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Simpe equation

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    What does it take for

    exp(ikL) - exp(-ikL) = 0

    ?

    3. The attempt at a solution

    cos(kL) + i sin(kL) - cos(kL) - i sin(kL) = 0

    0 = 0

    This suggests that kL can be anything, but this is not the right answer.
     
  2. jcsd
  3. Nov 14, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are using exp(ikL)= cos(kl)+ i sin(kL), right?

    So exp(-ikL)= cos(kL)- i sin(kL) because cosine is an even function and sine is an odd function.

    However, cos(kL)+ i sin(kL)- (cos(kL)- sin(kL)) is NOT 0.

     
  4. Nov 14, 2008 #3

    Mark44

    Staff: Mentor

    Re: Simple equation

    The equation above should be:
    cos(kL) + i sin(kL) - cos(-kL) - i sin(-kL) = 0
    or
    cos(kL) + i sin(kL) - cos(kL) + i sin(kL) = 0
    So 2 sin(kL) = 0, which says that kL = n*pi, where n is an integer.
     
  5. Nov 14, 2008 #4
    Re: Simple equation

    Here's the whole problem:

    There are two harmonic waves on a string:

    y1=Aexp(i(kx-wt))
    y2=Bexp(i(-kx-wt))

    (1) y(x=0)=0
    (2) y(=L) = 0

    The total wave is

    (3) y=exp(-iwt)[Aexp(ikx) + Bexp(-ikx)]

    (1) gives A = -B, so that (3) can be written

    y = exp(-iwt)[Aexp(ikL) - Aexp(-ikL)] = 0

    For (2) to be true, we must have exp(ikL) - exp(-ikL) = 0, that is

    cos(kL) + i sin(kL) - cos(-kL) - i sin (-kL) = 0 ---> 2sin(kL) = 0 ---> kL = n*pi

    My book is wrong, then?
     
    Last edited: Nov 14, 2008
  6. Nov 14, 2008 #5

    Mark44

    Staff: Mentor

    Books have been known to have wrong answers. Your equation is satisfied if kL = pi, or 3*pi, or 5*pi, and so on, values that your book's solution doesn't include.
     
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