443_Physics ProblemsTechnical Physics

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

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Chapter 15 445 P15.12 m = 100 . k g , k = 25 0 . N m, and A = 300 c m . At t = 0, x =− c m (a) ω == = k m 25 0 500 . . r a d s so that, T = 22 126 π . s (b) vA max .. . == × = 300 10 0150 2 m r a d s m s bg aA max . × = 2 0750 r a d s m s 2 (c) Because x . cm and v = 0 at t = 0, the required solution is xA t cos or xt .c o s . a f cm v dx dt t 15 0 5 00 .s i n . af cm s a dv dt t 75 0 5 00 o s . 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. 2 628 T and max . = 6 28 0 100 0 628 s m m s . P15.14 (a) max = A v v max ωω (b) t v t = − F H G I K J sin sin 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 the half-maximum points: KU if += + 0 1 2 1 2 1 2 2 +=+ kA mv kx 1 2 650 0100
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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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