Waves in a vertical hanging string

[Summer95]

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

 

[haruspex]

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]

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]

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?

Thinking about this some more, I'm not comfortable with simply treating it as a normal traveling 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?