Question on waves where a rope attached to ceiling.

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SUMMARY

The discussion focuses on the dynamics of a uniform rope of mass m and length L hanging from a ceiling, specifically analyzing the tension and wave speed. For part (a), the tension is derived as T = Mg, leading to the wave speed equation v = √(gy). In part (b), the time for a transverse wave to travel the length of the rope is established as T = 2√(L/g). The participants clarify the relationship between wave speed and tension, emphasizing the need for correct application of Newton's second law.

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Homework Statement


A uniform rope of mass m and length L hangs from a ceiling. The mass of the rope below a point P is M.

(a) Write down Newtons second law for the segment of the rope of mass M and find the tension ata distance y above the lower end of the rope. Show that the speed of a transverse wave on the rope is a function of y, the distance from the lower end, and is given by v = sqr(gy).

(b) Show that the time a transverse wave takes to travel the length of the rope is given by T = 2sqr(L/g)

***

(a) v = sqr(tension/[mu]), where tension= Mg, and [mu] = M/y --- I think that's right.



b) i do not have a clue for this bit am i meant to use speed= distance over time. If so the speed can't be the speed calculated in a) as then i do not get that answer.
Thanks.



The Attempt at a Solution

 
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You have v = √(g/y). If x=x(t) is the distance of a wave crest from the bottom of the rope at time t, what differential equation can you write down for x?
 

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