Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kater's Pendulum?

  1. Sep 29, 2010 #1
    Why is it that, when using a Kater's Pendulum where the pivot points are fixed and one weight is adjustable, it is acceptable to discard the factor of the difference between the two distances of the individual pivot points from the center of gravity (1/(l1-l2)) in the following formula?

    T^2 = [(T1^2 + T2^2)/2] + [{(T1^2-T2^2)/2} * {(l1+l2)/(l1-l2)}] --where T^2 is the period of an equivalent ideal pendulum, squared, and, T1 and T2 are the respective periods of corresponding ends of this reversible pendulum, then, l1 and l2 are the lengths from the respective end of the pendulum to the center of gravity.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Kater's Pendulum?
  1. A Pendulum Experiment (Replies: 13)

  2. Spring Pendulum (Replies: 1)

  3. Asimov pendulum (Replies: 1)

Loading...