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443_Physics ProblemsTechnical Physics

# 443_Physics ProblemsTechnical Physics - Chapter 15 P15.12...

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Chapter 15 445 P15.12 m = 1 00 . kg, k = 25 0 . N m, and A = 3 00 . cm . At t = 0, x = − 3 00 . cm (a) ω = = = k m 25 0 1 00 5 00 . . . rad s so that, T = = = 2 2 5 00 1 26 π ω π . . s (b) v A max . . . = = × = ω 3 00 10 5 00 0 150 2 m rad s m s b g a A max . . . = = × = ω 2 2 2 3 00 10 5 00 0 750 m rad s m s 2 b g (c) Because x = − 3 00 . cm and v = 0 at t = 0, the required solution is x A t = − cos ω or x t = − 3 00 5 00 . cos . a f cm v dx dt t = = 15 0 5 00 . sin . a f cm s a dv dt t = = 75 0 5 00 . cos . a f cm s 2 P15.13 The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s. ω π = = 2 6 28 T . s and v A max . . . = = = ω 6 28 0 100 0 628 s m m s b ga f . P15.14 (a) v A max = ω A v v = = max ω ω (b) x A t v t = − = F H G I K J sin sin ω ω ω Section 15.3 Energy of the Simple Harmonic Oscillator P15.15 (a) Energy is conserved for the block-spring system between the maximum-displacement and
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