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Energy Integral Lemma-Differential Eq. HW

  1. Oct 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the energy integral lemma to show that motions of the free undamped mass-spring oscillator my"+ky=0 objey

    m(y')^2 = ky^2=constant


    2. Relevant equations

    E(t)=1/2 y'(t)^2 - F (y(t))

    3. The attempt at a solution
    I am not sure how to even start this problem.I would like to know just what the problem is asking for.I do not have a problem doing the math to it.

    Thank you
     
  2. jcsd
  3. Nov 1, 2008 #2

    gabbagabbahey

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    If I remember correctly, the energy integral lemma says that the quantity E(t)=1/2 y'(t)^2 - F (y(t)) is a constant when y(t) satsifies the DE y''(t)=f(y) and f(y) has no explicit dependence on y' or t and [itex]F(y)=\int f(y)dy[/itex].... So, I would start by placing your ODE in the form y''(t)=f(y)...what is f(y) in this case? Does it have any explicit dependence on y' or t? If not, then E(t)=1/2 y'(t)^2 - F (y(t)) is a constant.
     
  4. Nov 1, 2008 #3
    so is my f(y)=-ky/m then i get the F(y) of that expression and plug into E(t)=
     
  5. Nov 1, 2008 #4

    gabbagabbahey

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    Yes, what does that give you?
     
  6. Nov 2, 2008 #5
    1st of all i apologize since i don't know how to write equation symbol and such..

    here is what i get

    y=-ky/m

    -k/m integral (y)= -k/m (y^2)/2

    (1/2)y'^2 - k/m(y^2)/2=k

    then i am trying to solve for y'^2 so
    i end up with

    my'^2 =2km +ky^2

    and what i am suppose to get is m (y'^2) +ky^2 =constant.
     
  7. Nov 2, 2008 #6
    so i realized i messed up my negative there since the original lemma has a negative and my f(y) has a negative so i end up with


    my'^2 + ky^2 =2km

    is it okay to re-write that expression as my'^2 + ky^2=constant since usually the values of k and m are #'s?

    so it would be my'^2 + ky^2 = constant

    ??
     
  8. Nov 2, 2008 #7
    Okay,I figured it out my question out.I don't know why i was adding an extra K.Thanks for guiding me through it.

    I have another question if you could just help me with setting it up.
    it involves the energy lemma .

    the question is
    Use the energy integral lemma to show that pendulum motions obey

    (theta prime)^2 /2 - (g/l)cos(theta) = C

    what is my f(y) in this case?
     
  9. Nov 2, 2008 #8
    okay so i think i figured out this one too:

    i did

    (theta)"^2/2= [(g/l)cos(theta) ]'

    (theta)"^2/2= -g/l sin(theta)

    f(y)= -g/l sin(theta)

    then after going through all the steps i ended up with

    (theta)'^2 /2 - (g/l)cos(theta)=C which is the original DE.


    let me know if this is okay.i am not sure if taking the 2nd derivative of the equation is allowed.
     
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