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The Steady-state motion of a forced oscillator

  1. Feb 3, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Solve for the steady-statem otion of a forced oscillator if the forcing is F_0*sin(wt) instead of F_0*cos(wt). Use complex representations.

    2. Relevant equations
    e^(i*x) = cos(x)+isin(x)

    3. The attempt at a solution

    I assume, First, steady-state means without damping.
    Next, we have the equation
    m*x'' = -kx + F_0*sin(wt)
    where x'' is the second derivative of x position with respect to time. Also known as acceleration.

    So, we have:

    z'' + w_0^2*z = F_0/m * (-i*e^(iwt)) where Re(z) = x

    So far so good? I think I changed it correctly, normally when you do F_0*cos(wt) as forcing, you just switch it to e^(iwt)

    Now, if everything is looking good up to here, we take z = A*(-ie^[i(wt-φ)])

    After taking the second derivative of this and inputting it into our equation, our equation looks like:

    i*w^2*A*e^[i(wt-φ)]-w_0^2*A*i*e^[i(wt-φ)] = F_0/m * (-i*e^(iwt))

    dividing through by -i*e^(iwt) nets us

    F_0/m = A(w_0^2 - w^2)*e^(-iφ)
    Simplifying: F_0/m *e^(iφ) = A(w_0^2-w^2)

    But now I seem to have run into a problem. The real part of this is F_0/m*cos(φ)
    But this is the exact same solution to the forcing with F_0*cos(wt).
    I thought the problem wants me to reach Re(z) = F_0*sin(wt)

    Any help? Did I go wrong somewhere?
     
  2. jcsd
  3. Feb 4, 2016 #2

    RJLiberator

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    Wait,

    Did I reach the answer here:

    I think that's all the question is looking for.
     
  4. Feb 4, 2016 #3

    RJLiberator

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    Hey guys, sorry to bump this with an extra post.
    But only 2 hours until class starts and was interested in if my answer was looking good, as it is a hard problem for me to understand fully.
     
  5. Feb 4, 2016 #4

    ehild

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    If you take the imaginary part of the solution, the driving function is Fo sin(wt)
     
  6. Feb 4, 2016 #5

    RJLiberator

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    I'm not sure what you mean by that.

    The problem states:
    So, isn't that already assumed.
     
  7. Feb 4, 2016 #6

    ehild

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    You solve the differential equation for the case F=F0eiwt, and you get the steady-state solution as z=Aeiwt. In principle, the amplitude can be complex, but you get that A=F0/(w02-w2), which is real. Taking the imaginary part, the real-life solution is

    y=F0/(w02-w2)sin(wt). sine function instead of cosine.
     
  8. Feb 4, 2016 #7

    RJLiberator

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    Hm.
    Few concerns:

    The forcing part of it is F_0*sin(wt)

    so we have
    m*a = -kx + F_0*sin(wt)

    If we are to convert this to complex numbers we have
    m*d^2z/dt^2 +kz = F_0 * (-i*e^(iwt))

    Why do you say z = Ae^(iwt) and not what I think should be (-i*e^(iwt)) as my part seems to provide a real part of sin(wt) instead of an imaginary part.
     
  9. Feb 4, 2016 #8

    ehild

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    F_0*sin(wt) is not the same as F_0 * (-i*e^(iwt)), but F_0*sin(wt)=Im(F_0 e^(iwt))
     
  10. Feb 4, 2016 #9

    RJLiberator

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    Okay, thatttt makes a lot of sense to me.

    So using z = Ae^[i(wt-Φ)] is the correct choice, but instead of Re(z) =x we let Im(z) = x.

    Solution would be then:
    (iw)^2*A*e^[i(wt-Φ)]+w_0^2*A*e^[i(wt-Φ)] = F_0/m * e^(iwt)

    With Im(z) = x.

    Would that be the correct way to display the solution (I guess here I am getting lost on what the problem actually wants.)
    It seems like this would be the steady-state motion of a forced oscillator. But this is still the same equation as with cos(wt) just where cos(wt) is with Re(z).

    Or, do I input for z = sin(wt) and solve?
     
  11. Feb 4, 2016 #10

    ehild

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    Solve this equation for A, and then give the solution as A*sin(wt-Φ)
    But you get the same solution if you just input x=Asin(wt-Φ). Φ will be 0 or pi.
     
  12. Feb 4, 2016 #11

    RJLiberator

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    A = (F_0/m) * (1/(w^2-w_0^2) * sin(Φ)

    :D ?
     
  13. Feb 4, 2016 #12

    ehild

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    And what is Φ?
     
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