Average value of a function

  • #1

Homework Statement



"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."



Homework Equations



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

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)
 

Answers and Replies

  • #2
CompuChip
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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 [itex]\omega[/itex], set up the integration from t = 0 to t = <one period further>, and make a substitution (in the mathematical sense, aka change of variables) [itex]\theta = \omega t - \phi[/itex].
 
  • #3
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 [itex]\omega[/itex], set up the integration from t = 0 to t = <one period further>, and make a substitution (in the mathematical sense, aka change of variables) [itex]\theta = \omega t - \phi[/itex].
but if you do that substitution wouldn't you be left with a 1/w when you substitute dt with dθ
 
  • #4
tiny-tim
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Hi Eric! :smile:
but if you do that substitution wouldn't you be left with a 1/w when you substitute dt with dθ
Yes. :smile:

(but why didn't you just integrate wrt t in the first place? :confused:)
 
  • #5
^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
CompuChip
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Nope.

When you have an oscillator with angular frequency [itex]\omega[/itex], what is its period (it should be a standard formula, so if you can't derive it look it up).
 
  • #7
Nope.

When you have an oscillator with angular frequency [itex]\omega[/itex], what is its period (it should be a standard formula, so if you can't derive it look it up).
Oh it's 2[itex]\pi[/itex]/[itex]\omega[/itex] which makes sense because the [itex]\omega[/itex] cancel and I'm left with 2[itex]\pi[/itex] 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[itex]\pi[/itex]/[itex]\omega[/itex]
 
  • #8
CompuChip
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Yes, exactly.
[In fact, you can set it up from any value t = t0 you like, to [itex]t = t_0 + 2\pi/\omega[/itex], you will see that t0 cancels out in the final result).

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

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