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 Discussion Response Post to:
the
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Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Apr 2008 21:13:16 IST
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a)If cos(10(pi)x) is zero, then for any value of t,the position at x is a node.
cos@ is zero at 2npi +- pi/2
10pix = 2npi+-pi/2
x = n/5 +- 1/20
Put n =1 and use minus sign.
b)Similarly for antinode use cos@ = npi
c) You did it!
d) k = 2pi/lambda
10pi = 2pi/lambda, result follows
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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