result is correct, but you could have used both x(0) and v(0) to calculate the tangent of the phase constant, and decide about the quadrant by examining the sign of the sign and cosine.
    It is possible that the SHM was defined as x(t)=cos(ωt + ϕ). Try it.


    ehild
     
    Last edited: Apr 20, 2012
  4. Apr 20, 2012 #3
    So instead of using -37 use 5 instead? I tried that and got 1.5324 radians and it's still wrong, lmao. Hmm..
     
    Last edited: Apr 20, 2012
  5. Apr 20, 2012 #4

    ehild

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    How did you get 1.5324 radians?

    If x=Acos(ωt+φ), v=-Aωsin(ωt+φ), X(0)=5=Acosφ, v(0)=-37=-Aωsinφ:

    tanφ=7.4/(8π)...

    Both sine and cosine are positive so the angle is in the first quadrant.

    ehild
     
  6. Apr 20, 2012 #5
    Ah, thanks I see where I made the error. It was .2863rad. Thanks alot!
     
  7. Apr 20, 2012 #6
    Actually now I am having trouble with the last part, lol. The question is to determine the position at t = 5.00s .

    I did this:

    Position = s(t)
    = A·sin[ ωt + ϕ ]
    = ( .052122 m )sin[ (25.133 s⁻¹)( 5.0 s ) + .2893 ]

    And got .01478 m, but that is wrong. Any help?
     
  8. Apr 20, 2012 #7

    ehild

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    But we have figured out that the position was x=Acos( ωt + ϕ) with ϕ=0.2893...


    ehild
     
  9. Apr 20, 2012 #8
    Wow, I didn't even realize, hah. Thanks again for all the help!
     
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