- #1
teroenza
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Homework Statement
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.
Homework 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
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