How to Derive the Differential Equation for Forced, Damped Oscillations

[qspeechc]

Homework Statement

Hi. The problem is question 1(a) in the file below:
http://www.mth.uct.ac.za/Courses/MAM24678/mod2od/Project1_07.pdf

The Attempt at a Solution

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; 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.

Please, any help on deriving the differential equation would be great

Any help is much appreciated thanks.

(P.S. I need this pronto please !)

Ok, what I did. First I said:
Y = y + p + l
l is the relaxed length of the spring (a constant), and p is the displacement from the equilibrium position of the mass on the spring. If you re-write:
p = Y - y - l
then find the equation of motion of the mass on the spring:
m[d^2(p)]/dt^2 = -kp -c(dp/dt)

And plugging in p = Y - y -l, but this does not give the correct answer. I really do not know how to get the differential equation, help please!

 

[qspeechc]

I'm wondering- is this in the correct section? Should I have posted this elsewhere?

 

[qspeechc]

Ok, I'll just write out the question here:

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.

 

[qspeechc]

Please, anyone?