Transforming E^2(x,t) to A_y^2 + A_z^2 in Harmonic Waves

[Athenian]

Homework Statement

Show that $I = \frac{1}{2} \Big( \frac{\epsilon}{\mu} \Big)^{\frac{1}{2}} (A_y^2 + A_z^2)$

Relevant Equations

$$I = \frac{1}{T} \int_0^T \Big( \frac{\epsilon}{\mu} \Big)^{\frac{1}{2}} E^2 (x,t)dt$$
$$E_y (x,t) = A_y cos(kx-\omega t)$$
$$E_z (x,t) = A_z cos(kx-\omega t + \phi)$$
Note that $\phi$ is the phase difference
$$E^2 = E_y^2 + E_z^2$$

To begin with, I am trying to understand how does $E^2 (x,t)$ transform to $A_y^2 + A_z^2$. And, noting that the already established equation of $E^2 = E_y^2 + E_z^2$, I would assume that $E^2 (x,t)$ somehow ends up to being $A_y^2 + A_z^2$. However, noting that $E^2 = (A_y cos(kx-\omega t))^2 + (A_z cos(kx-\omega t + \phi))^2$, I can't see how this can be equal to $A_y^2 + A_z^2$ yet. In other words, I am having a hard time with the math if I am going about this in the right direction.

In short, any help toward helping me understand and ultimately solve the question would be greatly appreciated. Thank you for your help!

 

[mjc123]

However, noting that E2=(Aycos(kx−ωt))2+(Azcos(kx−ωt+ϕ))2, I can't see how this can be equal to Ay2+Az2 yet.

It isn't. But integrate it from t = 0 to 2π/ω.

 

[Athenian]

Thank you for your reply. Here's what I did.

Since
$$E^2=(A_y cos(kx−\omega t))^2+(A_z cos(kx−\omega t+ \phi))^2$$

I can continue per your instructions (if I understood it correctly) by writing the following:
$$\int_0^\frac{2 \pi}{\omega} (A_y cos(kx−\omega t))^2+(A_z cos(kx−\omega t+ \phi))^2 dt = \frac{\pi A_y^2}{\omega} + \frac{\pi A_z^2}{\omega}$$

The above answer is what I got when I plugged the values into Wolfram|Alpha. However, despite integrating, my answer does not exactly equate to $A_y + A_z$. Am I doing something wrong here or am I on the right track?

Also, I noticed you substituted $\frac{2\pi}{\omega}$ (or $\frac{1}{f}$) for $T$. However, isn't $T$ (time) much larger compared to the time period of oscillation (i.e. $T >> \frac{1}{f}$)? Or, despite that, it's still fine to integrate from $0$ to $\frac{2\pi}{\omega}$. Anyway, I just wanted to make sure I am having my understanding completely accurate and coreect.

Once again, thank you so much for your kind and prompt assistance!

 

[Athenian]

Please ignore my confusion in the above reply. It was pretty late and I wasn't exactly thinking straight despite my efforts. Anyway, a good night's rest was all I needed to get my mind on the right track.

In short, my partial answer of $\frac{\pi A_y^2}{\omega} + \frac{\pi A_z^2}{\omega}$ was correct. However, I did not put in mind that $\frac{1}{T}$ before the integral would also become $\frac{1}{\frac{2\pi}{\omega}}$ instead.

Multiplying and simplifying everything, I got the desired solution of $I = \frac{1}{2} \Big( \frac{\epsilon}{\mu} \Big)^{\frac{1}{2}} (A_y^2 + A_z^2)$.

However, my original question re-quoted below still stands. If somebody can help answer this conceptual or math-based question to reinforce my understanding, I would greatly appreciate it. Thank you!

Also, I noticed you substituted $\frac{2\pi}{\omega}$ (or $\frac{1}{f}$) for $T$. However, isn't $T$ (time) much larger compared to the time period of oscillation (i.e. $T >> \frac{1}{f}$)? Or, despite that, it's still fine to integrate from $0$ to $\frac{2\pi}{\omega}$?

 

[mjc123]

The answer is strictly true only for T = an integral multiple of 1/f. However, if T >> 1/f the error is small. If we write T = n/f + d, then the 1/T factor mean the error due to d is very small if n is very large.