P17_019 - ω = 2 πf = 2 π(30 Hz = 190 rad s and the angular wave number is k = 2 π/ = 2 π(0 40 m = 16 m − 1 According to the graph the

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19. (a) We read the amplitude from the graph. It is about 5 . 0cm. (b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x =55cm ,so λ =55cm 15 cm = 40 cm = 0 . 40 m. (c) The wave speed is v = p τ/µ , where τ is the tension in the string and µ is the linear mass density of the string. Thus, v = s 3 . 6N 25 × 10 3 kg / m =12m / s . (d) The frequency is f = v/λ =(12m / s) / (0 . 40 m) = 30 Hz and the period is T =1 /f =1 / (30 Hz) = 0 . 033 s. (e) The maximum string speed is u m = ωy m =2 πfy m =2 π (30 Hz)(5 . 0 cm) = 940 cm / s=9 . 4m / s. (f) The string displacement is assumed to have the form y ( x, t )= y m sin( kx + ωt + φ ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. The amplitude is y m =5 . 0 × 10 2
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Unformatted text preview: ω = 2 πf = 2 π (30 Hz) = 190 rad / s, and the angular wave number is k = 2 π/λ = 2 π/ (0 . 40 m) = 16 m − 1 . According to the graph, the displacement at x = 0 and t = 0 is 4 . × 10 − 2 m. The formula for the displacement gives y (0 , 0) = y m sin φ . We wish to select φ so that 5 . × 10 − 2 sin φ = 4 . × 10 − 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 φ = 0 . 93 rad. The expression for the displacement is y ( x, t ) = (5 . × 10 − 2 m) sin £ (16 m − 1 ) x + (190 s − 1 ) t + 0 . 93 ¤ ....
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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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