Waves in a vertical hanging string

  • Thread starter Summer95
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  • #1
Summer95
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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

  • #2
haruspex
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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 :)
 
  • #3
DEvens
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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?
 
  • #4
haruspex
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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 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?
 

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