Calculate average power in one cycle.

[yungman]

Homework Statement

This is part of a bigger problem, I simplify to just:

$$P=k sin^2 \omega t$$

How do I integrate over one cycle?

The Attempt at a Solution

$$\left\langle P \right\rangle = \int_0^T k\; sin^2 (\omega t) d t \;\hbox { where }\; T= \frac c f = \frac {2\pi}{ \omega}$$

$$\left\langle P \right\rangle = k \int_0^{ \frac {2\pi}{ \omega}} sin^2 (\omega t) d t = \frac 1 {\omega} \int sin^2 u du = \frac k 2 u = \frac k 2 t|_0^{\frac {2\pi}{ \omega}} = \frac {k\pi}{\omega}$$

I know the answer is

$$\left\langle P \right\rangle = \frac k 2$$

Please show me how to get the integration over one cycle.

Thanks

Alan

 

[hikaru1221]

The average of f(x) of period T: $$<f(x)> = \frac{1}{T}\int ^{r+T} _r f(x)dx$$ with r is arbitrarily chosen

Another way to deal with your particular function is:
$$P = ksin^2wt = \frac{k}{2}(1-cos2wt)$$

 

[stevenb]

Homework Statement

This is part of a bigger problem, I simplify to just:

$$P=k sin^2 \omega t$$

How do I integrate over one cycle?

The Attempt at a Solution

$$\left\langle P \right\rangle = \int_0^T k\; sin^2 (\omega t) d t \;\hbox { where }\; T= \frac c f = \frac {2\pi}{ \omega}$$

$$\left\langle P \right\rangle = k \int_0^{ \frac {2\pi}{ \omega}} sin^2 (\omega t) d t = \frac 1 {\omega} \int sin^2 u du = \frac k 2 u = \frac k 2 t|_0^{\frac {2\pi}{ \omega}} = \frac {k\pi}{\omega}$$

I know the answer is

$$\left\langle P \right\rangle = \frac k 2$$

Please show me how to get the integration over one cycle.

Thanks

Alan

Your answer of kT/2 makes more sense. Wouldn't you expect the integral (area under the curve) to be proportional to T? The only issue is whether you should be integrating from 0 to T/2 rather than 0 to T, because sin^2 has double the frequency of sine, as pointed out by hikaru's trig identity. But, if you want to integrate over the period T for the unsquared sinewave, then you are correct.

EDIT: Oh wait ! You want the average power, so you have to divide by T. That's what the issue is. So k/2 is correct.

 

[yungman]

Thanks, I forgot the 1/T.

Alan