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441_Physics ProblemsTechnical Physics

# 441_Physics ProblemsTechnical Physics - Chapter 15 P15.5(a...

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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 x A t = sin ω and v v t i = cos ω Since f = 1 50 . Hz, ω π π = = 2 3 00 f . Also, A = 2 00 . cm, so that x t = 2 00 3 00 . sin . cm a f π (b) v v A i max . . . = = = = = ω π π 2 00 3 00 6 00 18 8 a f cm s . cm s The particle has this speed at t = 0 and next at t T = = 2 1 3 s (c) a A max . . . = = = = ω π π 2 2 2 2 00 3 00 18 0 178 a f cm s cm s 2 2 This positive value of acceleration first occurs at t T = = 3 4 0 500 . s (d) Since T = 2 3 s and A = 2 00 . cm, the particle will travel 8.00 cm in this time. Hence, in 1 00 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 x t v t i i a f = + F H G I K J cos sin ω ω ω implies velocity v dx dt x t v t i i = = − + ω ω ω sin cos and acceleration a dv dt x t v t x t v t x i i i i = = − = − + F H G I K J F H G I K J = − ω ω ω ω ω ω ω ω ω 2 2 2 cos sin cos sin (a) The acceleration being a negative constant times position means we do have SHM, and its angular frequency is ω . At t = 0 the equations reduce to x x i = and v v i = so they satisfy all the requirements. (b) v ax x t v t x
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