- #1
jbrussell93
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Homework Statement
Imagine that we have two waves of the same amplitude, speed and frequency over-lapping in some region of space such that the resultant disturbance is
[itex]\psi(y,t) = Acos(ky+\omega t) + Acos(ky-\omega t +\pi)[/itex]
Using complex exponentials show that
[itex]\psi(y,t) = -2Asin(ky) sin(\omega t)[/itex]
Homework Equations
[itex]Ae^{i\theta} = Acos(\theta) + Aisin(\theta) [/itex]
The Attempt at a Solution
I actually have the solution worked out, but my issue occurred once consulting the solutions manual. I ended up with:
[itex]\psi(y,t) = A(2icos(ky)sin(\omega t)-2sin(ky) sin(\omega t))[/itex]
with the real part being the solution:
[itex]\psi(y,t) = -2Asin(ky) sin(\omega t)[/itex]
Then the solutions manual says "Had we begun with sine waves, i.e.
[itex]\psi(y,t) = Asin(ky+\omega t) + Asin(ky-\omega t +\pi)[/itex]
the treatment would have been identical up until the last step, where, this time, the imaginary part would be taken to give:"
[itex]\psi(y,t) = 2Acos(ky)sin(\omega t)[/itex]
I'm confused as to why we can just choose the imaginary part because we happen to have started with sine functions. I thought the real part was always used and the imaginary part ignored. I think I'm missing some intuition with converting sine functions into complex exponentials... I understand that
[itex]Acos(\theta) = Ae^{i\theta}[/itex] because we can ignore the [itex]Aisin(\theta)[/itex] term,
but what if we start with [itex]Asin(\theta)[/itex]? How is this changed to an exponential?