- #36
TFM
- 1,026
- 0
Okay, so:
[tex] A^2\int{sin^2(\frac{n\pi}{L}x)dx = 1 [/tex]
Using:
[tex] sin^2 (kx) = \frac{1}{2} - \frac{cos(kx)}{2} [/tex]
Gives:
[tex] A^2\int{\frac{1}{2} - \frac{cos(kx)}{2}}dx = 1 [/tex]
[tex] A^2[{\frac{x}{2} - \frac{k}{2} sin(kx)}]^L_0 = 1 [/tex]
[tex] A^2[{\frac{L}{2} - \frac{k}{2} sin(kL)}] = 1 [/tex]
[tex] k = n\pi/L [/tex]
[tex] A^2[{\frac{L}{2} - \frac{n\pi}{2} sin(n\pi)}] = 1 [/tex]
sin of n*pi = 0
[tex] A^2[{\frac{L}{2}] = 1 [/tex]
[tex] A^2{\frac{L}{2} = 1 [/tex]
[tex] A^2 = {\frac{2}{L} [/tex]
gives A to be:
[tex] A = \sqrt{\frac{2}{L}} [/tex]
Is this better now?
[tex] A^2\int{sin^2(\frac{n\pi}{L}x)dx = 1 [/tex]
Using:
[tex] sin^2 (kx) = \frac{1}{2} - \frac{cos(kx)}{2} [/tex]
Gives:
[tex] A^2\int{\frac{1}{2} - \frac{cos(kx)}{2}}dx = 1 [/tex]
[tex] A^2[{\frac{x}{2} - \frac{k}{2} sin(kx)}]^L_0 = 1 [/tex]
[tex] A^2[{\frac{L}{2} - \frac{k}{2} sin(kL)}] = 1 [/tex]
[tex] k = n\pi/L [/tex]
[tex] A^2[{\frac{L}{2} - \frac{n\pi}{2} sin(n\pi)}] = 1 [/tex]
sin of n*pi = 0
[tex] A^2[{\frac{L}{2}] = 1 [/tex]
[tex] A^2{\frac{L}{2} = 1 [/tex]
[tex] A^2 = {\frac{2}{L} [/tex]
gives A to be:
[tex] A = \sqrt{\frac{2}{L}} [/tex]
Is this better now?
Last edited: