# Propagation of waves on a string

1. Jan 17, 2012

### maxwellcauchy

1. Why do waves propagates on a string? I mean, when we create a disturbance on one end, it affects all the string. How can it be?

2. How can there be a continuous flow of energy along a medium when the particles of the medium simply oscillate about their equilibrium positions?

3 Intuitively, when the string is tied to a free end (using massless ring), why does the ring moves upward as high as 2x the amplitude of the wave?

Please give some references also. Thank you!

Last edited: Jan 17, 2012
2. Jan 17, 2012

### serverxeon

waves are by definition, a method of transmitting energy simply with object oscillating about a fixed position.

I guess there are no 'why' to you questions. That is how the world is, and we seek to model these behaviours.

What I can answer is that energy is transmitted simply because the medium is linked. I guess a crude way is to visualise a particle 'pulling' another adjacent particle to move along with it.

3. Jan 17, 2012

### Ken G

This is a very deep question about waves. The string must have a tension, and the tension provides a restoring force, but it is a restoring force with an interesting aspect-- it depends not on the displacement of one point on the string (like a chain of disconnected harmonic oscillators would, but they would not carry a wave), but rather on the curvature of the string (you can see this by considering a finite segment and noting that if it curves, the tension at both ends of the segment does not cancel, but instead provides that restoring force). So the dependence on string curvature is what gives rise to the "wave equation" mathematically, but physically that is just the "connectedness" element that maxwellcauchy referred to. It's not purely local to a point on the string, it's the way that point knows about its neighbors, and that is what allows there to be propagation. A little connectedness goes a long way!

It is the same for other waves. Consider sound waves in air-- there it is not the displacement of the air particles from their equilibrium that matters, it is the curvature in that displacement. A fixed displacement doesn't do anything, and a displacement that varies linearly with distance only causes the pressure to either go up or down but it is a curvature (second derivative) in the displacement as a function of distance that causes a pressure gradient, and the pressure gradient is what creates forces on the air that causes it to oscillate. The nonlocal character of the curvature is also what allows the wave to propagate.
Another deep question about waves! Again the answer is the dependence on curvature, which introduces a nonlocal character such that what one part of the medium does depends on what its neighbors are doing. This crosstalk between parts of the medium allows energy in one part to get transferred to another part-- the different parts can do work on each other if the motion of one part creates forces on the other parts. These forces are transverse to the wave, if we have transverse waves on a string, so the energy itself is due to transverse motions. However, that energy is transferred longitudinally along the string-- there is longitudinal transport of transverse motion.
This I don't understand-- perhaps you are confusing the amplitude with the total range of displacements, where the former is half the latter?

4. Jan 18, 2012

### maxwellcauchy

I'll change my question, When the pulse struck the massless ring on the free end, the ring will go up and then back to y=0 and reflect the pulse without change of sign. Is it true that the ring will go up as high as 2*y ( twice the amplitude) ?

5. Jan 19, 2012

### Ken G

Yes, I see what you are saying. The ring cannot support any transverse forces, so this means the string always has to come in perpendicular to the rod the ring is on. The way to do this is if the wave reflects off the end in phase with what comes in. That will also double the amplitude of what comes in. Note that if the end is fixed, the displacement is what needs to be zero. That is accomplished if the reflected wave reverses its amplitude.

6. Jan 19, 2012

### maxwellcauchy

Thank you Ken G!