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440_Physics ProblemsTechnical Physics

440_Physics ProblemsTechnical Physics - 442 Oscillatory...

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442 Oscillatory Motion Section 15.2 Mathematical Representation of Simple Harmonic Motion P15.2 (a) x t = + F H G I K J 5 00 2 6 . cos cm a f π At t = 0, x = F H G I K J = 5 00 6 4 33 . cos . cm cm a f π (b) v dx dt t = = − + F H G I K J 10 0 2 6 . sin cm s b g π At t = 0, v = 5 00 . cm s (c) a dv dt t = = − + F H G I K J 20 0 2 6 . cos cm s 2 e j π At t = 0, a = 17 3 . cm s 2 (d) A = 5 00 . cm and T = = = 2 2 2 3 14 π ω π . s P15.3 x t = + 4 00 3 00 . cos . m a f a f π π Compare this with x A t = + cos ω φ b g to find (a) ω π π = = 2 3 00 f . or f = 1 50 . Hz T f = = 1 0 667 . s (b) A = 4 00 . m (c) φ π = rad (d) x t = = = 0 250 4 00 1 75 2 83 . . cos . . s m m a f a f a f π *P15.4 (a) The spring constant of this spring is k F x = = = 0 45 9 8 0 35 12 6 . . . . kg m s m N m 2 we take the x -axis pointing downward, so φ = 0 x A t = = = = cos . cos . . . cos . . ω 18 0 12 6 84 4 18 0 446 6 15 8 cm kg 0.45 kg s s cm rad cm 2 (d) Now 446 6 71 2 0 497 . . rad rad = × + π . In each cycle the object moves 4 18 72 a f = cm , so it has moved 71 72 18 15 8 51 1 cm cm m a f a f + = . . . (b) By the same steps, k = = 0 44 9 8 0 355 12 1 . . . . kg
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