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Math Help - Another Application Problem

  1. #1
    Super Member
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    Another Application Problem

    Hi

    The following question i am having trouble completing

    The equation of motion of a mass on the end of a vertical spring is
    \frac{d^2y}{dt^2} +4y = 0
    (where y is the displacement of the mass from the static equilibirum position). Given that initially ( i.e at time t=0) the displacement (y) is 3 and the velocity (y') is 0, find the solution to the equation in the form y=Ccos(3t-\theta).

    This is what i have done:

    r^2+4 = 0
    r=\sqrt{4}j
    y=Acos(2t)+Bsin(2t)

    y(0)=3
    3 = Acos(0)+Bsin(0)
    A=3

    y'=-2Asin(2t)+3Bcos(2t)
    y'(0)=0
    0=-2Asin(0)+2Bcos(0)
    B=0

    y=3cos(2t)

    what should i do next in order to get it in the form of y=Ccos(3t-\theta).

    P.S
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Paymemoney View Post
    Hi

    The following question i am having trouble completing

    The equation of motion of a mass on the end of a vertical spring is
    \frac{d^2y}{dt^2} +4y = 0
    (where y is the displacement of the mass from the static equilibirum position). Given that initially ( i.e at time t=0) the displacement (y) is 3 and the velocity (y') is 0, find the solution to the equation in the form y=Ccos(3t-\theta).

    This is what i have done:

    r^2+4 = 0
    r=\sqrt{4}j
    y=Acos(2t)+Bsin(2t)

    y(0)=3
    3 = Acos(0)+Bsin(0)
    A=3

    y'=-2Asin(2t)+3Bcos(2t)
    y'(0)=0
    0=-2Asin(0)+2Bcos(0)
    B=0

    y=3cos(2t)

    what should i do next in order to get it in the form of y=Ccos(3t-\theta).

    P.S
    Nothing it can't be done, because the solution has period \pi and the target form has period 2\pi/3
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  3. #3
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    well according to my book's answer it can be done.

    This is what the answer is: y=5cos(2t-53^{\circ} 08') or y=5cos(2t-0.9271^{c})
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  4. #4
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    Which is not the form the question wanted it in. Seems like a typo in the question to me.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Paymemoney View Post
    well according to my book's answer it can be done.

    This is what the answer is: y=5cos(2t-53^{\circ} 08') or y=5cos(2t-0.9271^{c})
    Asserting that it can be done and then showing it not done! Nice, I will stick with as asked it can't be done. If it had asked for the solution in the form y=A\cos(\omega t+\phi) of course my answer would be different.

    Also, don't mix degrees and radians (here you should only be using radians)

    CB
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  6. #6
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    checked the answers with lecturer and he said that the book's incorrect.
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