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Unformatted text preview: Ï‰ = 2 Ï€f = 2 Ï€ (30 Hz) = 190 rad / s, and the angular wave number is k = 2 Ï€/Î» = 2 Ï€/ (0 . 40 m) = 16 m âˆ’ 1 . According to the graph, the displacement at x = 0 and t = 0 is 4 . Ã— 10 âˆ’ 2 m. The formula for the displacement gives y (0 , 0) = y m sin Ï† . We wish to select Ï† so that 5 . Ã— 10 âˆ’ 2 sin Ï† = 4 . Ã— 10 âˆ’ 2 . The solution is either 0 . 93 rad or 2 . 21 rad. In the Frst case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select Ï† = 0 . 93 rad. The expression for the displacement is y ( x, t ) = (5 . Ã— 10 âˆ’ 2 m) sin Â£ (16 m âˆ’ 1 ) x + (190 s âˆ’ 1 ) t + 0 . 93 Â¤ ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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