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Simple Pendulum lab not so simple

  1. Apr 24, 2006 #1
    This lab I would LOVE to use as TP. I'm sure its me being stupid and not really the lab. The problem I have it trying to set up a graph for this. I dont know if I did it right. My numbers seem way off. I'm supposed to make a period^2 vs length graph. Part of what I don't get is if I'm using the saxe x-value the isnt it going to be just a vertical line? Could someone just check out to see if its right?
     

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  3. Apr 24, 2006 #2

    Hootenanny

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    What are you varying (independant variable)? And what are you measuring (dependant variable)? What is the purpose of the lab?

    ~H
     
  4. Apr 24, 2006 #3

    Kurdt

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    If you keep the x value the same you will get a single value for the period and thus only a single point on the graph. I believe the purpose of this lab would be to determine the relationship between period and length and thus you must vary one (which one I'll leave for you to decide).
     
  5. Apr 24, 2006 #4

    andrevdh

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    Why do you say that your values seem to be wrong? Which values are you referring to (the periods I guess since your lenghts are probably right).
     
  6. Apr 24, 2006 #5

    HallsofIvy

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    You don't say either in your question or on your graph what x and y are supposed to represent.
    You do say "I'm supposed to make a period^2 vs length graph."
    Okay, so I presume that you are timing the periods for several different pendulums of several different lengths. Are you taking the length of the pendulum as x and the square of the period as y or vice-versa?
     
  7. Apr 24, 2006 #6

    andrevdh

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    You are suppose to have the x-values in one column and the corresponding y-values in the column next to it in Excel. Then select both columns by dragging over them before drawing the graph. Your data points do not have x-coordinates in the graph! Which means there was some thing wrong with your data selection before you drew the graph.
     
    Last edited: Apr 24, 2006
  8. Apr 24, 2006 #7
    x=length
    y=T^2, period^2

    For each length, we had five time trials. The slope of the line is to determine gravity of the pendulum which should be 9.81. The values are completeley off. When I calculate some of the values using the formula for the period of a simple pendulum, I get close. I guess what i'm askiing is they way I have that spreadsheet set up-is it right? Are the values really that far off or am I doing something wrong?
     
  9. Apr 24, 2006 #8

    andrevdh

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    The values do'nt seem to be in SI units either.
     
  10. Apr 25, 2006 #9
    I just wanted to say thanks for the help. I guess everyone was having diffculity with the lab so he gave us some pointers and extended the deadline. Thanks again!
     
  11. Apr 25, 2006 #10

    Integral

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    It looks like the length of your pendulum, plus the radius of the ball is in the range d4:d9. This appears to be in cm, you need it in meters. Excell hint. you do this formula (=d4+c4) where c4 contains the radius, if you instead use (= d4 +$c$4) you can then drag to fill the rest of your data and the value in c4 will be added to each. This will help avoid errors like in cell d8.

    It looks like the time for each trial is in columns in the range b13:g19, you need to compute the average of these numbers (by column)

    Now the relationship between Length and period is

    [tex] T = 2 \pi \sqrt{ \frac L g} [/tex]

    so your line is
    [tex] L = {(\frac T {2 \pi})}^2 g [/tex]

    You will need the factor of 2 pi if you want a slope of 9.8.

    It would be easiest if you placed the column containing

    [tex] { (\frac T {2 \pi})}^2 [/tex]

    adjacent to a column containing the corresponding lengths. I got a slope of 9.71.
     
    Last edited: Apr 25, 2006
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