View Full Version : Simpe equation
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.
HallsofIvy
Nov14-08, 03:24 PM
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
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.
0 = 0
This suggests that kL can be anything, but this is not the right answer.
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
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.
0 = 0
This suggests that kL can be anything, but this is not the right answer.
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?
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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