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Math Help - Speed Bump Dynamics

  1. #1
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    Speed Bump Dynamics

    so far i have found the problem for when the tire is before the bump and during but i am having trouble with finding after it always turns out to be equal to the before so im looking for someone to check my work. to save time i scanned the problem and all of my work and put it on photobucket so any help would be great.
    here are the links
    http://i1199.photobucket.com/albums/...238problem.jpg
    http://i1199.photobucket.com/albums/.../mathstuff.jpg
    http://i1199.photobucket.com/albums/...mathstuff0.jpg
    http://i1199.photobucket.com/albums/...mathstuff2.jpg

    thanks for the help
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  2. #2
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    Your choice of a particular function for the method of undetermined coefficients is incorrect. You're going to have to have the function

    y_{p}=A\cos\left(\dfrac{\pi V t}{L}\right)+B\sin\left(\dfrac{\pi V t}{L}\right).

    In solving the DE for the region

    -\dfrac{L}{2V}<t<\dfrac{L}{2V},

    you're not applying the initial conditions at the correct location. It needs to be at

    -\dfrac{L}{2V}, not -\dfrac{\pi}{2V}.

    Try carrying those corrections through and see what happens.

    Very interesting and fun problem, by the way! Thanks for posting!
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  3. #3
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    ok

    I know I didn't show the canceling out when I solved the problem but it states that L=pi so if you were to plug that into your suggestion it would give you the same thing unless canceled out wrong
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  4. #4
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    Oh, gotcha. Ok, time to peruse your solution further, then. Hang on a bit.
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  5. #5
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    In the mathstuff0.jpg file, where you're applying the initial condition to the derivative, I would simplify a bit:

    y'\left(-\dfrac{\pi}{2V}\right)=0=-C_{1}\left(\sin\left(-\dfrac{\pi}{2V}\right)+\dfrac{\cos^{2}\left(-\dfrac{\pi}{2V}\right)}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}

    =-C_{1}\left(\dfrac{\sin^{2}\left(-\dfrac{\pi}{2V}\right)+\cos^{2}\left(-\dfrac{\pi}{2V}\right)}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}

    =-C_{1}\left(\dfrac{1}{\sin\left(-\dfrac{\pi}{2V}\right)}\right)+\dfrac{V}{1-V^{2}}\dots

    I think your simplifications are going awry somewhere, because the solution for this region that I get is

    y(t)=\dfrac{\cos(V t)-V\sin\left(t+\dfrac{\pi}{2V}\right)}{1-V^{2}}.

    Also note that, once you've got this solution, you'll need to obtain the initial conditions for the far right region by plugging in \pi/(2V) into this solution. That's why you're not going to get zero for the far right region: you don't have zero initial conditions (well, most likely you don't).

    Make sense?
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  6. #6
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    got it

    I eventually solved it just took some time and determination and one thing I had wrong in my work was a negative sign was off in C1 so thanks again for the help
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  7. #7
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    You're welcome for whatever help I gave!
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