Propagation of standing wave

Question

We know wave is just a transmission of energy. In case of mechanical wave, this transmission of energy requires a medium so that the particles of medium can help in transfer of energy.

Let us think of a string of particles which are equally spaced out that is uniformly distributed on the string. Name those particles alphabetically from A to Z from one end of the string. Now I want to create a wave with this string. So with the help my two fingers I lifted the 1st particle that is te A particle. Now because of the attraction force between A and B, B moves upward a little bit. Similarly C and D. also moved a little bit upward. When I released the particle A, due to elasticity (ignoring gravity) it tries to move to its equilibrium position (ignore resistances). But the B particle moves a bit more upward at a height where A was initially when it was lifted, due to inertia. In this way a wave is formed and propagates along the string. Now suppose I stopped the particle by holding it firmly. So it is certain that since I resisted the oscillation of E, the wave energy will not be able to reach F. I guess I got it right upto this. I thought of this because initially B wasn't oscillating but when I started to lift A, B also initiated oscillation. So it seems that the spontaneous oscillation of a particle depends on the precedent one unless there is any external force applied. I don't exactly know whether this is correct or not but it seemed intuitively correct.

Now moving on to the main part of the question. In a standing wave intially a travelling wave progresses and then refelcts back to superimpose with the 1st wave (the traveling wave not the reflected one which will be referred as the 2nd wave from now on). Both of these waves are continuous( basically the 1st wave). In case of standing wave we know some nodal points are formed where there is no oscillation. Ao my question is that if a nodal point is formed then how does energy propagate to the succeeding particle? I mean previously we noticed that wave energy is not able to progress towards a particle when the preceeding particle isn't oscillating (considering there is no external forces applied on the particle we are dealing with).

Please help me figure out this problem. I have asked a bunch of people but could not get any satisfying answer.

If there is anything wrong about fundamentals that I have known then please help me to understand where I am wrong.

Answer 1

Sometimes a figure is worth a thousand words:

Answer 2

(a) Consequences of Linearity

[The routine treatment of standing wave formation in this section does not directly address the propagation of waves through nodes, but the argument in the next section depends on it.]

Particle displacements, $\xi$ in a 1-dimensional linear medium (such as a stretched string) obey the partial differential equation $$\frac{\partial^2 \xi }{\partial^2 t} = v^2\ \frac{\partial^2 \xi }{\partial^2 x}\ \ \ (\text{eq 1)}$$
It is easy to show that solutions to this equation include $$\xi =A\cos 2\pi \left(f t - \frac {x}{\lambda}\right)\ \ \ (\text{eq 2)}$$ $$\xi = A\cos 2\pi \left(f t + \frac {x}{\lambda}\right) \ \ \ (\text{eq 3)}$$ $$ \xi =A\cos 2\pi \left(f t - \frac {x}{\lambda}\right)\ \ +\ \ A\cos 2\pi \left(f t + \frac {x}{\lambda}\right)\ \ \ (\text{eq 4)}$$ in which $f\lambda=v$

Eq. 1 represents a progressive wave (W+, let us call it) propagating in the $+x$ direction. Eq. 2 represents a progressive wave (W–) propagating in the $-x$ direction. Equation 3 shows that another displacement, $\xi $ that fits eq.1 is the vector sum of the displacements due to the two progressive waves – each unmodified by the presence of the other.

The identity $\cos \theta + \cos \phi = 2\left(\cos \frac12 (\theta - \phi)\right)\left( \cos \frac12 (\theta + \phi)\right)$ enables us to write eq 4 as $$\xi =2A\left(\cos \frac {2\pi}{\lambda} x\right) \left(\cos 2\pi f t \right)$$ The $\cos 2\pi f t$ factor represents in-phase vibrations for all $x$. The amplitude is $2A\cos \frac {2\pi}{\lambda} x $, and this factor dictates zero amplitude and a phase reversal when $x=(n+\frac12)\frac{\lambda}{2}$. Thus two progressive waves of equal amplitude and frequency, with the same polarisation, travelling in opposite directions, will propagate unaffected by each other and give us a standing wave according to the definition:

“A standing wave is a collection of oscillating particles with amplitude varying periodically with distance along an axis, going through zeros at ‘nodes’. The oscillations are in phase between adjacent nodes, but there is a phase reversal at each node.”

(b) Dissipating the paradox: how can progressive waves propagate through nodes, as there is never any displacement at a node?

We have shown a consequence of the linearity of eq 1 to be that progressive waves propagate simultaneously in the same medium as if each were the only wave. But a standing wave is simply the vector sum of two suitable progressive waves, W+ and W–. So as W+ propagates unaffected by W–, it propagates unaffected by the standing wave, because W– is all there is in the medium apart from W+ itself when there is a standing wave!

Since W+ and W- propagate as if the standing wave weren’t there, each progressive wave is unaffected by nodes! In the progressive wave picture, nodes arise, and are maintained, instant by instant, through the vector addition of displacements due to the waves themselves. They are not pre-established features with which the progressive waves have to contend.

Answer 3

Ok I see what youre asking. How can energy go through a node that doesnt move, when holding a part of the string to not move would STOP a wave? Right?

What you have to understand is that the model is based on a theoretically perfect string, which doesnt exist. So in the mathematical limit, the two waves are going opposite directions and each passes the full energy through the string. In real life there would be horizontal as well as vertical motion, and there would be at least SOME elasticity (meaning the string bends and stretches ever so slightly at the node when it transmits energy through the node, and this could happen with standing waves but could not happen if you held the string very firmly).

In real life, we have two factors: 1. the waves not being perfectly timed to each other and hence the node moving a teeny bit but very quickly, and 2. the stretching and bending of the string lets the energy pass. Holding the string eliminates both in real life.

Answer 4

Now because of the attraction force between A and B, B moves upward a little bit. Similarly C and D. also moved a little bit upward.

This is not the complete description. If you move A upwards, the string is no longer a straight line and in addition to the transversal movement (perpendicular to the string), there is also a longitudinal movement (along the string). Without any elasticity (in solids) or viscosity (in liquids) or a tendency to equalise density (in gases), mechanical waves are not possible.

In case of standing wave we know some nodal points are formed where there is no oscillation. … if a nodal point is formed then how does energy propagate to the succeeding particle?

First at all half waves between two walls are easy to evoke. The transversal kinetic energy together with the tension in the string (the string has and has to have the behavior of a spring) allows a half wave oscillation. On the ends of the string periodically the tension changes from min to max and back.

Now, the nodes with the tensions from the left and right are in equilibrium, if and only if the string has the same amplitude with opposite directions of these amplitudes for adjacent half-waves.

I mean previously we noticed that wave energy is not able to progress towards a particle when the preceeding particle isn't oscillating …

That is correct. In standing waves, there is no propagation of energy. With each half-wave, the kinetic energy is converted into tension energy and back. If you stop a half-wave with your hands at one moment and hold the string exactly up to the nodes, the remaining half-waves will continue to vibrate.