Why Is the Time Average of the Cosine Squared Term in the Poynting Vector 1/2?

[ronaldoshaky]

Hello.

I am reading in my book about the Poynting vector for monochromatic plane waves. It includes a cosine term: cos^2 (kz - omega t + phi). My book states that the time average of this term is 1/2. Can anyone explain this? I don't understand how they work that out.

Thank you

 

[tiny-tim]

Hello ronaldoshaky!

(have an omega: ω and a phi: φ and try using the X² tag just above the Reply box )

Use one of the standard trigonometric identities …

cos²x = 1/2 (1 + cos2x) ​

 

[ronaldoshaky]

Hi tiny-tim.

Does finding the time average have something to do with integrating the cos^2 term?

Thanks again

 

[gabbagabbahey]

The time average of any periodic function, $f(t)$, with period $T$ is given by

$$\langle f\rangle_t=\frac{\int_{t_0}^{t_0+T}f(t)dt}{\int_{t_0}^{t_0+T}dt}=\frac{1}{T}\int_{t_0}^{t_0+T}f(t)dt$$

Apply that to your $\cos^2$ term

 

[tiny-tim]

Hi ronaldoshaky!

Does finding the time average have something to do with integrating the cos^2 term?

"integrating" is a very technical word to use …

can't you tell the average of cosx (or of cos²x = (1 + cos2x)/2) just by looking at the graph?!

 

[ronaldoshaky]

Thanks to all who replied. I will do both the graph and the integration. This has helped me a lot!