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Mx''+dx'+kx=constant*t^2*cos(constant*t^2) ?

  1. Nov 19, 2009 #1
    mx''+dx'+kx=constant*t^2*cos(constant*t^2) ???

    Hey folks out there.

    Im having a problem with this differential equation.

    The mathematical model beneath is supposed to describe a nacelle with blades, standing og a tower, as shown in the embedded picture. The stiffness k of the beams are know, aswell as the damping d and the mass m. The unbalance c is also know. What i need to describe, is the first 20 seconds (t=0..20), where the motor starts, and the angular velocity of the nacelle varies linear from 0 up to 10 Hz. The angular velocity is therefore given by omega=Pi [rad/s^2]*t

    So now the equations:

    mx''+dx'+kx=p(t) where

    p(t)=c*Pi^2*t^2*cos(Pi*t^2), t=0..20

    All constants are known. I only need to describe the solution for t=0..20.

    I seem to get very complex results - and it should indeed be solveable. Any help - links or solutions is very very much appreciated.



    Attached Files:

  2. jcsd
  3. Nov 20, 2009 #2
  4. Nov 20, 2009 #3
    Re: mx''+dx'+kx=constant*t^2*cos(constant*t^2) ???

    Your ODE cam be solved in Maple exactly. General solution is not so complicated

    ans := x(t) = exp(1/2/m*(-d+(d^2-4*k*m)^(1/2))*t)*_C2+exp(-1/2*(d+(d^2-4*k*m)^(1/2))/m*t)*_C1+1/64*c/(d^2-4*k*m)^(1/2)*(((1-I)*d*(d^2-4*k*m)^(1/2)-(4+4*I)*Pi*m^2+(2-2*I)*k*m+(-1+I)*d^2)*erf(1/8*((1-I)*(d^2-4*k*m)^(1/2)+(4+4*I)*Pi*t*m+(-1+I)*d)*2^(1/2)/Pi^(1/2)/m)*exp(1/8*(-4*d*Pi*t*m-I*d^2+2*I*k*m+d*(d^2-4*k*m)^(1/2)*I+4*(d^2-4*k*m)^(1/2)*Pi*t*m)/Pi/m^2)+erf(1/8*2^(1/2)/Pi^(1/2)*((-1-I)*(d^2-4*k*m)^(1/2)+(-4+4*I)*Pi*t*m+(1+I)*d)/m)*((-1-I)*d*(d^2-4*k*m)^(1/2)+(4-4*I)*Pi*m^2-(2+2*I)*k*m+(1+I)*d^2)*exp(-1/8*(4*d*Pi*t*m-I*d^2+2*I*k*m+d*(d^2-4*k*m)^(1/2)*I-4*(d^2-4*k*m)^(1/2)*Pi*t*m)/Pi/m^2)-((1+I)*d*(d^2-4*k*m)^(1/2)+(4-4*I)*Pi*m^2-(2+2*I)*k*m+(1+I)*d^2)*erf(1/8*((1+I)*(d^2-4*k*m)^(1/2)+(-4+4*I)*Pi*t*m+(1+I)*d)*2^(1/2)/Pi^(1/2)/m)*exp(1/8*(-4*d*Pi*t*m+d^2*I-2*I*k*m+d*(d^2-4*k*m)^(1/2)*I-4*(d^2-4*k*m)^(1/2)*Pi*t*m)/Pi/m^2)-((-1+I)*d*(d^2-4*k*m)^(1/2)-(4+4*I)*Pi*m^2+(2-2*I)*k*m+(-1+I)*d^2)*erf(1/8*2^(1/2)/Pi^(1/2)*((-1+I)*(d^2-4*k*m)^(1/2)+(4+4*I)*Pi*t*m+(-1+I)*d)/m)*exp(-1/8*(4*d*Pi*t*m+d^2*I-2*I*k*m+d*(d^2-4*k*m)^(1/2)*I+4*(d^2-4*k*m)^(1/2)*Pi*t*m)/Pi/m^2)-8*(d^2-4*k*m)^(1/2)*m*cos(Pi*t^2)*2^(1/2))*2^(1/2)/m^2;

    It'll be better if you assign the initial conditions, for example


    and then solve the ode with given ic


  5. Nov 20, 2009 #4
    Re: mx''+dx'+kx=constant*t^2*cos(constant*t^2) ???

    Thanks, everyone.

    So when i put in my initial conditions, and the values of m, d and k, i get out this from Maple 13:

    ans := x(t) = (1/125650000)*exp(-(3/4)*t)*Pi^2*sqrt(359)*((Int(cos((7/4)*sqrt(359)*_z1)*_z1^2*cos(Pi*_z1^2)*exp((3/4)*_z1), _z1 = 0 .. t))*sin((7/4)*sqrt(359)*t)-(Int(sin((7/4)*sqrt(359)*_z1)*_z1^2*cos(Pi*_z1^2)*exp((3/4)*_z1), _z1 = 0 .. t))*cos((7/4)*sqrt(359)*t))

    , which is quite simple actually. However, how do i get rid of the "_z1" - can I in some way tell maple that is only has to solve the equation form t=0 to t=20?

    THanks again!

  6. Nov 20, 2009 #5
    Re: mx''+dx'+kx=constant*t^2*cos(constant*t^2) ???

    That _z1 is the variable of integration. See those "Int" bits in there?

    If you say value(...) on your right-hand-side Maple will do the integrals, and the answer is not so short afterward.
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