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Homework Help: 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

    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.
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