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T^2 vs. L and T^2 vs M' Graphs

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    What is the expected slope of the line? What was the actual slope of the line of best fit? Calculate the gravity constant g from the slope of your graph

    2. Relevant equations
    k = (Mg) / (y_0 - y)

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

    3. The attempt at a solution

    I understand how to acquire gravity using the second equation for T^2 vs L. But I don't have a clue what my expected slope should be for either graph. My calculated slope for T^2 vs L is 3.223 and for T^2 vs. M' 3.616
     
  2. jcsd
  3. Nov 2, 2011 #2

    Delphi51

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    Homework Helper

    If you graph y vs x and the formula is y = mx,
    then the slope is expected to be m.
    If you graph T² vs L and the formula is T² = (4pi²/g) x L
    then the expected slope is (4pi²/g). The calc for g would then be
    g = 4pi²/slope

    I don't see a formula relating T² and M'. What is M'?
    It looks like you might be doing a pendulum experiment?
     
  4. Nov 2, 2011 #3
    So that means my expected slope would be ~4? This was a pendulum experiment for T^2 vs L and and oscillating spring for T^2 vs M'. M' is the mass of our hook + spring + added weight, while M is just the mass of our weight added to the hook and spring.

    Through my notes I found the equation T^2 = (4pi^2m)/k. So if I replace y=T^2 and x=m does that mean my slope is (4pi^2)/k?
     
  5. Nov 2, 2011 #4

    Delphi51

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    Homework Helper

    Yes, that is the idea. Good luck.
     
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