489_Physics ProblemsTechnical Physics

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

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern