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Using a simple pendulum to determine g

  1. Aug 31, 2008 #1
    Using a simple pendulum to determine "g"

    1. Using a simple pendulum to determine "g"
    Basically, we were asked to do an experiment where we measured with a stopwatch for 20 complete oscillations for a pendulum bob swinging through a small arc (10 degrees). We increased the length of the string attached to the bob by 10 cm every time from 20cm to 1m.
    Then, we were supposed to graph length to time^2 and figure out a value of "g" and give a straight line graph in the form of y=mx + c

    Basically, i recorded data like this
    [​IMG]

    Using Time^2, and Length, I plot the graph and drew a line of best fit in the form of y=mx + c. I got a slope (m) of 14.371

    2. Relevant equations

    I used g = 4pi^2 / m

    Where m is the slope of the graph.

    3. The attempt at a solution

    Using the above equation, I subbed in all the numbers to find "g" for the first point of data.

    So, g = 4pi^2/14.371
    g = 2.747

    My question is, as we all know "g" should be 9.8 meters per second ^2. I am getting a figure of 2.747 meters per second ^2. Which is awfully off what it should be right?

    I am wondering if one of my equations are set up wrong, or did I measure something wrong?
     

    Attached Files:

  2. jcsd
  3. Aug 31, 2008 #2

    Doc Al

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    Staff: Mentor

    Re: Using a simple pendulum to determine "g"

    What's the actual relationship in terms of length and period?

    How does your measured value of "time" relate to the period of the pendulum?
     
  4. Aug 31, 2008 #3
    Re: Using a simple pendulum to determine "g"

    Well, I know that length and period are used in this equation:

    T^2 = 4pi^2L / g.

    my measure of Time (s) is of 20 oscilliations of the pendulum.

    and Period = time / # of cycles.

    So is it what I am supposed to do is use Period ^2 instead of Time^2 ?

    Am I supposed to use T^2 = 4pi^2L / g and bring it over to y = mx + c, where Period^2 is at the y-axis, Length at the x-axis?

    Sorry, I'm still a little confused as to how to do this question.
     
  5. Aug 31, 2008 #4

    Doc Al

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    Staff: Mentor

    Re: Using a simple pendulum to determine "g"

    Right. Here, T is the period, not your measured 20 oscillations.

    Yep. Or rewrite the equation above in terms of measured time, instead of period.
     
  6. Aug 31, 2008 #5
    Re: Using a simple pendulum to determine "g"

    In my school i am doing an experiment on pendulum to plot a l-t^2 graph.
     
  7. Aug 31, 2008 #6
    Re: Using a simple pendulum to determine "g"

    Uh yeah, I think I was confused over using Time and Period. It should be period.

    I got 1088.48 cm / s^2

    I was wondering how to calculate the percentage error?

    Is it ( 1 - (980 / 1088.48) ) * 100

    Anyways, thanks for your clarification and help Doc Al
     
  8. Aug 31, 2008 #7

    LowlyPion

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    Homework Helper

    Re: Using a simple pendulum to determine "g"

    It's best to express your error as a percentage of the correct value.
     
  9. Feb 14, 2011 #8
    Re: Using a simple pendulum to determine "g"

    I damn confused i have to make a L-T graph. which will be a parabola but with my readings a staright line is coming....
     
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