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Simple Harmonic Motion of a pendulum

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Hello, I'm doing a simple pendulum lab where I measure the difference in period as I increase the length of the pendulum. Simple enough. However, I'm having trouble analyzing my graphs of a linear versus a curve fit.

    Can someone explain why the curved/nonlinear fit is the proposed fit (versus linear), and what do the values "A" and "B" represent in the curve fit equation of Ax^B.

    Thank you
     
  2. jcsd
  3. Aug 10, 2010 #2

    rock.freak667

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    What are you plotting on your graph?

    How does the period T relate to the length L ?
     
  4. Aug 10, 2010 #3
    Sorry, I didn't clarify. Period in seconds is on the y axis and Length in meters is on the x axis
     
  5. Aug 10, 2010 #4

    rock.freak667

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    Do you know how to derive the relationship for small angles? From your notes on the simple pendulum do you have a formula for the period T?
     
  6. Aug 10, 2010 #5
    1) I didn't think angles were important in this particular graph because the angle was kept constant at 15 degrees for each trial

    2) T = 2pi sqrt(L/g) ?
     
  7. Aug 10, 2010 #6

    rock.freak667

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    Yes right!. T=2π√(L/g)

    Now if we use the rules of indices, (a/b)n=an/bn and knowing that √a=a1/2, what does the equation become?
     
  8. Aug 10, 2010 #7
    See I don't remember the rule of indices :S. Haven't taken math since high school; which was a while ago.

    I'm thinking the equation becomes T = L^1/2 ? Even though I'm not exactly sure what that means lol. Im sure that Length and Period are positively correlated. Or am I wrong?
     
  9. Aug 10, 2010 #8

    rock.freak667

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    If you have y=(2x/3)2, this is the same as y=(2/3)2x2 and this is the same as y= (22/32)x2.

    So what does your equation become?
     
  10. Aug 10, 2010 #9
    Do you mean, what does T=2π√(L/g) become?

    T = 2pi/(sqrt)g x L^1/2 ?
     
  11. Aug 11, 2010 #10

    rock.freak667

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    Good, so we now have this:


    T=(2π/√g)L1/2

    So if you are plotting period 'T' (y-axis) against length 'L' (x-axis), you will get something like this: y=(2π/√g)x1/2 right? Now doesn't this look like the equation y=AxB? So what are the values of 'A' and 'B'?
     
  12. Aug 11, 2010 #11
    Hmm, A = 2.01 and B = 0.5

    What does this all mean?
     
  13. Aug 11, 2010 #12

    rock.freak667

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    Well if you are plotting T against L, you will not get a straight line or drawing a best fit straight line will not fit most of the points.
     
  14. Aug 11, 2010 #13
    Ok, that makes sense.

    My values for the curve fit when I graph the data were: A=2.087 and B =0.49

    whereas for the graph that looked at the linear fit (y=mx+b) gave a slope of 1.65 and a correlation of .99

    So I know that ideally the data should follow the non-linear curve; so does this mean that the curve fit line is a better interpretation of my data? (meaning that I did everything correct?)
     
  15. Aug 11, 2010 #14

    rock.freak667

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    Right those are close to the values you should ideally get.

    I am not sure how you got r2=0.99, unless you did not have a lot of data points. Could you post a picture of the linear and non-linear plots?

    Given what you plotted, then yes the curve fit is better.

    Had you plotted something like T vs. √x or T2 vs. x, the linear fit is better.
     
  16. Aug 11, 2010 #15
    I don't think the linear fit graph really matters that much since my values match the curved fit! Thanks for everything. I really appreciate it!

    One last thing, when we're looking at the curved fit equation, is the value for gravitational acceleration 3.1? What does this say compared to the accepted value of 9.8?
     
  17. Aug 11, 2010 #16
    What formula are you using for your "curve fit"? There are many least squares minimizations, let's make sure you are using the right one. :) Given that you have used the right equation, and you really are getting 3.1m/s^2, that's really bad and you should do your experiment over again.
     
  18. Aug 11, 2010 #17
    I'm not even sure if I calculated that right. I just used the equation T=(2π/√g)L1/2, where √9.8 = 3.1

    I guess I did that wrong? What does it mean when they want to know the value of g based off the curved fit equation?
     
  19. Aug 12, 2010 #18

    rock.freak667

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    Your value of 'A' is not acceleration due to gravity. But A=2π/√g, so if you solve for 'g', you should get around 9.8
     
  20. Aug 12, 2010 #19
    Oh yeah, you need to, as rockfreak has pointed out, go further and actual solve for g given your slope value. I assumed that's what you did.

    Just because you found the slope, doesn't mean you're done and that's your answer. In fact, if you continue in physics, there will be lots of times when you "linearize" a problem as you have here so that you can find the linear least squares of that particular linearization. It's a transformation into the form that's linear. In the end, all you did was find the best fit for what you linearized. So when all is said and done, you have to un-linearize, un-transform, your best fit. In this case, it is saying that g=(2π/A)^2.
     
  21. Aug 12, 2010 #20
    Oook, my bad. I calculated my value for g to be 9.04 which is a little off, but is definetely closer than my first attempt. Thanks so much for all your help!
     
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