P17_058 - x direction(d The wavelength is = 2 π/k ≈ 31 m or 31 cm(e We obtain u = dy dt = − ωy m cos kx − ωt = ⇒ u m = ωy m so that

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58. We use Eq. 17-2, Eq. 17-5, Eq. 17-9, Eq. 17-12, and take the derivative to obtain the transverse speed u . (a) The amplitude is y m =2 . 0 mm. (b) Since ω = 600 rad/s, the frequency is found to be f = 600 / 2 π 95 Hz. (c) Since k = 20 rad/m, the velocity of the wave is v = ω/k = 600 / 20 = 30 m/s in the
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Unformatted text preview: + x direction. (d) The wavelength is λ = 2 π/k ≈ . 31 m, or 31 cm. (e) We obtain u = dy dt = − ωy m cos( kx − ωt ) = ⇒ u m = ωy m so that the maximum transverse speed is u m = (600)(2 . 0) = 1200 mm/s, or 1 . 2 m/s....
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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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