441_Physics ProblemsTechnical Physics

441_Physics ProblemsTechnical Physics - Chapter 15 P15.5...

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Chapter 15 443 P15.5 (a) At t = 0, x = 0 and v is positive (to the right). Therefore, this situation corresponds to xA t = sin ω and vv t i = cos Since f = 150 . H z , ωπ π == 23 0 0 f . Also, A = 200 c m , so that xt = 300 .s i n . cm af (b) A i max .. . == = = = 200300 600 188 cm s . cm s The particle has this speed at t = 0 and next at t T 2 1 3 s (c) aA max . = = 2 2 2 2 00 3 00 18 0 178 cm s cm s 22 This positive value of acceleration first occurs at tT 3 4 0500 s (d) Since T = 2 3 s and A = c m , the particle will travel 8.00 cm in this time. Hence, in 100 3 2 s = F H G I K J T , the particle will travel 8 00 4 00 12 0 . cm cm cm += . P15.6 The proposed solution x t v t i i =+ F H G I K J cos sin implies velocity v dx dt v t ii + ωω sin cos and acceleration a dv dt v t x t v tx i i = + F H G I K J F H G I K J =− 2 cos sin cos sin (a) The acceleration being a negative constant times position means we do have SHM, and its angular frequency is . At t = 0 the equations reduce to xx i = and i = so they satisfy all the requirements. (b) v a x x tv t x t v t i i i i 2 2 2 −= + + F H G I K J F H G I K J sin cos cos sin cos sin bg ej va x x
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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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