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476_Physics ProblemsTechnical Physics

# 476_Physics ProblemsTechnical Physics - 478 P16.14 Wave...

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478 Wave Motion P16.14 (a) See figure at right. (b) T = = = 2 2 50 3 0 125 π ω π . . s This agrees with the period found in the example in the text. y (cm) t (s) 10 0 0.1 0.2 —10 FIG. P16.14 P16.15 (a) A y = = = max . . 8 00 0 080 0 cm m k = = = 2 2 0 800 7 85 1 π λ π . . m m a f ω π π π = = = 2 2 3 00 6 00 f . . a f rad s Therefore, y A kx t = + sin ω a f Or (where y t 0 0 , b g = at t = 0 ) y x t = + 0 080 0 7 85 6 . sin . b g b g π m (b) In general, y x t = + + 0 080 0 7 85 6 . sin . π φ b g Assuming y x , 0 0 b g = at x = 0 100 . m then we require that 0 0 080 0 0 785 = + . sin . φ b g or φ = − 0 785 . Therefore, y x t = + 0 080 0 7 85 6 0 785 . sin . . π b g m P16.16 (a) 0.0 –0.1 –0.2 0.1 0.2 y (mm) x (mm) t = 0 0.2 0.4 FIG. P16.16(a) (b) k = = = 2 2 0 350 18 0 π λ π . . m rad m T f = = = 1 1 12 0 0 083 3 . .
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