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cpmiller
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Homework Statement
Carry out the addition of two waves Ψ=Ψ1+Ψ2 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.
Homework Equations
Re{e (iΘ) } = cos Θ = [e (iΘ) + e (-iΘ) ]/2
Im{e (iΘ) } = sin Θ = [e (iΘ) - e (-iΘ) ]/(2i)
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:
Ψ1+Ψ2 = 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!
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