# Thread:

1. ## 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 !)

2. 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).

3. 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.

4. That makes two of us. I'll put this on my "subscribed" list and check back on it later.

-Dan

5. 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.

6. 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.

7. 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.

8. 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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