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Math Help - Forced, Damped Oscillations

  1. #1
    Junior Member qspeechc's Avatar
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    Forced, Damped Oscillations

    Hi people. Here's the file: http://www.mth.uct.ac.za/Courses/MAM...roject1_07.pdf

    Question 1(a) is the one I have a problem with. I just don't know what he's getting at. Is y(x) the function that describes the road? And comparing y(t) and y(x) implies, to me, that x=vt, so it has a constant velocity with respect to the x-axis (in the direction of the x-axis); a very odd thing to do...
    Is Y then the vertical displacement of the vehicle from the x-axis? So the car is like a mass on a spring on the road? I have no idea how he derived that Differential Equation.

    Any help is much appreciated thanks.

    (P.S. I need this pronto please !)
    Last edited by qspeechc; April 7th 2008 at 09:41 AM.
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  2. #2
    Junior Member qspeechc's Avatar
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    Even a simple: "Yes I agree, the question is confusing and badly written. I do not know what he's on about" would be very nice thank-you (my prof is Russian).
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  3. #3
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    Unfortunately...that link isn't working for me. I'm not sure if that's why you're not getting replies, but I'm definitely unable to reach the URL you provided.
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  4. #4
    Forum Admin topsquark's Avatar
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    That makes two of us. I'll put this on my "subscribed" list and check back on it later.

    -Dan
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  5. #5
    Junior Member qspeechc's Avatar
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    Hmm, works fine for me, but that's not the point. Oh, yes, it isn't working, it was a moment ago. Argh! This is frustrating! It appears the entire site is down.

    Ok, the question is:

    Suppose that a car oscillates vertically as if it were a mass m on a single spring with constant k, attached to a single dashpot (dashpot provides resistance) with constant c. Suppose that this car is driven along a washboard road surface with an amplitude a and a wavelength L (Mathematically the 'washboard surface' road is one with the elevation given by y=asin(2*pi*x/L).)

    (a) Show that the upward displacement of the car Y satisfies the equation:

     m\ddot{Y} + c\dot{Y} +kY = c\dot{y} + ky

    where y(t) = asin(2*pi*v*t/L)
    and v is the velocity of the car.
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  6. #6
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    As far as I can tell, your original assessment of the problem is correct. y(t) and y(x) are very much related, except for the obvious distinction of one being depending on t and the other being dependent on x, but the two equations for y you have listed both make sense considering that x=vt.

    That would lead to me to believe that Y is the displacement, and y(t) is just the height of the surface itself, because the washboard surface you're working with is sinusoidal as described by the y(t) function.
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  7. #7
    Junior Member qspeechc's Avatar
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    This is how I understand the question:

    The car can be regarded as a mass on top of a spring. The car moves along a road in the xy-plane, the road is described by y(x). The x-component of the velocity of the car is constant and x=vt (this is very odd). The distance from the x-axis to the mass on the spring is Y.

    From that we must deduce the differential equation, but how? The LHS, with the RHS=0 looks like the equation of the free decay of a damped oscillator, so the whole equation looks like that of a driven, damped oscillator. I cannot see how the road described by y(x) would give a driving force like the RHS.

    What I tried to do is this:
    Y = y + p + l

    where l is the equilibrium length of the spring with the mass (a constant), and p the extension of the spring.
    Re-arrange for p:
    p = Y - y - l

    and find the equation of the free decay of a damped oscillator in terms of the variable p, but this does not give the right answer.
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  8. #8
    Junior Member qspeechc's Avatar
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    Nevermind trying to solve for Y- the solution is extremely ugly, and he still wants us to find the amplitude as a function of velocity, puh!
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