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Complex exponential description of SHM

  1. Jun 18, 2012 #1
    Hey,

    I'm currently reading a textbook which is attempting to derive the equation for a standing wave from first principles. I understand most steps with the exception of one.

    It derives a sinusoidal function {x = A \sin \omega t} from a second order ODE, but then immediately interchanges this to {exp{\pm i \omega t}}. Presumably the \pm indicates that there are two solutions, both of which may take the {cos \theta \pm i sin \theta} form from Euler's formula.

    My question is how is this format compatible with a single sinusoidal function {x = A \sin \omega t}. No matter how I manipulate it, I always end up with the sum of a sine and cosine, and the sine generally comes out to be imaginary. I imagine there's some simple feature that I'm overlooking, but can't really see it at the moment. Any help would be greatly appreciated!
     
  2. jcsd
  3. Jun 18, 2012 #2

    tiny-tim

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    welcome to pf!

    hey savvvvvvvvvvy! welllllllllcome to pf! :smile:

    (use #s :wink:)
    difficult to tell without seeing the book,

    but i'll guess that the coefficients of eiωt are allowed to be imaginary (or complex)
     
  4. Jun 18, 2012 #3

    AlephZero

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    Remember that the general equatinn for SHM is ##x = A \sin \omega t + B \cos \omega t##. There must be some reason why your book choose to set ##B = 0##.

    If you have an expression like ##x = C_1e^{i\omega t} + C_2 e^{-i\omega t}##, the constants ##C_1## and ##C_2## are complex numbers. To be precise, it should be written ##x = \Re(C_1e^{i\omega t} + C_2 e^{-i\omega t})## where ##\Re## means "the real part of".

    To make the complex number form the same as the sine function, ##-i e^{i\omega t} + ie^{-i\omega t} = 2 \sin \omega t##.
     
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