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Math Help - General solution of this equation

  1. #1
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    General solution of this equation

    I need to get a general solution of the following:

    ma +5kx = -mgsin(alpha) +11klo

    m = mass, k stiffness, g gravity, lo natural length of spring
    I think that I need it in the form of:

    ab^2 + ab + c = 0

    so i can get the auxiliary equation and therefore find the general solution but Im not sure how to rearrange to get in this form

    cheers!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    I need to get a general solution of the following:

    ma +5kx = -mgsin(alpha) +11klo

    m = mass, k stiffness, g gravity, lo natural length of spring
    I think that I need it in the form of:

    ab^2 + ab + c = 0

    so i can get the auxiliary equation and therefore find the general solution but Im not sure how to rearrange to get in this form

    cheers!
    Well,
    a = \frac{d^2x}{dt^2}, so

    m \cdot \frac{d^2x}{dt^2} + 5kx = -mg~sin(\alpha) + 11kl_0
    if that's what you are looking for.

    -Dan

    [Pet Peeve]
    g is "the acceleration due to gravity." Gravity is a force, not an acceleration.
    [/Pet Peeve]
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  3. #3
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    Right I understand that bit but its the next part Im stuck on!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    m \cdot \frac{d^2x}{dt^2} + 5kx = -mg~sin(\alpha) + 11kl_0
    Quote Originally Posted by thermalwarrior View Post
    Right I understand that bit but its the next part Im stuck on!
    The auxiliary equation for the homogeneous solution is
    mq^2 + 5k = 0

    So
    q = i \sqrt{\frac{5k}{m}}

    which makes the general solution to the homogeneous equation
    x(t) = A~sin \left ( t \sqrt{\frac{5k}{m}} \right ) + B~cos \left ( t \sqrt{\frac{5k}{m}} \right )

    Now find a particular solution and apply your initial conditions.

    -Dan
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  5. #5
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    i think i understand now. but where does the t variable come into it?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    i think i understand now. but where does the t variable come into it?
    The general solution of your auxiliary equation is q_1, q_2, so the general solution to your homogeneous equation is
    x(t) = Ae^{q_1t} + Be^{q_2t}

    I'm not sure exactly what you are confused about? t is simply your independent variable in terms of which you are solving for x.

    -Dan
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