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Solving for normal frequencies with a gravity term

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Hey guys, I have been making progress in classical mechanics but I'm having a tough time figuring out how to solve for the normal frequencies for this set of equations of motion:
    http://imgur.com/nQKPNsu

    In order to account for the mass gravity term here the book suggests setting new primed variables to create new simpler equations of motion like so
    http://imgur.com/vQBGInD

    I'm comfortable with everything here except that I really don't know how or why the defined those primed variables the way they did. I don't know how they fit in to the new equations of motion or resemble the old unprimed ones. Any help here would be aweomse. Thank you.


    2. Relevant equations
    http://imgur.com/nQKPNsu
    http://imgur.com/vQBGInD
    3. The attempt at a solution
    I tried adding the old equations of motion together to try and figure out if they found a relationship between the y variables and then exploited it when defining the primed ones but I had no luck there.
     
  2. jcsd
  3. Dec 6, 2015 #2

    mfb

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    Staff: Mentor

    You can solve the given transformations for y1 and y2 (trivial as it is just a sum), and plug those into the original equations, then you'll see how it works.

    Unrelated: Defining y''A = y'1 + y'2 and y''B = y'1 - y'2 would decouple the equations.
     
  4. Dec 6, 2015 #3
    Awesome, thanks; I found expressions for the unprimed y's in terms of the primed ones and plugged them back in to get out the expressions they have in terms of the primed variables except with a '+mg' still tagged on.. Also, I'm still unsure about how you would go about figuring out the expressions for the primed variables in the first place?
     
  5. Dec 7, 2015 #4
    All they're doing is solving for the values the y's would have if the mass-spring system were at equilibrium (zero y'' terms). They then represent the actual y's by these equilibrium values plus the deviations from the equilibrium values. They then solve for the deviations. This gets rid of the gravitational terms in the differential equations.

    Chet
     
  6. Dec 7, 2015 #5

    mfb

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    The +Mg have to disappear, that is the whole point of the substitution.
    How to find them: Usually by guessing in a clever way, with some experience. Looking for equilibrium positions is always a good start.
     
  7. Dec 7, 2015 #6
    Okay, so maybe take [itex]\frac{\partial}{\partial t}[/itex] of the expression and set this to zero to find y'?
     
  8. Dec 7, 2015 #7

    mfb

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    Staff: Mentor

    Where do you want to find y (and what does that mean) how?
     
  9. Dec 7, 2015 #8
    I guess to find the equilibrium position we would take the derivative of the potential with respect to y and set that to zero. However, this doesn't leave any y terms..
     
  10. Dec 7, 2015 #9
    If you want to find the equilibrium position, just set the accelerations to zero, and solve for the corresponding y's.
     
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