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Homework Help: Why integrating sin^2(wt) gives the half-period?

  1. Jul 5, 2012 #1
    In an exercise with included solution I can't understand how integrating sin^2(ωt) gives T(period)/2

    [itex]\int[/itex]sin^2(ωt)dt = Period/2

    I posted the whole problem below, because I had more doubts, but understood them typing up the problem.

    I appreciate any help.


    The probelm

    A circular coil, r=10 and Ω=1.5Ω, rotates around its diameter with a constant ω0 in a uniform and constant magnetic field B that forms an angle of θ=∏/3 with the axis of rotation of the coil.
    Knowing that the maximum current Imax=0.15A, and that the component of B parallel to the axis of rotation is Bparall=1.0T, find
    1) intensity of B
    2) angular velocity ω0 of the coil
    3) energy needed per rotation to keep the angular velocity ω0

    The solution included on teh book
    my problem in in red

    1) B=Bparallel/cosθ=2Bparallel=2.0T
    I did the same, so no problem here.

    Bperp(responsible for the current in the coil)=Bparallel*tanθ=Bparallel√3

    2) fem=-dphi/dt=∏*r^2*ω*Bperp*sin(ωt), max fem when sin(ωt)=1,

    3)To get the work, i integrate over one turn, so over the period T=2∏/ω
    W= [itex]\int[/itex]R*I^2*dt = R*Imax^2[itex]\int[/itex]sin^2(ωt)dt = R*I^2*Period/2
    I don't get how do you integrate sin^2(wt) and get T/2?
  2. jcsd
  3. Jul 5, 2012 #2


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    Staff: Mentor

    How is ω related to T?
  4. Jul 5, 2012 #3

    Simon Bridge

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    Science Advisor
    Homework Helper

    [itex]\omega = \frac{2\pi}{T}[/itex] ... like you said above.
    Your limits of integration are 0-2π?

    (note: π is the lower case of ∏.)
    Last edited: Jul 5, 2012
  5. Jul 5, 2012 #4

    D H

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    Staff Emeritus
    Science Advisor

    Do you know what the integral of [itex]\sin^2x[/itex] is? If you don't, use the cosine double angle formula [itex]\cos 2x = \cos^2x - \sin^2x[/itex] and the Pythagorean identity [itex]\sin^2x + \cos^2x = 1[/itex] to express [itex]\sin^2x[/itex] in terms of [itex]\cos 2x[/itex]. From there it should be easy sailing.
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