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Spring oscillation question

  1. Jan 6, 2005 #1
    Im not sure exaclty what to do with a part of this problem

    A 5 kg block is fastened to a vertical spring that has a spring constant of 1000 newtons per meter. A 3 kg block rests on top of the 5 kg block. The blocks are pushed down and released so that they oscillate.

    Determine the magnitude of the max acceleration that the blocks can attain and still remain in contact at all times.

    There are other parts of the problem but nothing before this that would help with this part.

    Now im pretty sure that the max acceleration would be g, the accleration due to gravity but I dont know how to go about it to prove with formulas Am i even correct in this assumption?
    THanks for the help
     
    Last edited: Jan 6, 2005
  2. jcsd
  3. Jan 6, 2005 #2

    dextercioby

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    Post your work.My answer is
    [tex] a_{max}=\frac{m}{m+M}g [/tex]
    ,where "m" is the mass of the smaller body,and "M" is the mass of the heavier body.
    Numerically,for [itex] g\sim 10ms^{-2}[/itex],[itex]a_{max}\sim 3.75ms^{-2} [/tex]

    Daniel.
     
  4. Jan 6, 2005 #3
    well first i just assumed it would be g.
    Now i keep hitting dead ends trying various methods.

    i tried using amax=Aw^2 and then got stuck without an amplitude.
    I tried Fnet=ma and then plugged in kx=(m+M)amax but dont have an x value. So im really not sure where to go with this
     
  5. Jan 6, 2005 #4

    dextercioby

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    What's the condition of equilibrium for the small body (being attached to the big one) ???Why would it lose contact with the larger one?

    Daniel.
     
  6. Jan 6, 2005 #5
    the upward force from the spring is greater than its weight?
     
  7. Jan 6, 2005 #6

    dextercioby

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    Exactement.Next question.What's the connection between the maximum acceleration and maximum distance from the equilibrium point??

    Daniel.
     
  8. Jan 6, 2005 #7
    max acceleration = amplitude times angular frequency squared
    amplitude would be the max distance from the equilibrium point for the max acceleration right?
     
  9. Jan 6, 2005 #8
    would it be that at the point where amax without disattachment occurs Fg=Fs so mg/k=x
     
  10. Jan 6, 2005 #9

    dextercioby

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    Exactly,u're on the right track.Put it all together now and post your work.

    Daniel.
     
  11. Jan 6, 2005 #10
    Fnet=ma
    Fs-Fg=(m+M)amax
    amax=(kx-(m+M)g)/(m+M)
    amax=k((m+M)g)/k)-(m+M)g)/(m+M)
    amax=0/(m+M)

    obviously i screwed something up
    also why wouldnt the max acceleration of the blocks in order to remain together just be the acceleration due to gravity? Wouldn't the blocks have to be accelerating faster than that in order to keep the block that is not attached moving after the block attached to the spring has reached the max x value and starts mvoing back down the spring?
     
  12. Jan 6, 2005 #11

    dextercioby

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    Obviously one of us is wrong.It may be me,or it may be.
    The second law of dynamics for the small body is
    [tex] m\vec{a}=m\vec{g}+\vec{F}_{el,max}=\vec{0} [/tex]
    ,because the small body must be imponderable (its's accleration in the reference system of the big body it is attached to is zero).
    [tex] F_{el,max}=kA [/tex]

    From these 2 eq.u find:
    [tex] A=\frac{mg}{k} [/tex]

    The maximum acceleration is
    [tex] a_{max}=\omega^{2}A=\frac{k}{M+m}\frac{mg}{k}=\frac{m}{M+m}g [/tex]

    Daniel.

    PS.We need a third party... :tongue2:
     
  13. Jan 6, 2005 #12
    yea i think we need a 3rd party bc one of my friends in my class is telling me its just g and that im thinking about it too much so i dont know
     
  14. Jan 6, 2005 #13
    Just as a different point of view(i'm not sure if it's right or not):

    We can try using energy to solve this: We know the max height is 2x, due to the oscillation.

    so

    kx^2/2 = mg(2x)

    kx = 4mg = Fspring.

    Fspring/m = acceleration = 4g.
     
  15. Jan 6, 2005 #14

    dextercioby

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    I'm sorry,but it's not valid.There are two bodies implied in the interaction with the spring.U cannot make abstraction of the bigger one.
    Besides,your result is absurdly large.

    Daniel.
     
  16. Jan 6, 2005 #15
    i think im gonna go with amax=g but ill let u guys know what it turns out to be when i get the hw back monday
     
  17. Jan 6, 2005 #16
    Actually my answer should've been 3g, forgot about the net force. As for dexter... i don't understand why my answer would be wrong.
     
  18. Jan 7, 2005 #17

    learningphysics

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    Why didn't you use (m+M)a=(m+M)g + Fel=0 with Fel=kA leading to A=(m+M)g/k?

    With this you get max acceleration=g.
     
  19. Jan 7, 2005 #18

    dextercioby

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    Yes,you're right,i was an idiot,i forgot that those two bodies are joint together and the 2nd law should apply on the whole body.

    I appologize again.I was really tired and semi-drunk... :tongue2:

    Daniel.
     
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