# Waves in a vertical hanging string

I want to make sure I have this right before I move on to the rest of this problem!

1. Homework Statement

A string with mas M and length L is hanging freely and the mass is uniformly distributed along its length.

## Homework Equations

a: Fend the tension in the string as a function of x, M, L , and g

b: Show that speed of transverse wave v is x dependent and find v as a function of x, M, L, and g.

c:

## The Attempt at a Solution

a: $F_{T}=\frac{L-x}{L}Mg$

b: $v=\sqrt{\frac{F_{T}}{\mu }}=\sqrt{\frac{\frac{L-x}{L}Mg}{\frac{M}{L}}}=\sqrt{\frac{L(L-x)g}{L}}=\sqrt{(L-x)g}$

## Answers and Replies

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haruspex
Homework Helper
Gold Member
a: $F_{T}=\frac{L-x}{L}Mg$

b: $v=\sqrt{\frac{F_{T}}{\mu }}=\sqrt{\frac{\frac{L-x}{L}Mg}{\frac{M}{L}}}=\sqrt{\frac{L(L-x)g}{L}}=\sqrt{(L-x)g}$
Looks right. (Assuming x is measured from the top :)

DEvens
Gold Member
It's an interesting question to think about. It seems to be saying that the speed goes to zero at the bottom end of the string. Does that mean that a wave in the string will stop dead at the bottom end of the string? Does that seem to be what really would happen in a vertically hanging string?

haruspex
Thinking about this some more, I'm not comfortable with simply treating it as a normal travelling wave with a speed that changes along the length. Writing out the differential equation and applying separation of variables, the time-dependent part is SHM, but for the x-dependent (after normalising somewhat) I get xX"+X'+X=0. That has the series solution $\Sigma \frac{x^i}{{i!}^2}$. Anyone see how to get that into a known closed form?