Resultant of two waves -- Different amplitude+phase difference

[CricK0es]

Homework Statement

a.) Find the resultant of the two waves: Phasor method
$$E1 = 4cos(ωt)$$ & $$E2 = 3cos(ωt+\frac{\pi}{2})$$

b.) By algebraically writing
$$E1 = 4cos(ωt+α-α)$$ & $$E2 = 3cos(ωt+α+\frac{\pi}{2}-α)$$

Choose α to make the wave have only a cosine term

Homework Equations

The Attempt at a Solution

I'll show my working for the algebraic attempt so far, then try and see how to do it via phasors later.

$$E0 = E1 + E2$$
$$E0 = 4cos(ωt+α-α)+3cos(ωt+α+\frac{\pi}{2}-α)$$
My ability to use LaTeX ran out... sorry, I just couldn't get it to work. I'll try to keep it clear... ;(

[ 4 . Cos(α) + 3 . Cos( π/2 - α) ] . Cos(ωt + α) + [ 4 . Sin(α) - 3 . Sin( π/2 - α) ] . Sin(ωt + α)

Make the wave have only a cosine term ∴

[ 4 . Sin(α) - 3 . Sin( π/2 - α) ] = 0

4. Sin(α) = 3. Cos(α)
tan(α) = 3/4

That's pretty much as far as I got. I wasn't sure what to do after this. Any help would me much appreciated.

Many thanks...
[/B]

 

[Charles Link]

The technique they are trying to teach you in part "b" is somewhat clumsy IMO, and I will show you what I do with expressions like these. You can probably work it to the method they are trying to teach you in part "b". You need to first get the sum expressions in the form $y=Acos(\omega t)+Bsin(\omega t)$. $\\$ (In the case of $E2=-3sin(\omega t)$ , by use of the $cos(\theta +\phi)$ identity.) $\\$ Anyway, once you have the form as shown, factor as follows: $y=\sqrt{A^2+B^2} [\frac{A}{\sqrt{A^2+B^2}}cos(\omega t)+\frac{B}{\sqrt{A^2+B^2}}sin(\omega t)]$. $\\$ Let $cos(\phi)=\frac{A}{\sqrt{A^2+B^2}}$ and $sin(\phi)=\frac{B}{\sqrt{A^2+B^2}}$. $\\$ Then $y=\sqrt{A^2+B^2}cos(\omega t-\phi)$ . $\\$ (Notice how the identity $cos(\omega t-\phi)=cos(\omega t)cos(\phi)+sin(\omega t) sin(\phi)$ was used here. Notice also that $\phi=tan^{-1}(\frac{B}{A})$.) $\\$ This comes up often enough in engineering and/or physics that the procedure is something worth memorizing. Perhaps the $\phi$ that I found is the $\alpha$ that they want you to find, but their method appears somewhat clumsy to me.

 

[ehild]

Instead of memorizing formulas, you can derive the amplitude and phase angle of the sum.
You want to make 4cos(ωt)+3cos(ωt+π/2)=Acos(ωt+θ)
Expand the cosines:
4cos(ωt) - 3sin(ωt) = Acos(ωt)cosθ - Asin(ωt)sin(θ).
Collect the terms with cos(ωt) and the ones, with sin(ωt):
[4-Acos(θ)]cos(ωt) + [ -3 + A sin(θ)] sin(ωt)
The equation should be true for all values of t, so the coefficients of both cos(ωt) and sin(ωt) have to be zero.
Acos(θ) - 4 = 0, Asin(θ)-3=0, or
Acos(θ) = 4 and
Asin(θ) = 3
You get A² by squaring the last two equations and adding them (and using that sin²(θ)+cos²(θ)=1). Dividing both equations with A, you get both the sine and cosine of the phase angle theta. Yes, you are right, tan(θ)=3/4. Find theta (in radians).

 

[CricK0es]

Thank you both. Your responses were very helpful and I think I've got it now. I have another question, but I'll post that as a separate thread.