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Resonance questions

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

    1) The equation of the particle on one dimension given by;

    d^2y/dt^2 + 2 dy/dt + 6y = 10sinωt

    a) Find the displacement as a function of time.
    b) find T= ? A=? (period,amplitude)

    2) Show that the amplitude of the vibration damping halved on time 1,39/γ.
    3)On one LC circuit;
    C = 100mikroF and resonance frequency is 2MHz.Find the value of coil. (L)

    2. Relevant equations


    General form = d^2y/dt+γdy/dt+y=F0cosωt
    And I don't know can we use this formula but maybe can help;
    A=-F0/m /(ω^4+γ^2ω^2)^1/2

    d^2θ/dt^2 = -W0^2q


    3. The attempt at a solution


    I just tried to use the formula for amplitude but there's no mass.And I don't have an idea about 2 and 3.
     
  2. jcsd
  3. Dec 2, 2012 #2

    haruspex

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    Do you know how to solve the homogeneous equation y'' + ay' + by = 0?
     
  4. Dec 2, 2012 #3
    y=c1e^r1x+c2er^2x (distinct) if we have two root,
    y= c1e^rx+c2xe^2x (repeated) if we have one root,
    y= e^θx*(c1cos(Bx)+c2sin(Bx)
     
  5. Dec 2, 2012 #4
    d^2y/dt^2 + 2 dy/dt + 6y = 10sinωt

    I found imaginer roots ;

    r1 = (-1-10i)
    r2 = (-1+10i)

    But matlab says that roots are ;

    -1.0000 + 2.2361i
    -1.0000 - 2.2361i

    So, what? is it a non-lineer homogenous dif eq.?

    IF we use the general solution formula ;

    y= e^θx*(c1cos(Bx)+c2sin(Bx)

    where can we use our roots?


    Edit : -1.0000 + 2.2361i
    -1.0000 - 2.2361i are roots.I have forgot to take the square root of ∇
     
    Last edited: Dec 2, 2012
  6. Dec 2, 2012 #5
    y'' + 2y' + 6y = 10sinωt

    r1=(-1 + 2,236)i
    r2=(-1-2,2236)I

    Yp = Asinωt + Bcosωt
    Y' = ωAcosωt -ωBsinωt
    Y'' = -ω^2Asinωt - ω^2Bcosωt

    (-ω^2Asinωt-ω^2Bcosωt)+2ωAcosωt-2ωBsinωt+6Asinω+6Bcosωt=10sinωt

    -ω^2A-2ωB+6A=10
    -ω^2B+2ωA+6B=0

    A = 10/(-ω^2)+6b
    B=0

    Yp=10/(-ω^2)+6bsinωt

    y= e^θx*(c1cos(Bx)+c2sin(Bx)+10/(-ω^2)+6bsinωt
     
    Last edited: Dec 2, 2012
  7. Dec 2, 2012 #6

    haruspex

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    So can you answer all the questions now?
     
  8. Dec 2, 2012 #7
    No.Because my expression is so complex :D
     
  9. Dec 2, 2012 #8

    haruspex

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    Did you verify that it satisfies the equation?
     
  10. Dec 2, 2012 #9
    No? How can I do it? Where can I use roots.
     
  11. Dec 2, 2012 #10

    haruspex

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    Your solution for the roots of the homogeneous equation was correct and leads to
    y = Ae(-1+i√5)t+Be(-1-i√5)t
    You can rewrite that as
    y = e-t(C cos(αt) + D sin(αt)) where α=√5.
    Don't confuse that α with the given ω.
    Now we just have to find a particular solution for the inhomogeneous equation. Clearly this will be of the form A cos(ωt) + B sin(ωt) (different A and B from before).
    you correctly obtained
    2A-2ωB+6A=10
    2B+2ωA+6B=0
    but I don't understand where you went from there. How did you deduce B = 0?
    Wrt q 2, what is γ in this context?
     
  12. Dec 2, 2012 #11
    2A-2ωB+6A=10
    2B+2ωA+6B=0 I can't solve it.I had a mistakes.I just figure now.There's no γ in my context.
     
  13. Dec 2, 2012 #12

    haruspex

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    You still can't solve it? ω is a given constant here. It's just a pair of linear simultaneous equations in A and B.
    Q2 reads:
    2) Show that the amplitude of the vibration damping halved on time 1,39/γ.​
     
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