1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    Last edited: Dec 25, 2013
  2. jcsd
  3. Dec 25, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?

  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.

  9. Dec 26, 2013 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted