# Homework Help: Summing Waves Using Complex Notation

1. Nov 20, 2009

### cpmiller

1. The problem statement, all variables and given/known data

Carry out the addition of two waves Ψ=Ψ12 where
Ψ1 = Asin[(k+δk)x - (w+δw)t]
Ψ2 = Acos[(k-δk)x - (w-δw)t]
by means of the complex-number representation and interpret the result.
(Hint: You may find it convenient to rewrite the sine as a cosine by introducing a phase shift in the argument.

2. Relevant equations

Re{e (iΘ) } = cos Θ = [e (iΘ) + e (-iΘ) ]/2
Im{e (iΘ) } = sin Θ = [e (iΘ) - e (-iΘ) ]/(2i)

3. The attempt at a solution

I rewrote Ψ1 using a Pi/2 as a phase shift. So I got

Ψ1 = Acos[(k+δk)x - (w+δw)t + (Pi/2)]

Then I applied the above relevant equation for cosine being the real part of the complex equation to get
Ψ1 = A Re{e i[(k+δk)x - (w+δw)t + (Pi/2)] }
Ψ2 = A Re{e i[(k-δk)x - (w-δw)t] }

So summing yields:

Ψ12 = A Re{ei[(k+δk)x - (w+δw)t + (Pi/2)] + ei[(k-δk)x - (w-δw)t] }

Pulling out the like terms of eikx and e-iwt yields:

Ψ = A Re { [e i[δkx - δwt + Pi/2] + e-i[δkx - δwt] * ei(kx-wt) }

"Simplifiying" yields:

Ψ = A Re { [eiPi/2 * ei[δkx - δwt]+ e-i[δkx - δwt]] * ei(kx-wt) }

So we have

Ψ = A Re { [-e i[δkx - δwt] + e -i[δkx - δwt] ] * e i(kx-wt) }

Now I'm not sure what to do with the part in the bolded brackets. I could call it 2i Sin (δkx - δwt) which would be an imaginary part, so taking the real value of an imaginary part yields 0. Which would physically mean that my waves would destructively interfere, but I'm not sure that I can stick the 2i in like that. I wouldn't have a problem putting a 2 in, but I'm not sure if it's okay to multiply by i, or if I need to have the i already there.

Thanks for wading through all this algebra with me!

Last edited: Nov 21, 2009
2. Nov 20, 2009

### tiny-tim

Welcome to PF!

Hi cpmiller! Welcome to PF!

(try using the X2 tag just above the Reply box )
No, Re {[iB] ei(kx-wt)} ≠ Re {[iB]} Re {ei(kx-wt)}

3. Nov 21, 2009

### cpmiller

Re: Welcome to PF!

Thanks for the response I went back and used your suggestion to try to make my original post a bit more "user friendly." Your answer helped a lot and I managed to figure the problem out this morning!

4. Nov 21, 2009

### tiny-tim

ooh, that's neat! :tongue2:

Please accept this present of a pi and an omega for future efforts: π ω