489_Physics ProblemsTechnical Physics

# 489_Physics ProblemsTechnical Physics - Chapter 16 P16.59...

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Chapter 16 491 P16.59 v T = µ where Tx g = , the weight of a length x , of rope. Therefore, vg x = But v dx dt = , so that dt dx gx = and t dx gx g xL g L L == = z 0 1 2 0 1 2 P16.60 At distance x from the bottom, the tension is T mxg L Mg = F H G I K J + , so the wave speed is: v TT L m xg MgL m dx dt =+ F H G I K J = . (a) Then td t x g MgL m dx tL + F H G I K J L N M O Q P zz 0 12 0 t g xg MgL m x = + = = 1 1 2 0 bg t g Lg MgL m MgL m F H G I K J F H G I K J L N M M O Q P P 2 t L g mM M m = +− F H G I K J 2 (b) When M = 0 , as in the previous problem, t L g m m L g = F H G I K J = 2 0 2 (c) As m 0 we expand m M M m M m M += + F H G I K J −+ F H G I K J 11 1 2 1 8 2 2 to obtain t L g Mm M m M M m = + F H G G I K J J 2 1 2 1 8 23 2 ej e j t L g m M mL Mg F H G I K J = 2 1 2 P16.61 (a) The speed in the lower half of a rope of length L is the same function of distance (from the bottom end) as the speed along the entire length of a rope of length L 2 F H G I K J . Thus, the time required
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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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