# Average value of a function

1. Aug 23, 2009

### Eric_meyers

1. The problem statement, all variables and given/known data

"A particle of mass m executes one-dimensional simple harmonic oscillation
under the action of a conservative force such that its instantaneous x coordinate
is x(t) = a cos(ωt − φ).

Find the average values of x, x^2, x', and (x')2 over a single cycle of the oscillation.
Find the average values of the kinetic and potential energies of the
particle over a single cycle of the oscillation."

2. Relevant equations

Average value of a function = integral of a function / interval of the integral

3. The attempt at a solution

So I just want to make sure I can set the integral up like this:

integral from 0 to 2π (A cos(θ)) dθ

basically is it alright to convert A cos(wt - phi) ----> A cos (θ)

2. Aug 23, 2009

### CompuChip

Yes, a single cycle is one where the phase advances by exactly 2 pi. In fact you can take any integration interval from x to x + 2pi.

If you want to do it completely officially, you can relate the period to $\omega$, set up the integration from t = 0 to t = <one period further>, and make a substitution (in the mathematical sense, aka change of variables) $\theta = \omega t - \phi$.

3. Aug 23, 2009

### Eric_meyers

but if you do that substitution wouldn't you be left with a 1/w when you substitute dt with dθ

4. Aug 23, 2009

### tiny-tim

Hi Eric!
Yes.

(but why didn't you just integrate wrt t in the first place? )

5. Aug 23, 2009

### Eric_meyers

^Well because I didn't understand how to write my interval in terms of t, but I guess I can just set up my integral from 0 to t + 2*pi??

6. Aug 24, 2009

### CompuChip

Nope.

When you have an oscillator with angular frequency $\omega$, what is its period (it should be a standard formula, so if you can't derive it look it up).

7. Aug 24, 2009

### Eric_meyers

Oh it's 2$\pi$/$\omega$ which makes sense because the $\omega$ cancel and I'm left with 2$\pi$ which is just the same behavior I started with at t = 0

So I could then set up my integral from t = 0 to t = 2$\pi$/$\omega$

8. Aug 24, 2009

### CompuChip

Yes, exactly.
[In fact, you can set it up from any value t = t0 you like, to $t = t_0 + 2\pi/\omega$, you will see that t0 cancels out in the final result).

Now do the integration, performing a change of variables (substitution) to $\theta \equiv \omega t - \phi$. You will need to take the Jacobian factor into account ($d\theta = \cdots \, \dt$) as well as the change of integration limits. You will see how things work out rather nicely.