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Using a pendulum to determine g using T = 2π√(l/g)

  1. Nov 30, 2010 #1
    I can do the solution, I do not understand the theory!
    Here it is:

    Using a pendulum to determine g using T = 2π√(l/g)
    (that little n looking thing is pi)
    (given l and T)

    So, then we get
    T^2 = (4π^2/g) x l


    This is where I get lost.
    Supposedly, I am to make the equation T^2 = kl (where k is the group of constants)
    Then, I am to compare this formula with the general equation for a straight line y=kx.
    Thus, k = m (of a graph, where vertical T^2 and horizontal l is the axis)

    Why/how does k = m ?
    k being (4π^2/g)
    and m being the gradient of my graph?
    ______________

    Thank you!
     
  2. jcsd
  3. Nov 30, 2010 #2
    y = force
    x = displacement from center

    Comparison of straight line to Hooke's law

    y=mx -----> F=-kx

    This is because small oscillations about a point obey Hooke's law, which is a linear relationship
     
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