1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Dimensional Analysis Pendulum Equation

  1. Jun 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by
    T = 2π sqrt3a.gif ℓ/g
    where ℓ is the length of the pendulum and g is the acceleration due to gravity, in units of length divided by time squared. Show that this equation is dimensionally consistent. (You might want to check the formula using your keys at the end of a string and a stopwatch.

    2. Relevant equations
    T=[t] possibly [S] because we are talking about a time period. l=L g=m/s^2 or [L/LT^2]

    3. The attempt at a solution
    I am attempting dimensional analysis on this problem but I get lost after a certain point because of the square root and two paths of solving the problem. This is my work thus far and would like insight as to where I am going wrong or what steps I may have missed.
    [S]= square root ([m]/[m/s^2] and the m's would cancel and were left with square root of 1/s^2 which becomes [S] so it's dimensionally correct because [S]=[S] and 2 pie is constant so it's irrelevant to the dimensional analysis. Or is not giving me the T=[S] in the problem which would then change it because [t]= square root [L/L/T^2] where L's cancel and are left with square root of 1/T^2 which becomes T so [t]=[t]. Am I doing this correctly?
    Last edited by a moderator: Apr 18, 2017
  2. jcsd
  3. Jun 21, 2016 #2


    User Avatar

    Staff: Mentor

    Hi khaledS. smiley_sign_welcome.gif

    I would like to see a pair of parentheses ( ) incorporated into your formula for T, to remove an ambiguity.
  4. Jun 22, 2016 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    When doing dimensional analysis, use L, T, M etc., not SI units like m, kg and sec. Dimensional analysis is independent of the units used. So your second method is OK, the first is not.
    You did that part wrong. You are left with sqrt (T^2) = T. But I think you have the right idea. Just don't cheat! :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted