How Does Wave Propagation Affect Rope Tension and Displacement?

[TwinGemini14]

http://i662.photobucket.com/albums/uu347/TwinGemini14/showme5.gif

Consider a triangular wave pulse shown traveling to the right on a rope with a speed v. Its height is h. The total length of the pulse is 3L, as shown in the figure. The vertical displacement is much exaggerated. The solid line in the figure is a snapshot of the wave at t = 0 and shows that the leading edge has just arrived at point A, which is indicated by the blue dot. The mass per unit length of the rope is m.

This is a snapshot of the actual rope at t = 0.


1) What is the tension (T) in the rope?

A) T = (v^2)(m^2)
B) T = (v^2)m
C) T = v(m^2)

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I believe the answer to be B. I used the formula: v^2 = sqrt(F/m) and derived the answer.
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2) At t = 2L/v, what is the displacement of point A from its t = 0 position?

A) 0
B) h/3
C) h/2
D) h
E) 3h/2

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I am not exactly sure here, but think the answer is C. When point 'A' is at the top of the peak of the wave, it will take L length. When another L passes, 'A' will be half way back to the bottom because it will take 2L to get completely back to its original position. So the answer is C) h/2
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3) Which of the following is the y versus t (vertical displacement vs. time) graph for point A.

A) http://i662.photobucket.com/albums/uu347/TwinGemini14/showme4.gif

B) http://i662.photobucket.com/albums/uu347/TwinGemini14/showme4B.gif

C) http://i662.photobucket.com/albums/uu347/TwinGemini14/showmeC.gif

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I believe the answer is C. I used similar logic as for question 2. It will go up to the peak of wave in L length of the wave. It will then decend in 2x that time. So C fits this description.
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Can somebody please look over these answers and verify my attempt? I would really appreciate it. Thanks in advance!

 

[Redbelly98]

Looks good for all 3.