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Gravitational Acceleration of Pendulum

  1. Jan 3, 2014 #1
    I am doing a lab report for IB Physics SL and I am supposed to use the slope of the period of a pendulum graphed against the length to find gravitational acceleration. I am trying to use the equation T=2∏√(l/g) but I'm not getting the right answer when I solve for g. (the answer is in s^2/m which obviously is wrong. I'm probably making a stupid mistake, so any help would be nice. Thanks.
     
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  3. Jan 3, 2014 #2

    berkeman

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    Welcome to the PF.

    Can you show your work in detail? It's probably just a simple algebra mistake. Good job checking the units, BTW. That's a good way to look for mistakes... :smile:
     
  4. Jan 3, 2014 #3
    Sure, my slope was .0135 s/cm or 1.35 s/m.

    I then plugged in 1.35 for T and 1 for l (this may be the completely wrong way of approaching this, but I thought it made sense).

    So my equation looked like this:

    1.35s = 2∏√(1m/g) and as I'm typing this I realized what I did wrong haha
    so that becomes
    .0462s^2 = 1/g and this is where I accidentally forgot it was 1/g and not g.

    But even then, this solves to g = 21.65 m/s^2 which isn't right, is it? I would think the answer should be 9.8 m/s^2, so something is wrong or I had some massive error in my experiment.
     
  5. Jan 3, 2014 #4

    berkeman

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    Can you post your data?
     
  6. Jan 3, 2014 #5
    Length | Period

    15.5 | 1.03
    28.2 | 1.17
    41.0 | 1.38
    50.0 | 1.54
    57.0 | 1.67
    67.5 | 1.75
    79.1 | 1.89
    94.7 | 2.07

    Length is in cm and period is in seconds. (These are also averages of eight different trials for each length)

    The slope of the best fit line, like I said, is .0135 s/m
     
  7. Jan 3, 2014 #6

    berkeman

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    How do your data compare to these other graphs?

    https://www.google.com/search?hl=en....0....0...1ac.1.32.img..5.19.1247.8_g03pcRNS8

    .
     
  8. Jan 3, 2014 #7
  9. Jan 3, 2014 #8

    berkeman

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    If you do the calculation using the data from one of those graphs, what do you get for g? If you get the right answer using those example graphs, and not the right answer using your data, then that would imply that there was something causing errors in your data. This is a lab report, so it's okay (usually) to say that you see there is a problem with the data, and then make some intelligent guesses at what the source of the errors could be...
     
  10. Jan 3, 2014 #9
    It's hard to get the slope of some of the graphs, but I used one and got around 18 m/s^2, which is not ridiculously far off from my answer. However, shouldn't the answer be 9.8 m/s^2? That is the gravitational acceleration constant, right?
     
  11. Jan 3, 2014 #10

    berkeman

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    If you plot T versus √L you should get a straight line...
     
  12. Jan 3, 2014 #11

    vela

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    T and ##l## aren't related linearly. You have
    $$T = \frac{2\pi}{\sqrt g} \sqrt{l},$$ so as Berkeman noted, it's T and ##\sqrt{l}## that are related linearly, with the slope corresponding to ##2\pi/\sqrt{g}##. I tend to avoid radicals, so I'd square the equation to get
    $$T^2 = \frac{4\pi^2}{g} l$$ and plot appropriately.

    By the way, depending on how I fit your data, I got 9.5 m/s^2 or 8.5 m/s^2 for ##g##. It looks like there's a bit of systematic error in your data that you need to attempt to explain.
     
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