# Can you correctly measure wave properties but get wrong wavelength?

bkraabel

## Homework Statement

You and a friend each have one rope. You tie the two ropes together and stand as far apart as possible, each holding one end of the new longer rope and pulling to put it under tension. You then begin moving your arm in such a way as to produce a harmonic wave with a wavelength of 1.0 m. Your friend looks at the waves as they reach her arm. Is it possible that she measures a wavelength of (a) 0.8 m, (b) 1.0 m, or (c) 1.2 m?

## Homework Equations

For standing wave with BOTH ends fixed: $\lambda=\frac{2 \ell}{n-1}$ where $n$ is the number of nodes (including the node at each end) and $\ell$ is the length of the rope.

If one end of the rope is driven, this end is an antinode, so the wavelength would be so you get $\lambda/4+(n-1)\lambda/2=\ell$ , which simplifies to $\lambda=\frac{4\ell}{2n-1}$

We also have the following relation between wave speed $c$, frequency and wavelength: $c=\lambda f$.

## The Attempt at a Solution

I assume the friend does not more her arm, so her end of the rope is a node. Your end of the rope is an antinode. The friend "looks at the waves as they reach her arm." What does that mean? She measures the wave speed? Frequency? Amplitude? All of those? Does she count the nodes between the two ends of the rope? That would seem to not be consistent with "looking at the waves as they reach her arm." So I assume she measures wavespeed and frequency and calculates wavelength. If she does that correctly, she should get the correct wavelength, so the answer would be
(a) no, (b) yes, (c) no
However, others argue (I don't know why) that the answer is
(a) yes, (b) yes, (c) yes

My question is: What argument(s) can you give to support the second answer (all yes)??

Thanks for any ideas.

## Answers and Replies

Mentor
Your "relevant equations" all assume that there is a single wavelength everywhere. You cannot use them here.

What does that mean? She measures the wave speed? Frequency? Amplitude? All of those?
She can probably observe all, but only the wavelength is interesting here.
Does she count the nodes between the two ends of the rope?
She measures the distance between nodes at her part of the rope. For the answer, it does not matter if there is a standing wave or not.
So I assume she measures wavespeed and frequency and calculates wavelength.
Okay, that is possible.
If she does that correctly, she should get the correct wavelength
Sure, she will get the wavelength at her part of the rope.
so the answer would be
(a) no, (b) yes, (c) no
Why? You started with the (wrong) conclusion here.

What do you know about the speed of waves in a rope?
What do you know about the frequency you and your friend will measure?

bkraabel
The problem statement says "moving your arm in such a way as to produce a harmonic wave with a wavelength of 1.0 m." Does that not means there is a single wavelength-frequency combination on the rope? Where would the other frequencies come from?
thanks, Brett

Staff Emeritus
Homework Helper
Are the two ropes identical? Does the knot have an effect?

Mentor
The problem statement says "moving your arm in such a way as to produce a harmonic wave with a wavelength of 1.0 m." Does that not means there is a single wavelength-frequency combination on the rope? Where would the other frequencies come from?
thanks, Brett
You have a well-defined wavelength at your arm. That does not mean that the wavelength has to be the same everywhere.

The frequency will stay constant, but the propagation speed of the wave does not have to be (why? -> see the question in my previous post). What happens if you have the same frequency, but different propagation speeds?

bkraabel
Ah - I guess the two ropes are not identical. I took "one rope" as meaning a single rope. So the wavelength can change if the ropes have different linear densities.

Mentor
Ah - I guess the two ropes are not identical. I took "one rope" as meaning a single rope. So the wavelength can change if the ropes have different linear densities.
Exactly.