451_Physics ProblemsTechnical Physics

# 451_Physics ProblemsTechnical Physics - Chapter 15 P15.39 e...

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Chapter 15 453 P15.39 T = 0250 . s , Im r == × × −− 23 3 2 20 0 10 5 00 10 .. kg m ej e j (a) I 500 10 7 k g m 2 (b) I d dt 2 2 θ κθ =− ; κ ω π IT 2 κω × F H G I K J I 27 2 4 2 316 10 . . . Nm rad FIG. P15.39 Section 15.6 Damped Oscillations P15.40 The total energy is Em v k x =+ 1 2 1 2 22 Taking the time-derivative, dE dt mv dx dt kxv 2 2 Use Equation 15.31: md x dt kx bv 2 2 dE dt vk xb v k v x =−− + af Thus, dE dt bv < 2 0 P15.41 i 15 0 . t ° 1000 550 bg . xA e bt m = 2 x x Ae A e i bt m bm 2 1000 2 15 0 = . . ln . . . . 15 0 100 2 2 100 10 31 F H G I K J = ∴= × b m b m s P15.42 Show that e t bt m 2 cos ωφ is a solution of = kx b dx dt m dt 2 2 (1) where F H G I K J k m b m 2 2 . (2) e t bt m 2 cos (3) dx dt Ae b m tA e t bt m bt m F H G I K J +− + 2 cos sin ω ωφ (4) dt b m Ae b m e t bt m bt m
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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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