1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Physical Pendulum

  1. May 11, 2010 #1

    bon

    User Avatar

    1. The problem statement, all variables and given/known data

    Ok worked out the time period of a physical pendulum (T) to be (in terms of constant a, and variable length x):

    T^2 = 4pi (1/2a^2 + x^2)/(gx)

    Now asked how i could use a measurement of T(x) to measure g.


    2. Relevant equations



    3. The attempt at a solution

    I guess i have to vary x and measure T and plot some sort of graph, but given the relation above I can't see what the easiest way to determine g would be...

    thanks
     
  2. jcsd
  3. May 11, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Normally, if you have to plot a graph to get 'g', you'd try as best as you could to make the equation into the form Y=MX+C.

    Eg. T2=4π2(L/g), you write this as T2=(4π2/g)L

    So that Y=T2 and X=L, plot Y vs. X.
     
  4. May 11, 2010 #3

    bon

    User Avatar

    Thanks but I knew this - I'm just not sure how to do this in this case..
     
  5. May 11, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    [tex]T^2 = 4 \pi \frac{\frac{1}{2}a^2 + x^2}{(gx)} [/tex]


    [tex]T^2 =\frac{4 \pi}{g} \frac{\frac{1}{2}a^2 + x^2}{x} [/tex]
     
  6. May 11, 2010 #5

    bon

    User Avatar

    Thanks but you still have the x at the bottom..
     
  7. May 11, 2010 #6

    rock.freak667

    User Avatar
    Homework Helper

    I meant X=((1/2)a2+x2)/x
     
  8. May 11, 2010 #7

    bon

    User Avatar

    Oh rite i see. Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Physical Pendulum
  1. Physical pendulum hole (Replies: 5)

  2. Physical pendulum (Replies: 1)

  3. Physical pendulum (Replies: 1)

Loading...