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

[blanxink]

In an exercise with included solution I can't understand how integrating sin^2(ωt) gives T(period)/2

$\int$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.

YOU CAN SKIP THE PROBLEM

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,
ω0=R*I/(∏*r^2*Bperp)

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

 

[gneill]

In an exercise with included solution I can't understand how integrating sin^2(ωt) gives T(period)/2

$\int$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.

YOU CAN SKIP THE PROBLEM

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,
ω0=R*I/(∏*r^2*Bperp)

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

How is ω related to T?

 

[Simon Bridge]

How is ω related to T?

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


(note: π is the lower case of ∏.)

 

[D H]

I don't get how do you integrate sin^2(wt) and get T/2?

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

 

[Nickie]

The reason why integrating sin^2(ωt) gives the half-period is because the integral of a function over a certain period represents the total area under the curve of that function within that period. In this case, the period T represents the time it takes for the coil to make one full rotation. By integrating sin^2(ωt) over this period, we are essentially finding the total area under the curve of the function within one rotation of the coil.

Since sin^2(ωt) is a periodic function with a period of T/2 (half of the full rotation period), the area under the curve of this function within one full rotation is equal to half of the total area under the curve of the function within the period T. Therefore, integrating sin^2(ωt) over one rotation gives us T/2, which represents the total energy needed per rotation to maintain the angular velocity of the coil.

I hope this explanation helps to clarify the concept for you. If you have any further doubts, please feel free to ask.