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Kinematics (rectilnear motion) spring.

  1. Dec 25, 2013 #1
    the 14 in. spring is compressed to an 8 in length, where it is released from rest and accelerates black A. the acceleration has an initial value of 400ft/sec^2 and then decreases linearly with the x-movement of the black, reaching zero when the spring regains its original 14 in. length. calculate the time for the black to go to 3 in.

    just wondering If I am on the right track. I believe that is the right function for acceleration of a displacement. I am unsure about initial velocity though. Yes I made the adjustment to a=-240x+3360
    [URL=http://s1341.photobucket.com/user/nebula-314/media/20131225_150014_zpsac8609aa.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131225_150014_zpsac8609aa.jpg[/URL][/PLAIN]
     
    Last edited: Dec 25, 2013
  2. jcsd
  3. Dec 25, 2013 #2

    haruspex

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    How do you get x = -6in. for the compressed length?
     
  4. Dec 26, 2013 #3
    yea idk I took x to be in the positive direction to the right, thus giving me a negative X.

    I want to use the equation of acceleration as a function of position a=f(s). and.

    [itex]\int^{V}_{V0}[/itex]vdv=[itex]\int^{S}_{s0}[/itex]f(s)ds

    to derive me velocity as function of position then, use that to further drive the equations for time.

    I am just having trouble with the initial acceleration function.
     
  5. Dec 26, 2013 #4

    haruspex

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    The spring has a relaxed length of 14 in. When it is compressed to being only 8 in. long, how much has it been compressed by?
    Your diagrams on the RHS of the OP are correct, but the graph on the left is wrong. I think you have been inconsistent in what you mean by 'x'. In the lower diagram you have x as the change in length, which starts at 0 and becomes -6. But in the graph you have x changing from 14 to -6.
     
  6. Dec 26, 2013 #5
    Ok my new strat, a= [itex]\frac{-2400x}{7}[/itex]+4800
    (general integration setup
    [itex]\int^{V}_{0}[/itex]vdv=[itex]\int^{X}_{0}[/itex] ([itex]\frac{-2400x}{7}[/itex]+4800)dx

    gives me V= [itex]\sqrt{\frac{-2400x^2}{7}+9600x }[/itex]

    then, velocity as a function of position.

    [itex]\int^{t}_{0}[/itex]dt=[itex]\int^{3}_{0}[/itex][itex]\frac{dx}{ \sqrt{\frac{-2400x^2}{7}+9600x }}[/itex]


    put that into wolfram
    gives me .0360 sec, book says .0370 sec
     
    Last edited: Dec 26, 2013
  7. Dec 26, 2013 #6
    This time I used the position of 8 in as x=0, which allows me to generally set up an integration for velocity where initial position is zero. does this look correct?

    20131226_150252_zps042894e0.jpg
     
  8. Dec 26, 2013 #7
    got it guys finally!!!, Yes you were right haruspex I was having HUGE difficulty as to what was being meant by "X". Thanks for the help haruspex.

    [URL=http://s1341.photobucket.com/user/nebula-314/media/20131226_161422_zps90413be0.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131226_161422_zps90413be0.jpg[/URL][/PLAIN]
     
  9. Dec 26, 2013 #8

    haruspex

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    I confirm the answer. My method is rather different.
    We know it will be SHM with an amplitude of 6 in.
    Just plug the known facts in to find omega, then find the time at x = -3in.:
    ##x = -6 \cos(\omega t)##
    ##\ddot x = 6 \omega^2 \cos(\omega t)##
    At t = 0:
    ## 4800 = 6 \omega^2 ##, ##\omega = 20\sqrt 2##
    After 3 in. of expansion:
    ## x = 3 = -6 \cos(\omega t_3)##
    ## \omega t_3 = \pi/3##
     
  10. Dec 26, 2013 #9
    Yea I saw those equations else where, but I wanted to use the method the book was giving me.
     
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