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 T xg = µ , the weight of a length x , of rope. Therefore, v gx = But v dx dt = , so that dt dx gx = and t dx gx g x L 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 T TL m xg MgL m dx dt = = = + F H G I K J = µ . (a) Then t dt xg MgL m dx t L = = + F H G I K J L N M O Q P z z 0 1 2 0 t g xg MgL m x x L = + = = 1 1 2 1 2 0 b g 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 1 2 1 2 t L g m M 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 M m M + = + F H G I K J = + + F H G I K J 1 1 1 2 1 8 1 2 2 2 to obtain t L g M m M m M M m = + + F H G G I K J J 2 1 2 1 8 2 3 2 e j 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
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