Changing wavelength with changing tension

[Mr Davis 97]

Homework Statement

A rope is hanging vertically straight down. The top end is being vibrated back and forth, and a standing wave with many loops develops on the rope, analogous (but not identical) to a standing wave on a horizontal rope. The rope has mass. The separation between successive nodes is (a) everywhere the same along the rope (b) greater near the top of the rope than near the bottom (c) greater near the bottom of the rope than near the top.

Homework Equations

$\displaystyle f_n = \frac{n}{2L} \sqrt{\frac{T}{ \mu }}$, where $T$ is tension, $\mu$ is linear density of the rope, $L$ is the length of the rope, and $n$ is the harmonic number.

The Attempt at a Solution

The solution is (b) greater near the top of the rope than near the bottom. However, I can't see why. I tried to reason that as we go up the rope, tension increases because of the added mass below polling down on the point. This would increase T as we go up the rope. This means that frequency would increase. Wavelength is inversely proportional to frequency, and proportional to the the distance between nodes, so I thought that as frequency increases, wavelength decreases, and the distance between nodes decreases. However, this is wrong because the answer is that the distance increases as we go up the rope. What am I doing wrong?

 

[Henryk]

This would increase T as we go up the rope. This means that frequency would increase

that is where you are going wrong.
We are talking about a standing wave pattern for which the frequency is fixed !.
This is not an exact calculation, but will do for the purpose of this question. The question you should ask first, given the frequency, what is the distance between nodes given the string mass density (constant) and tension. In other words, take your equation, set n =1. L is now the distance between the nodes.
If f is fixed, then $L \sim \sqrt{T}$ i.e. larger at the top of the string