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Working out experimental periodic time from a wieghted spring in s.h.m

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A spring hanging from a lab stand has an equilibrium position of 66mm.
    Weights are added in 50g increments from 100g upto 250g, record the spring displacement and work out the spring constant.

    Part 2:
    With each mass in turn pull them down 20mm from their equilibrium point and record the avergae time taken to complete 20 oscillations. Plot the reults of mass against experimental periodic time and also against periodic time squared.



    2. Relevant equations

    For spring constant : k=mg/x
    Where m= mass, g= gravity, x= spring displacement

    For periodic time: T = 2p √m/k



    3. The attempt at a solution

    The first part im sure ive done correctly,

    Number Mass (kg) Spring displacement (m) Spring constant (N/m)
    1 0.05 0 n/a
    2 0.1 0.0195 50.31
    3 0.15 0.039 37.73
    4 0.2 0.061 32.16
    5 0.25 0.081 30.28

    Im struggling with working out periodic time. The equation ive been given has 2p in it. What is p representing?

    Any help on this will be very appreciated as its driving me mad.
    thanks
     
  2. jcsd
  3. Apr 19, 2012 #2

    tiny-tim

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    welcome to pf!

    hi shaun_598! welcome to pf! :smile:
    π (pi) :wink:

    (2π ≈ 360°)
     
  4. Apr 19, 2012 #3
    Re: welcome to pf!

    Thank you, that a massive help :smile:

    So T = 2π x √m/k

    So using table in the question
    100g mass and 50.31N/m constant

    T= 2π x √0.1/50.31
    = 0.28 (Not sure what units this should be? Seconds?)

    Have i done this correctly?

    thanks
     
  5. Apr 19, 2012 #4

    tiny-tim

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    yup, looks fine! :smile:

    (and N/m is an SI measurement, so so long as everything else is also SI, the result will be too, so yes, it's in seconds :wink:)
     
  6. Apr 19, 2012 #5
    Thank you very much :smile:
     
  7. Apr 19, 2012 #6
    What ive worked out here is theoretical periodic time?

    Im also asked to work out experimental periodic time.
    Take the 100g mass. It took an average of 8.5 seconds to complete 20 oscillations.
    One oscialltion is a complete up and down so as i undertsand it this is one period.
    So is this just as simple as 8.5/20
     
  8. Apr 19, 2012 #7

    tiny-tim

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    by "up and down" do you mean "up and down left and up and down right" (one complete cycle)?

    if so, yes 8.5/20 s :smile:
     
  9. Apr 19, 2012 #8
    I mean if you pull on the spring and start when its at the bottom of its travel, wait for it to got to extent of of its travel the oppoisite way and come back to its bottom travel. Thats what we said was one oscialltion.
     
  10. Apr 19, 2012 #9
    Why is the spring constant not a constant value in your table as expected? :confused:
     
  11. Apr 19, 2012 #10

    tiny-tim

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    oh yes, that's right :smile:

    (i was thinking it was a pendulum :rolleyes:)
     
  12. Apr 19, 2012 #11
    Im not sure about that yet, im looking into it. Its what the content of my report will be focused on.
    Anyone have any ideas?
     
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