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3-d non- isotropic oscillator

  1. Nov 30, 2005 #1

    ok. mass held by six springs and is located at the origin. Potential function is given by V = k/2 (x^2 + 4y^2 + 9z^2). at t = 0 the mass is given a push in the (1,1,1) direction imparting vo. find x(t) y(t) z(t) numerically if k = m(pi^2). part b: will it every get back to origin, if so what t does it return with v = vo.
    what does it mean solve numerically? t is in a cosine.
    i don't get it.

    what i did is n my post #2.
     
    Last edited: Dec 1, 2005
  2. jcsd
  3. Dec 1, 2005 #2

    Tide

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    Your vector expression for the acceleration is not correct.
     
  4. Dec 1, 2005 #3
    how?
    if F=-grad(V) then F=-k(xx+4yy+9zz)
    ma=-k(xx+4yy+9zz)

    so if k = m╥² then:

    a = -╥²(xx+4yy+9zz) [eqs 1.(1,2,3)]]

    you can also go like:

    a+(k/m)(xx+4yy+9zz) = 0 [eqs 2]

    x = Acos(╥t) ; y = Bcos(2╥t) ; z = Ccos(3╥t) ;where ╥ = √(k/m) [eqs 3.(1,2,3)]

    let xDoubleDot be d²x/dt² then

    xDD = -╥²Acos(╥t) ; yDD = -4╥²Bcos(2╥t) ; zDD = -9╥²Ccos(3╥t) [eqs 4.(1,2,3)]

    combining [eqs 1.1] and [eqs 4.1] gives:

    -╥²x = -╥²Acos(╥t) ; which reduces to:

    x = Acos(╥t) ; which is just [eqs 3.1], the x solution to the homogenous DE described by [eqs 2].

    so clearly my expression for the acceleration isn't wrong.
     
  5. Dec 1, 2005 #4
    Just to point out, I would not recomdend using x,y,z in bold, but rather i,j,k instead.
     
  6. Dec 1, 2005 #5
    oh. in the first one it should be (x+4y+9z)r
    guess i did write it down wrong.


    HOW DO I FIND THE AMPLITUDES?
    i know they come from IC. it isn't going to go through (1,1,1) right? it just gets pushed that way? it says that to show that vo has no cos sin buisiness and is nonzero and of equall magnitude in all directions?
     
  7. Dec 1, 2005 #6
    i don't much care for i j k. never used it yet, not about to start now.
    if it's so i don't confuse myself, i'm not worried. i actually get less confused like this.
    if there's some other kind of reason, please tell me.
     
  8. Dec 1, 2005 #7
    3-d non-isotropic oscillator

    ok. mass held by six springs and is located at the origin. Potential function is given by V = k/2 (x^2 + 4y^2 + 9z^2). at t = 0 the mass is given a push in the (1,1,1) direction imparting vo. find x(t) y(t) z(t) numerically if k = m╥². part b: will it every get back to origin, if so what t does it return with v = vo.
    what does it mean solve numerically? t is in a cosine.
    i don't get it.
    F = -kxx -4kyy -9kzz = ma
    a + (k/m)(x+4y+9z)r = 0
    ok so then is it like:
    a+╥²(x+4y+9z)r = 0
    take this and seperate and integrate? so there's no trig functions?
    how do i find the amplitudes of the trig functions? I know they come from IC. it isn't going to go through (1,1,1) right? it just gets pushed that way? it says that to show that vo has no cos sin buisiness and is nonzero and of equal magnitude in all directions?
    this thing is driving bonkers and it shouldn't be.
     
    Last edited: Dec 1, 2005
  9. Dec 1, 2005 #8
    well, its so that when taking derivatives you dont mistakenly take the derivative of a unit orthognal pointing vector. As for your acceleration term, I get:

    [tex] [ai+aj+ak] + k/m [xi+4yj+9zk] =0 [/tex]

    your sign for k/m is wrong, its a + not a -
     
    Last edited: Dec 1, 2005
  10. Dec 1, 2005 #9

    Tide

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    First, [itex](x - 4y - 9z)\vec r[/itex] is NOT the same as [itex]x \hat i - 4y \hat j - 9z \hat k[/itex].

    Second, the vector x is understood to mean [itex]x \hat i[/itex] so when you write x x people will interpret that as [itex]x^2 \hat i[/itex]. As an alternative you could write the unit vector as [itex]\hat x[/itex] where the circumflex or hat tells the reader that you're talking about a unit vector.

    With regard to your problem, you determine the amplitudes by applying the initial conditions (you know both the initial position and initial velocity).

    So, can you determine whether the mass will return to its starting postion? :)
     
  11. Dec 1, 2005 #10
    it's right in my second post.
    and that's what i've been working wiht.
     
  12. Dec 1, 2005 #11

    Tide

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    It was wrong in the first and I explained why.
     
  13. Dec 1, 2005 #12
    r is the unit vector i guess how you guys write it is (i+j+k).
    see that's why i use x y z. i know that it isn't the same as times i just wrote it short like. then the x gets timesd on the x and the like...
    how do you make it write math style?
    i can do circonflex but only on like ê â and î. i can do dot notation on those, maybe i should use hthose. does this thing recognise mathematica fonts?
     
  14. Dec 1, 2005 #13
    r is not (i+j+k)

    r is <i,j,k>/ |i,j,k|

    (for the unit pointing vector)
     
    Last edited: Dec 1, 2005
  15. Dec 1, 2005 #14

    Tide

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    No, r is not the unit vector. It's the position vector given as [itex]x \hat i + y \hat j + z \hat k[/itex]. If you want to specify a unit vector you can use [itex]\hat r[/itex]. I recommend adopting the conventions used by most people since otherwise both you and they will be confused.

    To use the math symbols you'll need to learn a little LaTex which you can find out about by clicking on any equation someone has posted. A popup will appear with a link leading you to some reference material.
     
  16. Dec 1, 2005 #15
    so:

    .5mv²+.5kr² = .5mvo²

    v²+╥²r² = vo²

    (xD)²+╥²x² = (xDo)²

    how do i solve that for x(t) numerically?
    the re-tracing part isn't bothering me.
    it's the solving for x(t) numerically. how?
     
  17. Dec 1, 2005 #16
    if i would have known how to put a hat on it i would have.
    sorry.
    and
    r hat = (i hat + j hat + k hat)
    is what i suppose i should have wrote.
     
  18. Dec 1, 2005 #17

    Tide

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    The potential energy is not [itex]\frac {1}{2}k r^2[/itex].

    You were on the right track in #3.
     
  19. Dec 1, 2005 #18
    sorry. i wrote it wrong.
    potential energy = [itex]\frac {1}{2}k (x^2+4y^2+9z^2)[/itex]

    that still doesn't really tell me anything about:
    what does solve numerically mean?
     
  20. Dec 1, 2005 #19

    Tide

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    WTG! You figured out the LaTeX!

    I am not sure what the authors of your problem mean by "solve numerically" but it usually means that you use approximation methods to find the solutions. However, in this case you don't need to resort to numerical methods since you can solve the equations directly and exactly. Does your textbook offer any guidance?
     
  21. Dec 1, 2005 #20
    from the post #5:
    xDo = yDo = zDo

    so at t = 0

    xDo = -╥A yDo = -2╥B zDo = -3╥C

    all = each other then A =2B =3C ?

    and i get the numerical value of the amplitude by....
     
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