488_Physics ProblemsTechnical Physics

488_Physics ProblemsTechnical Physics - 490 Wave Motion T...

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490 Wave Motion P16.55 (a) v T == × = µ 80 0 500 10 179 3 . . N kg 2.00 m ms ej (b) From Equation 16.21, P = 1 2 22 µω vA and ωπ λ = F H G I K J 2 v = F H G I K J = = F H I K = × 1 2 2 0 040 0 179 0160 177 10 177 2 2 3 2 2 23 2 4 3 π πµ vA v . . . .. kg 2.00 m m m s m W k W bg b g af P16.56 v T = and in this case Tm g = ; therefore, m v g = 2 . Now vf = implies v k = ω so that m gk = F H G I K J = L N M O Q P = 2 1 1 2 0250 980 18 0750 14 7 . . kg m ±ms s m kg 2 . *P16.57 Let M = mass of block, m = mass of string. For the block, Fm a = implies T mv r mr b 2 2 . The speed of a wave on the string is then v TM r r M m t r v m M t m M m r = = = θω 2 1 00032 00843 . . kg 0.450 kg rad P16.58 (a) µρ ρ = dm dL A dx dx A v TT A T ax b
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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