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AmTran_Bus

  • one year ago

Help!!!

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  1. AmTran_Bus
    • one year ago
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    If X(x)=0 and A cos Beta x + B sin Beta x Why does A=0 (meaning the A cos Beta x =0) and not the B sin Beta x ?

  2. Loser66
    • one year ago
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    I don't get the question. Can you please take a snapshot?

  3. AmTran_Bus
    • one year ago
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    |dw:1441568612734:dw|

  4. Loser66
    • one year ago
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    because X(0)=0 means x=0, so that beta x =0, hence sin(beta 0) =0, to get the whole thing =0, cos (0) =1, only one way is that A =0

  5. Loser66
    • one year ago
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    |dw:1441568809736:dw|

  6. Loser66
    • one year ago
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    |dw:1441568848437:dw|

  7. Loser66
    • one year ago
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    |dw:1441568877970:dw|

  8. Loser66
    • one year ago
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    Got it?

  9. AmTran_Bus
    • one year ago
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    Yes. So it is all basically because cos (1) is 1, so to get the term on the left =1, A must =0. Since Sin (0)=0, B automatically =0. Is that right?

  10. Loser66
    • one year ago
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    cos (0) =1, not cos(1) =1

  11. Loser66
    • one year ago
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    B is not =0

  12. AmTran_Bus
    • one year ago
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    Whoops, that is what I meant!

  13. Loser66
    • one year ago
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    Bsin beta x =0, because sin beta x =0, there is nothing to conclude about B

  14. AmTran_Bus
    • one year ago
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    Ok. Thanks a million. That makes sense. So when I introduce that X(l) = B sin Beta l =0 I can say that Sin Beta l=0 .

  15. AmTran_Bus
    • one year ago
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    To give integer values of integer pi for n

  16. AmTran_Bus
    • one year ago
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    That is, 1 pi, 2 2 pi, 3 pi. To keep A=0 and B=0, both trivial answers.

  17. Loser66
    • one year ago
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    I don't get what you mean by X(l) = B

  18. Loser66
    • one year ago
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    I want to see the original one. please, take a snapshot.

  19. AmTran_Bus
    • one year ago
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    X(l) = B sin Beta l =0 Or B sin Beta l =0 Instead of saying B=0, which is trivial, I can say

  20. freckles
    • one year ago
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    \[X(x)=Acos(\beta x)+B \sin(\beta x) \\ \text{ we have} X(0)=0 \\ \text{ so when } x=0 \text{ we have} X=0 \\ 0=A \cos(\beta \cdot 0)+B \sin(\beta \cdot 0) \\ 0=A \cos(0)+B \sin(0) \\ 0=A(1)+B(0) \\ 0=A+0 \\ 0=A \\ \text{ so that means our solution now looks like this } \\ X(x)=0\cos(\beta x)+B \sin(\beta x) \\ \text{ or } \\ X(x)=B \sin(\beta x) \\ \text{ now we are given } X(L)=0 \\ \\ \text{ tso this means when } x=L \text{ we have } X=0 \\ 0=B \sin( \beta L) \\ \text{ dividing both sides by } B \\ 0= \sin(\beta L) \\ \text{ now we know } \sin(x)=0 \text{ when } x=npi \\ \text{ so we have } \beta L =n \pi \\ \text{ and dividing both sides by } L \\ \text{ we have } \beta =\frac{ n \pi}{L} \\ \text{ so now our solution can be written as } \\ X(x)=B \sin ( \frac{ n \pi }{L}x)\]

  21. freckles
    • one year ago
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    I don't know if I wrote it clearer than the book or not

  22. AmTran_Bus
    • one year ago
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    @freckles YES thanks x1,000,000,000,000!!!!!

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