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!

Time average of SHO kinetic energy = (1/2)E

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I believe I understand the problem except I cannot get

    the time average of [sin(wt-d)]^2 = ½

    I can do the rest once this is established.

    2. Relevant equations

    Average of a function= <function>= 1/(period)* integral(f*dt) from 0 to the period

    Function to be averaged = [sin(wt-d)]^2

    Period of sin(t)^2 is pi, so period of [sin(wt-d)]^2 is wt-d=pi --> t=(pi+d)/w

    w=omega
    d = delta
    (This is for a problem on the average kinetic energy of a simple harmonic oscillator.) Taylor "Classical Mechanics" 5.12

    3. The attempt at a solution


    Used Identity sin(t)^2 = (1/2)(1-cos(2*t))

    <[sin(wt-d)]^2 > = w/(pi+d)(1/2) [ (pi+d)/w - integral cos(2wt-2d)dt from 0 to (pi+d)/w]

    Where I have already integrated for the first term in [ ]. My issue is that I need the last term in [ ] to be zero. Then I wold get the desired (1/2) result. To solve the integral

    I do a u substitution ,
    u = 2wt+2d
    dt= 1/(w*2)du
    Upper bound (pi+d)/w becomes 2*pi
    Lower bound (0) becomes -2d

    The lower bound is what prevents me from eliminating the term. When integrated I get
    1/(w*2) [ sin(2*pi) - sin(-2d)]

    I have tried and checked many times. Did I set up the problem wrong? It seems to be so close.

    Thank you
     
  2. jcsd
  3. Mar 5, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Your error is here: << ... from 0 to (pi+d)/w] >>

    The difference between the lower and the upper bound on your integration must be an integer number of periods One period would do nicely, which is T = 2π/ω.
     
  4. Mar 6, 2012 #3
    Thank you. I now realize my error.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Time average of SHO kinetic energy = (1/2)E
Loading...