P17_050 - kx ) sin( t ) = 0 . 04 m at x = 0 . 20 m . (b)...

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50. From the x = 0 plot (and the requirement of an antinode at x = 0), we infer a standing wave function of the form y = (0 . 04) cos( kx ) sin( ωt )w h e r e ω = 2 π T = π rad / s with length in meters and time in seconds. The parameter k is determined by the existence of the node at x =0 . 10 (presumably the frst node that one encounters as one moves from the origin in the positive x direction). This implies k (0 . 10) = π/ 2sothat k =5 π rad/m. (a) With the parameters determined as discussed above and t =0 . 50 s, we ±nd y =
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Unformatted text preview: kx ) sin( t ) = 0 . 04 m at x = 0 . 20 m . (b) The above equation yields zero at x = 0 . 30 m. (c) We take the derivative with respect to time and obtain u = dy dt = . 04 cos( kx ) cos( t ) = 0 at t = 0 . 50 s where x = 0 . 20 m. (d) The above equation yields u = . 126 m / s at t = 1 . 0 s. (e) The sketch of this function at t = 0 . 50 s for 0 x . 40 m is shown. 0.04 0.02 0.02 0.04 0.1 0.2 0.3 0.4 x...
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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.

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