# Homework Help: Time average of SHO kinetic energy = (1/2)E

1. Mar 4, 2012

### teroenza

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. Mar 5, 2012

### rude man

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.

3. Mar 6, 2012

### teroenza

Thank you. I now realize my error.