# A wave speed of a hanging chain

• Gravitino22
Gravitino22

## 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?
time it takes the wave to move along the full length is
given by

t=2$$\sqrt{\frac{L}{g}}$$

## Homework Equations

String waves speed: $$\frac{u}{T}$$$$\frac{\delta ^{2}y}{\delta t^{2}}$$= $$\frac{\delta ^{2}y}{\delta x^{2}}$$

## 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 :)

Last edited:

Mentor
Hint: Make use of the results of problem 2.

Gravitino22
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.

Mentor
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.

Gravitino22
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.