LinkBack URL
About LinkBacksEven 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).
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 !)
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:
where y(t) = asin(2*pi*v*t/L)
and v is the velocity of the car.
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.
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.
  |
