1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Average value of a function

  1. Aug 23, 2009 #1
    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. jcsd
  3. Aug 23, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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].
     
  4. Aug 23, 2009 #3
    but if you do that substitution wouldn't you be left with a 1/w when you substitute dt with dθ
     
  5. Aug 23, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Eric! :smile:
    Yes. :smile:

    (but why didn't you just integrate wrt t in the first place? :confused:)
     
  6. Aug 23, 2009 #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??
     
  7. Aug 24, 2009 #6

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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).
     
  8. Aug 24, 2009 #7
    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]
     
  9. Aug 24, 2009 #8

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Average value of a function
  1. Average Value (Replies: 3)

  2. Average value (Replies: 2)

  3. Average Value Function (Replies: 5)

  4. Average value quest. (Replies: 7)

  5. Average value of a wave (Replies: 18)

Loading...