Waves - I just need to confirm an idea

[DivGradCurl]

Hello everyone!

I've got a question from the book "Fundamentals of Physics/Halliday, Resnick, Walker" - 6th ed, page 395, #23P.

A uniform rope of mass m and length L hangs from a ceiling.

(a) 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)

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(a) Follows from the wave speed on a string definition - v = sqr([tau]/[mu]), where [tau] = mg, and [mu] = m/y --- I think that's right.

(b) I think that the solution might follow from the period of a simple pendulum --- T = sqr(L/g)... but I'm not quite sure how the 2 comes up... it's just a guess.

I hope you guys can give a hand.

Thanks a lot.

 

[M98Ranger]

Hi there! It looks like you are trying to solve a problem related to waves and their speed and time of travel. I am not familiar with the specific problem from the book you mentioned, but I can offer some general guidance and confirm your ideas.

For part (a), you are correct in using the wave speed on a string formula, v = sqr([tau]/[mu]), where [tau] is the tension in the rope and [mu] is the linear mass density. In this case, [tau] = mg, where m is the mass of the rope and g is the acceleration due to gravity. The linear mass density [mu] can be expressed as m/L since the rope is uniform. Substituting these values in the formula, we get v = sqr((mg)/(m/L)) = sqr(gL/m) = sqr(gy), where y is the distance from the lower end of the rope.

For part (b), your intuition is correct in using the period of a simple pendulum formula, T = 2π sqr(L/g). The 2 comes from the fact that the wave travels up and down the length of the rope, so we need to multiply by 2 to get the total time it takes to travel the length of the rope. Substituting this value in the formula, we get T = 2sqr(L/g).

I hope this helps and good luck with your problem solving! Let me know if you have any other questions or need further clarification.