Crapy Linear Algebra Text book?

AI Thread Summary
The discussion revolves around the modeling of oscillations in a spring using linear algebra, specifically the equation y(t) = C_1 sin(ωt) + C_2 cos(ωt). Participants clarify that this equation is equivalent to y(t) = C sin(ωt + φ), where C represents amplitude and φ is the initial phase. There is confusion regarding the constants C_1 and C_2, with the consensus that they can be expressed in terms of C and φ using the sine sum rule. The conversation highlights the importance of recognizing that both forms represent the same physical phenomenon, emphasizing the need for understanding the underlying mathematics in physics. Ultimately, the participants agree that the instructor's initial misunderstanding stemmed from not recognizing the equivalence of the two equations.
Cyrus
Messages
3,237
Reaction score
17
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 there's something that I am missing.

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

y(t) = C_1 sin( \omega t) + C_2 cos( \omega t)

y(t) = C_1 sin( \omega t) + C_2 cos( \omega t) ( my latex won't work for some reason? I can't 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 can't both have simulatious nonzero numbers. Do you agree with us? Or is there something else to the problem than meets the eye?
 
Last edited:
Physics news on Phys.org
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.
 
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.
 
krab said:
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.


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) .
 
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.
 
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.
 
O, I haven't taken differential equations yet. That must be the reason I did not recognize it. Your so smart tide :-)
 
Gokul43201 said:
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.


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 haven't heard of that one before.
 
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.
 
  • #10
but he's right no instructor would make that error.
 
  • #11
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.
 
  • #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.
 
  • #13
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!
 
  • #14
aww shucks :-P
 
  • #15
I'll try again when LaTex is up...




Warren?
 
  • #16
BobG said:
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.

To rephrase the instructor's comment 'its problably some "damm PHYSICIST" doing a problem without understanding what actually happens.'
 
  • #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. :-)
 
  • #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.
 
Back
Top