# Homework Help: A wave speed of a hanging chain

1. Jan 29, 2010

### Gravitino22

1. The problem statement, all variables and given/known data

Problem 5 from: http://www.swccd.edu/~jveal/phys274/images/hw01.pdf in case you dont 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$$\sqrt{\frac{L}{g}}$$
2. Relevant equations
String waves speed: $$\frac{u}{T}$$$$\frac{\delta ^{2}y}{\delta t^{2}}$$= $$\frac{\delta ^{2}y}{\delta x^{2}}$$

3. The attempt at a solution
Ive spent 2 hours trying to use the forumula for a string waves speed but I really dont understand the concept of solving the partial differential equations.

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

btw used delta for partial derivatives.

Thanks alot :)

Last edited: Jan 29, 2010
2. Jan 29, 2010

### Staff: Mentor

Hint: Make use of the results of problem 2.

3. Jan 29, 2010

### Gravitino22

Yes, ive 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 dont see where i would get a 2 from plugging that stuff in.

4. Jan 30, 2010

### Staff: Mentor

Show what you've tried.

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. Jan 30, 2010

### 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.