How can I find the superposition of two vectors with a phase shift of pi/4?

[Kazza_765]

I'm considering the superposition of 2 vectors.

$$E_x=E_1cos(wt)<br /> E_y=E_2cos(wt+/pi/4)$$

Trying to eliminate t. Its easy when the phase shift is 0 or pi/2 but I'm not sure how to go about it in this case. I can get Ey to be a function of cos(wt)-sin(wt), or cos(wt)sin(wt) just using trigonometric formulas, but I don't know where to go from there.

The end result should be an elipse with axis that are not aligned with the x-y axis. If its quite difficult then I won't worry about it. I just get the feeling that I've done this before and I should know it, but somethings not clicking. Thanks.


edit: For some reason the LATEX graphics don't appear for me. Just in case I put the formulas in wrong they should be:

Ex = E1 cos(wt)
Ey = E2 cos(wt + pi/4)

 

[Andrew Mason]

I'm considering the superposition of 2 vectors.

$$E_x=E_1cos(wt)<br /> E_y=E_2cos(wt+\pi/4)$$

Trying to eliminate t. Its easy when the phase shift is 0 or pi/2 but I'm not sure how to go about it in this case. I can get Ey to be a function of cos(wt)-sin(wt), or cos(wt)sin(wt) just using trigonometric formulas, but I don't know where to go from there.

The end result should be an elipse with axis that are not aligned with the x-y axis. If its quite difficult then I won't worry about it. I just get the feeling that I've done this before and I should know it, but somethings not clicking. Thanks.edit: For some reason the LATEX graphics don't appear for me. Just in case I put the formulas in wrong they should be:

Ex = E1 cos(wt)
Ey = E2 cos(wt + pi/4)

How about $$E = \sqrt{E_x^2 + E_y^2}$$?

As far as the latex problem, you have to use lower case for the tex and /tex commands.

AM

 

[Kazza_765]

How about $$E = \sqrt{E_x^2 + E_y^2}$$?

As far as the latex problem, you have to use lower case for the tex and /tex commands.

AM

Thanks for the reply, I don't think I was very clear about what I'm trying to do though.

If I have
$$E_x=E_1cos(wt)$$

$$E_y=E_2cos(wt)$$

then $$E_1y=E_2x$$ gives me my equation of motion for the particle.

If
$$E_x=E_1cos(wt)$$

$$E_y=E_2cos(wt + \frac{\pi}{2})$$

then
$$E_y=-E_2sin(wt)$$

$$\frac{x^2}{E_1}+\frac{y^2}{E_2}=1$$

And the vector moves in an ellipse.

I'm not sure how to get the same sort of equation (ie. no dependence on t), or if its possible, for

$$E_x=E_1cos(wt)$$

$$E_y=E_2cos(wt+\frac{\pi}{4})$$

$$E_y=\frac{E_2}{\sqrt{2}}(cos(wt)-sin(wt))$$

$$E_y^2=-E_2^2(cos(wt)sin(wt))$$

Not sure where I can go from here

 

[robphy]

There's a rotation [transformation] involved.
Complex numbers might simplify your calculation.

 

[Kazza_765]

I'm sorry but I'm still not sure how to get there.

If I write it as,

$$E_x=E_1e^{iwt}$$
$$E_y=E_2e^{iwt}e^{\frac{\pi}{4}}$$

I don't know where to go from here.

I feel like its staring me right in the face. I know what the answer should look like. I know the transform should be along the lines of [cosA,-sinA;sinA,cosA] and I could probably work out A by playing with E1 and E2, but its just not falling into place for me. Maybe I'm just having a bad day, hopefully a good night's sleep will help.

 

[Kazza_765]

Just for clarification, I'll explain the context in which I ask this question. For a uGrad assignment I have been asked to propose an exam question and provide a solution (The hard part about this assignment is thinking of a question that hasn't already been on previous exams).

Part of my question goes along the lines of:
Consider 2 EM waves, prpagating along the z axis, with angular frequency w and wave number k. Wave 1 has its E-field aligned with the x axis, wave 2 with the y axis. There is a relative phase difference of $$\phi$$ between them. Describe how the polarisation of the superposition of waves 1 & 2 varies in time.

For the case $$\phi=0$$ we have a plane wave with an E-field of magnitude $$\sqrt{E_1^2+E_2^2}$$. For all $$0 < \phi < \pi$$ we should have an elliptically polarized plane wave, but I only know how to demonstrate this for the case where $$\phi=\frac{\pi}{2}.$$

 

[Kazza_765]

Nevermind. Worked it out. Just wasn't thinking yesterday.