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Crapy Linear Algebra Text book?

  1. Sep 14, 2004 #1
    I was reading through my linear algebra book and came across a picture of a mass on a spring! We are not on that chapter but i decided to look at it anyways since we just finished oscillations in physics! :wink: . But it seems to me that the author of the linear alebgra book might not understand phyiscs very well, or theres something that Im missing.

    he says that the oscillation of the spring can be modeled by the linear equation:

    [tex] y(t) = C_1 sin( \omega t) + C_2 cos( \omega t) [/tex]

    y(t) = C_1 sin( \omega t) + C_2 cos( \omega t) ( my latex wont work for some reason? I cant post the formula up. its got a red x.!?)


    Where in the heck did this extra trig function come from!? I asked my physics teacher and he said, its problably some "damm mathematician" doing a problem without understanding what actually happens. We both agreed that either one of the weights, C1 or C2 has to be zero. They cant both have simulatious nonzero numbers. Do you agree with us? Or is there something else to the problem than meets the eye?
     
    Last edited: Sep 14, 2004
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  3. Sep 14, 2004 #2

    krab

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    You're wrong, and so is your physics teacher. The equation you gave is the same as y(t) = C sin( \omega t + \phi); you can use either C_1 and C_2 as constants, or C and \phi. As an exercise, use the sine sum rule to write C_1 in terms of C and \phi and same for C_2.
     
  4. Sep 14, 2004 #3

    Gokul43201

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    What krab is saying is "expand sin( \omega t + \phi) using the sum rule and you'll see that this is what your linear algebra book was talking about".

    What is \phi ? It is the initial phase, {or y(t=0) = C sin ( \phi)}.

    I can't imagine any physics teacher would say what you claim yours did. Either he misunderstood you or you misunderstood him.
     
  5. Sep 14, 2004 #4

    Ah i see what you mean. But then in the standard physics book equation,

    y(t) = C cos(wt + phi), C is the initial amplitude of displacement. In the linear alegbra book, C1 and C2 would not be the inital displacement then would it? It just seems rather silly to break apart the y(t) = C cos(wt + phi) into two separate parts where you now have two constants C1 and C2 that are unknown. We just looked at this problem for literaly 1 min when class ended. So we dident really look at it very hard, and overlooked what you guys are talking about. In the linear books equation, which trig function would be the initial displacement phi? or is that not how its broken down between the two. It seems from your text that one of the trig functions would equal the value for cos(phi) .
     
  6. Sep 14, 2004 #5

    Gokul43201

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    No, C is not the initial displacement; it is a scaling factor or amplitude. Without the C, the displacement can only take values in the range [-1,1].

    The initial displacement comes from \phi.
     
  7. Sep 14, 2004 #6

    Tide

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    In any case, you are talking about the solution of a second order ODE which must have TWO linearly independent solutions and either form will work - the two explicit trig functions (with two multipliers) or the additional phase in the single trig function with a multiplier. The two arbitrary constants can be determined by the initial conditions which typical relate to initical amplitude and velocity.
     
  8. Sep 14, 2004 #7
    O, I havent taken differential equations yet. That must be the reason I did not recognize it. Your so smart tide :-)
     
  9. Sep 14, 2004 #8

    Im sorry, but I said it is the initial amplitude of displacement. Isint that the same thing you said? What do you guys mean by expand by the sum rule? I havent heard of that one before.
     
  10. Sep 15, 2004 #9
    sin (X+ Y) = sinx cosy + cosx siny

    is that the sine rule you speak of? Y would be the phi, and x would be the omega t i suppose. so i would have sin (wt + phi) = sin (wt)cos(phi) +cos(wt) sin(phi)


    which is the same as what i posted,

    y(t) = C_1 sin( wt) + C_2 cos( wt), so C1 would have to be cos(phi) and c2 would have to be sin of phi. Very neat! Thanks krab. I learned this stuff way back in trig long ago and forget it all. How did you remember that!!? :-) amazing.
     
  11. Sep 15, 2004 #10
    but hes right no instructor would make that error.
     
  12. Sep 15, 2004 #11

    BobG

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    It's called not looking at the problem long enough to realize it's the same equation in a different form.

    Which way do you learn to express a trig equation in order to graph it? That's why the 'y(t) = C sin( \omega t + \phi)' equation is used so often that the instructor didn't recognize the equation you showed him.
     
  13. Sep 15, 2004 #12
    Yeah, I have about 3-4 different phyiscs books and NONE of them use the decomposed sum of sin version. Thats why i think we were thrown off when we saw it.
     
  14. Sep 15, 2004 #13

    Integral

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    Try looking in a math book presenting differential equations. When doing a solution of the second order DE, d2x = -kx dt2 that is the common form of the solution. In that form the constants A and B have clear meaning.
    I had a very detailed long post written out detailing this but hit the refresh button instead of spell check! BYE BYE post. you'll have to make do with the cliff note venison. Sorry!
     
  15. Sep 15, 2004 #14
    aww shucks :-p
     
  16. Sep 15, 2004 #15

    Integral

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    I'll try again when LaTex is up....




    Warren?
     
  17. Sep 16, 2004 #16

    HallsofIvy

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    To rephrase the instructor's comment 'its problably some "damm PHYSICIST" doing a problem without understanding what actually happens.'
     
  18. Sep 16, 2004 #17
    LOL I will tell him that one HallsofIvy, :-p, actually in the theory of self preservation of ones grade, I will just show him what you guys told me, and he will probabily realize his error. However, he does understand the PHYSICS, it was just the math redone in another form. :-)
     
  19. Sep 16, 2004 #18
    I just mentioned the sum of sin and he immediately said oh oh yes, cos sine minus sin cos etc. So he problably just did not recognize it at first.
     
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