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Adding three SHM oscillations of equal frequency?

  1. Dec 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle is simultaneously subjected to three SHM, all of the same frequency, and in the x direction. If the amplitudes are 0.25, 0.20, 0.15mm, respectively, and the phase difference between the 1st and 2nd is 45 degrees, and between the 2nd and 3rd is 30 degrees, find the amplitude of the resultant displacement and it's phase relative to the first (0.25mm amplitude) component.


    2. Relevant equations

    z = x + iy

    z1 = x1 + iy1

    x1 = 0.25 cos(wt +0)
    x2 = 0.20 cos(wt +45)
    x3 = 0.15 cos(wt +75)

    z1 = 0.25 [x1 + i sin(wt +0)]
    z2 = 0.2 [x2 + i sin(wt +45)]
    z3 = 0.15 [x3 + i sin(wt +75)]

    3. The attempt at a solution

    Firstly, I assumed there was no initial phase angle for z1 said at t=0

    x1 = 0.25
    x2 = 0.20 cos(45)
    x3 = 0.15 cos(75)

    therefore the Amplitude must be x = x1 + x2 + x3 (as at t=0 it must be at maximum displacement so x = A)

    I got x = 0.430244mm
    the answers says 52mm

    then i decided maybe at t=0 x is not equal to A (although I don't know why :redface:) so i tried to calculate the magnitude of z.

    z = (x^2 +y^2)^0.5

    i added up all the y components at t =0 and got y = 0.286310mm

    therefore z = 0.516

    now for the angle:

    I used arctan(y/x) = 56 degrees, the answer is ~33.5 degrees :cry:
     
  2. jcsd
  3. Dec 13, 2013 #2

    tiny-tim

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    hi applestrudle! :smile:
    expand, and add: you should get Acosωt + Bsinωt …

    then convert that into amplitude*cos(ωt + phase) :wink:
     
  4. Dec 13, 2013 #3

    haruspex

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    I'm not sure whether it helps to introduce an iy component. Maybe.
    The x1, x2, x3 already have the amplitude factor. I think you mean
    z1 = [x1 + 0.25 i sin(wt +0)]
    z2 = [x2 + 0.2 i sin(wt +45)]
    z3 = [x3 + 0.15 i sin(wt +75)]
    That's only the instantaneous magnitude of x at time 0. It might achieve a greater magnitude at some other time. The amplitude is the peak magnitude.
    Same problem.
    As tiny-tim posted, you can just expand each cos(ωt+ψ) as cos(ωt)cos(ψ) - sin(ωt)sin(ψ).
     
  5. Dec 13, 2013 #4

    gneill

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    The frequencies are all the same so you can treat the components as phasors and add accordingly.
     
  6. Dec 14, 2013 #5

    rude man

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    I would go with post #2 method.

    Add the coefficients of sin(wt), add the coeff. of cos(wt), giving
    a sin(wt) + b cos (wt) = c sin(wt + phi).
     
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