wave propagating along a string -- Newton's 3rd Law

[Jack3000]

I'm a bit confused about the action-reaction forces (ie. Newton's Third Law) when a wave propagates along a string. I think I'm confused about a point not mentioned elsewhere in the forum.

Let's imagine the string is made up of many tiny particles, and that it carries a wave propagating in the right to left direction. A person holding the right end of the string rapidly displaces that end upwardly, then rapidly displaces that end back to its original position. Then if one point somewhere along the string (let us call it "x") waves upward, it places a force on the point just to its left (x+dx). By Newton's third law, wouldn't this point (x+dx) push on point "x" with an equal and opposite force?

It seems to me that such an equal and opposite force woud tend to return the string just behind the wave, to its original position. But I thought that it was the downward movement of the hand that returns the string to its original position?

Thanks!

[arildno]

When you wish to understand how tension in a string can make a wave propagate, it is important to realize that the tension counteracts local curvature.
Another way of saying this, is that tension seeks to minimize the string length (i.e, seeking to reestablish the string as a straight line)
I'll post some calculations a bit later

[arildno]

"A person holding the right end of the string rapidly displaces that end upwardly, then rapidly displaces that end back to its original position. Then if one point somewhere along the string (let us call it "x") waves upward, it places a force on the point just to its left (x+dx). By Newton's third law, wouldn't this point (x+dx) push on point "x" with an equal and opposite force?"

Certainly!
(In the following, I will assume that the other end of the string is fixed to some wall)
In addition, (assuming, for the moment that x is at a local top), x exerts an upwards force on a point at (x-dx).
By Newton's 3.law, the point at (x-dx) also exerts a downward force on "x".

"It seems to me that such an equal and opposite force woud tend to return the string just behind the wave, to its original position."

No it does not!
As long as "x" is higher placed than "x+dx" and "x-dx", "x" experiences a net force downwards, i.e accelerates downwards, which means gaining downwards velocity.
What happens then at the moment at which "x", "x+dx", "x-dx" are all on a line?
Well, "x" has downwards velocity, wheras "x+dx" and "x-dx" have upwards velocity.
Even though at that moment "x" does not experience a net vertical force,
it will proceed downwards, and will begin experience a net, upwards force (i.e, slowing down).

"But I thought that it was the downward movement of the hand that returns the string to its original position?"

The above argument shows, that if you disturb a string , you will set up a wave signal that will continue indefinitely, that will reflect at the other end, return, and reflect again..
Essentially, the initially imparted kinetic and potential energies will swith about, according to conservation of energy.

If we have both ends fixed for a vibrating string, the vibration will eventually die out as a result of frictional forces/air resistance (i.e, damping).

[arildno]

I'm not altogether sure if I have answered what you wanted an answer to here..