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Oscilation Differentail equation

  1. Sep 5, 2004 #1
    Oscilation differential equation

    Hello, I have a question about oscilations. In physics lecture, we derived a way of relating circular motion to a spring via a projection. And this made good sense, and could be applied to a horizontal spring as well, if we then shifted that circle track so that it is now verticle instead of horizontal. From there the professor went on to say, why do we give a damm about this? Yes, he loves to cruse in class its quite funny. He makes it very lively, but he is very smart :-), plus hes an old guy so that tops the cake. He then goes to show that this produces the same differnetail in terms of other things, like a pendulum, a circular spring etc. But the thing that changes is the terms that affect omega, the period. Ie. square root I/k etc. And he says that because we solved the case with striaght motion, playing that clever projection trick, we have solved the differential, and we can apply this anwser to other cases as well. The only thing i have a problem with is that for a spring thats circular, like in wristwatches, the motion cannot be described as back and forth projection, since it ticks side ways and back in arcs. But I thought about this and considered, could we not graph this side ways funny motions as a graph of theta vrs time. Then we would again obtain a sinusoidal function. ( or cosine if we wanted.) and because of this, we could solve based on this graph instead of that projection which only seems to make "sense" in that linear spring case. In the linear case could we not do the same thing, and graph the curve as displacment vrs time. And that would again be another sinusoidal curve. And we could call the distance the sin(x), but we avoid constraining ourselves to that clever projection. I think it is really the same thing, only the graph of the sine wave is "stretched" so to speak with time, if we did not graph time linearly on the x axis, then it would make that nice circle, and be a special case which is our projection.

    P.S. When we talk about the amplitude, we usually talk about it in terms of hte displacement x from the origin of the springs natural equilibrium point. In the case of that coil spring, ( i think thats what you call those things.),it would not be meaningful to talk about the amplitude in terms of this displacement would it? Since the differential its d2theta/ dt2 it would be more meaninful to tlak about the amplitude in terms of the angle theta that its been displaced, not the arclength s it has been displaced, is this correct? That must be one of the things that is throwing me off with that projection analogy.
    Last edited: Sep 5, 2004
  2. jcsd
  3. Sep 5, 2004 #2
    maybe my question is not clear. I was wondering if there is a way to solve the differential equation based on the cosine graphs without having to use the projection "trick" they played. Or is that the only reasonable way to solve the problem.
  4. Sep 5, 2004 #3
    I used to solve some simple harmonic motion (SHM) problems using the cosine graph.
  5. Sep 5, 2004 #4
    I think I proved it tell me if im wrong. You will see that my proof is basically the exact same thing as in any physics book, but what I tried to do was eliminate the projection part of the proof, so that it is purely mathematical. The reason I did this was beacause of the fact that i am a visual learner to the extreem. When I saw that picture of the ball going around in a circle, I though to myself that it could never relate to a coil spring, because a coil spring would not translate back and forth, but arc to the left and right through some arc length, thus the projection method would not work in that sense. What I tried to do was prove it just by using the cosine function.

    To start, I used any old cosine function. You can picture a graph of one going up and down forever. I then said the distance the thing has displaced will equal:

    [tex] x= A cos(theta)[/tex] (A is the amplitude of the cosine curve)

    and theta is measured in radians. But the thing about theta is that it varies with time, so it is not constant, and I want to derive some way of finding theta as a function of time. So what i said was that:

    [tex] theta= omega *t + theta_0 [/tex].

    Here I called omega not the angular frequency, but just some proportionality constant. The reason I say so is because the angle and the time are not always going to be equal to eachother. I.e. at time pi/2 the angle does not have to be ninety degrees, it could be something else, say 45, so therefore omega is what I am just defining as some proporionality constant at the moment so that it will reflect the curvature of my graph if it were a graph of x vs t and not x vrs theta.

    then if i do the first and second derivatives, i find that

    [tex] d^2x/dt^2 = - omega^2 A cos(omega *t + theta_0 ) [/tex]

    so we can subsitute the value of x for [tex]A cos(omega *t + theta_0 ) [/tex]

    we can then see that if hookes law is to be obeyed, then [tex] a=-k/m x [/tex]

    so we can see that this proportionality constant must equal [tex] sqrt(-k/m)[/tex].

    This gives the exact same anwser as the book has done, the only difference is that I tried to avoid the projection method all together. I suppose this was exactly what the book did, but at times when they do the same proof I did, they talk about the projection along the way, here I avoid it completely. I think this is helpful becuase if I am doing a displacement in terms of a coil spring, where the motion cannot accuratly be depicted by the projection, I can simply replace the appropriate variables, x with theta, a with alpha, m with I (moment of inertia), and it will still give me the same result, indepent of the projection. If this is correct please let me know, if not I would like to konw where i went wrong.
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