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This spring question is has got me oscilating!

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A spring is such that it would be stretched 3 in. by a 6 lb. weight. Let the spring first be stretched 4 ins. and then a 12 lb. weight attatched and given an initial velocity of 8 ft/s find out how far the wight will drop.

    2. Relevant equations
    d2y/dx2+(k/w)y=o


    3. The attempt at a solution I can't figure out exactly what they r asking me here- do I get my k from the first sentence making k=24 and w=3/8 and use the the others as
    y(0)=1/3 and y'(0)=8, or are they asking me to use the 2nd weight and add it to the first and if so, how? is that an added force? Im really confused be the wording of this problem.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 12, 2012 #2

    tiny-tim

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    welcome to pf!

    hi hatsu27! welcome to pf! :smile:
    the clue is in the word "would" …

    it hasn't been stretched by a 6 lb. weight, but if it was, it would stretch 3 in

    so you only use that information to find k :smile:
     
  4. Apr 12, 2012 #3
    Ok from there I got my diferential eq. y" + 64 = 0 so my x(t) = c1cos(8t)+c2sin(8t) and from there my sinusoidal is ((sqrt10)/3)cos(8t+tan-1(3)) So to find out how far the wieght would drop i take the derivative and set it = to zero and solve for t yes? Does all this look good to u?
     
  5. Apr 12, 2012 #4

    tiny-tim

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    why so complicated? :redface:

    just use conservation of energy! :smile:
     
  6. Apr 12, 2012 #5
    not allowed-only allowed to use tools learned in my dif eq. class. we are specifically told to not use physics formulas
     
  7. Apr 12, 2012 #6

    tiny-tim

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    ohhh! :rolleyes:
    the method looks ok :smile: (i haven't checked the figures)

    though i don't think you'll have to find t, you may be able to read the answer off without that

    (btw, you haven't said what units you're using :redface:)
     
  8. Apr 13, 2012 #7
    The amplitude of cos is (√10)/3, so that should be how far the weight will drop in inches, if everything else is correct.
     
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