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Math Help - Application DE (dampened motion)

  1. #1
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    Application DE (dampened motion)

    I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

    I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

    then it ask: intaially the spring starts out at 1 meter below equilbrium

    so i put that x(0) = 1

    then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?
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    Quote Originally Posted by neven87 View Post
    I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

    I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

    then it ask: intaially the spring starts out at 1 meter below equilbrium

    so i put that x(0) = 1

    then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?
    First of all, what is "a 10 times dampening"?

    The general equation is
    x^{\prime \prime} + \beta x^{\prime} + \omega_0^2 = 0

    \beta is the damping constant. Are you trying to say that \beta = 10~N/ms?

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neven87 View Post
    I was given a spring in a 10 times dapmening with a 1 kg mass and a constand of 16N/m.

    I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

    then it ask: intaially the spring starts out at 1 meter below equilbrium

    so i put that x(0) = 1

    then i try to find the constants and i come out with 1 = c_1+C_2 which gets me no where. What am i missing ?
    Quote Originally Posted by topsquark View Post
    First of all, what is "a 10 times dampening"?

    The general equation is
    x^{\prime \prime} + \beta x^{\prime} + \omega_0^2 = 0

    \beta is the damping constant. Are you trying to say that \beta = 10~N/ms?

    -Dan
    I'm going to assume my previous post is correct since it gives the same solution to the harmonic motion differential equation as the one you posted.

    There is a second initial condition:
    x^{\prime}(0) = 0~m/s
    since the mass is released at 1 m.

    -Dan
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  4. #4
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    yes, you are correct, sorry the book only calls it beta...

    yes, but the book's answer is two parted and with the derviatve i still get no where when i sub x(0) = 0 in

    the answer is x(T)= 4/3e^-2t - 1/3e^-8t for the distance

    and -2/3e^-2t + 5/3 e^-8t for the velocity.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neven87 View Post
    I have come out with the equation that x(t) = C_1 (e^-8t) + C_2 (e^-2t)

    x(0) = 1
    Quote Originally Posted by topsquark View Post
    There is a second initial condition:
    x^{\prime}(0) = 0~m/s
    since the mass is released at 1 m.

    -Dan
    So we have that
    1 = C_1 + C_2

    Now,
    x^{\prime}(t) = -8C_1e^{-8t} - 2C_2e^{-2t}

    Thus
    0 = -8C_1 - 2C_2

    From the first equation I get that
    C_2 = 1 - C_1

    Inserting this into the second equation gives:
    0 = -8C_1 - 2(1 - C_1)

    0 = -8C_1 - 2 + 2C_1

    0 = -6C_1 - 2

    6C_1  = -2

    C_1 = -\frac{1}{3}

    Thus
    C_2 = 1 - \frac{-1}{3} = \frac{4}{3}

    Thus
    x(t) = -\frac{1}{3}e^{-8t} + \frac{4}{3}e^{-2t}
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  6. #6
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    D'oh, algerbra mistake....its not the DE that gets ya

    Thanks for slapping my math sense.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neven87 View Post
    D'oh, algerbra mistake....its not the DE that gets ya

    Thanks for slapping my math sense.
    No problem. I've found that most Calculus students have more problems with the algebra than the Calculus, so you're fairly typical.

    -Dan
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  8. #8
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    Well that is how it goes, i honestly didnt pay attention in high school. When you get to cal and need a refresher it is hard to go back to lower level classes ; like when i got to college and want to take pre cal and cal, my tutition wasnt going to be covered by fincial aid, so i work with what i have
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