A wave speed of a hanging chain

  • Thread starter Gravitino22
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  • #1
Gravitino22
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Homework Statement



Problem 5 from: http://www.swccd.edu/~jveal/phys274/images/hw01.pdf in case you don't understand my text.

A chain of linear mass density u, and length L is hang-
ing from a ceiling. There is a wave moving vertically
along its length. a) Is the propagation speed constant?
(Justify your answer.) b) Show that the amount of
time it takes the wave to move along the full length is
given by

t=2[tex]\sqrt{\frac{L}{g}}[/tex]

Homework Equations


String waves speed: [tex]\frac{u}{T}[/tex][tex]\frac{\delta ^{2}y}{\delta t^{2}}[/tex]= [tex]\frac{\delta ^{2}y}{\delta x^{2}}[/tex]

The Attempt at a Solution


Ive spent 2 hours trying to use the forumula for a string waves speed but I really don't understand the concept of solving the partial differential equations.

I know that the propagation speed is not constant because of gravity but i don't know how to apply that to the formula.

btw used delta for partial derivatives.

Thanks a lot :)
 
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Answers and Replies

  • #2
Doc Al
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Hint: Make use of the results of problem 2.
 
  • #3
Gravitino22
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Yes, I've tried that approach but i think the awnser lies in the differential equation that i posted which is where the velocity of the wave in a string is derived from. Unless iam overcomplicating myself and iam not seeing something.

Because i know that T=uLg and the time would be L/v but i still don't see where i would get a 2 from plugging that stuff in.
 
  • #4
Doc Al
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45,493
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Yes, I've tried that approach but i think the awnser lies in the differential equation that i posted which is where the velocity of the wave in a string is derived from. Unless iam overcomplicating myself and iam not seeing something.
Show what you've tried.

Because i know that T=uLg and the time would be L/v but i still don't see where i would get a 2 from plugging that stuff in.
Careful. The tension--and thus the speed--varies along the chain. So neither of those two expressions are correct.

Try this. Write the tension as a function of distance (x) from the bottom. Then set up and solve a simple differential equation, realizing that v = dx/dt.
 
  • #5
Gravitino22
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Ahhhh i finally got it. THANKS ALOT. My problem was that i didnt understand the concept well enough to understand that tension varies with the speed. Was easier than i thought.
 

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