# Standing Waves and nodes

I have a question about standing waves.

Is this what a standing wave is: a wave produced by a propagating wave and a reflected wave, resulting in a wave of zero propagation. A standing wave is produced, at a particular point, by the two propagating waves and it is simply a superposition of the two separate waves moving from out of phase to in phase stages (is this bit about phases correct)

see diagram: http://www.rmcybernetics.com/images/main/pyhsics/standing_wave.gif

The node is produced at a point with zero change in displacement / amplitude. It is a fixed point. This point remains fixed for two reasons
a) the superposition of the two points of the two waves ALWAYS equals zero
b) the waves cross at that point where the displacement of the point is zero.

An antinode are the moving peaks and or troughs of the wave. They form because
a) the superposition of the wave is either greater than or less than zero. BUT when the waves moving in opposing directions are in anti-phase, the two waves will cancel out producing an overall displacement of zero (antinodes can have a total displacmenent of zero but they can also reach maximum displacment as well)

is this all correct? is there anything else fundemental to the principal.

I presume standing waves DO exist in reality. The reason, when proven experimentally, that standing waves don't appear IMMEDIATLEY is due to the fact that it takes a little time for the waves to build up to maximum displacement and also we need to wait a little bit before the wave reflects. Correct?

A question on all of this: is it easy to identify from a still image at one point to identify where the nodes would be. Also what has this got to do with harmonics? (Fundemental harmonic, second harmonic (first overtone) ...)

Thanks

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The reason, when proven experimentally, that standing waves don't appear IMMEDIATLEY is due to the fact that it takes a little time for the waves to build up to maximum displacement and also we need to wait a little bit before the wave reflects. Correct?
This is not correct.

Take a rope, lie it loose along the ground and waggle one end.
You can induce a wave train to travel along the rope. That is you can create travelling waves in the rope.

However you cannot create standing waves with this arrrangement.

Now tie one end to a fixed point, pull it taught and waggle the other end up and down.
You are now creating immediate standing waves. There is no time lag or reflection involved.
Further there is no need to "build up waves to achieve maximum displacement". this happens automatically on the first stroke.

This is a YouTube video of standing waves on a flat metal plate covered by salt grains. The plate is being driven by a variable frequency vertical harmonic oscillator at the center.

[added] The salt grains move away from areas where the vertical acceleration (motion) is highest, and toward areas where there are nodes. If the vertical acceleration exceeds mg, the grains are airborne for part of the cycle. They in essence randomly walk away from regions of maximum vertical acceleration to regions of minimum acceleration (nodes) where the random walk velocity is zero.

Bob S

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The main question is: is the information I posted initially correct?

sophiecentaur
Gold Member
This is not correct.

Take a rope, lie it loose along the ground and waggle one end.
You can induce a wave train to travel along the rope. That is you can create travelling waves in the rope.

However you cannot create standing waves with this arrrangement.

.
This is not strictly true - despite what one may see when doing the experiment. Because there is a discontinuity at the loose end of the rope there must also be a reflection and a wave must travel back towards your hand. The only reason that a standing wave is not identifiable when the rope is on the ground is that friction with the ground takes away most of the energy of the wave. If you let the rope hang down (removing the friction with the ground) and do the same side-to-side wiggling you can see a standing wave with an antinode at the end (if you choose the right frequency). There is a zero in the tension at the end and a maximum in the tension at your hand. This corresponds to the situation in an open ended pipe with a vibrating air column (an odd number of quarter waves). The Q factor of the rope resonator is fairly low (the rope is lossy within itself because of the fibres) so the frequency is not critical and the last node may not be exactly a quarter wavelength from the free end but you still have a standing wave as long as the losses are not too great. It takes a finite time for the wave to build up to its final amplitude (about Q cycles will get you nearly there).

To sum up, you can only have no standing wave when there is no reflection or no net discontinuity

@ismith: I think most of what you say is pretty well right and that your basic insight is good.

sophiecentaur
Gold Member
A question on all of this: is it easy to identify from a still image at one point to identify where the nodes would be. Also what has this got to do with harmonics? (Fundemental harmonic, second harmonic (first overtone) ...)

Thanks
It is possible, with a suitable probe, to identify standing waves in mechanical and electromagnet resonating systems.

Harmonics are multiples of the fundamental frequency. Many practical resonators do not show resonances at precise harmonic frequencies - this is because of 'end effects' whereby the ends of the pipe / waveguide / crystal are not well defined. What you have with, say, an organ pipe, are called overtones. They lie close to the harmonic frequencies but are not necessarily the same value. The timbre of many musical instruments is due to this difference.
Multiple resonances can occur at the same time, of course.

Because there is a discontinuity at the loose end of the rope there must also be a reflection and a wave must travel back towards your hand
What is discontinuity and why does it form at the end of a loose rope

sophiecentaur
Gold Member
A discontinuity is where the medium the wave is flowing through changes. For instance, however clean you make a piece of glass, it will still reflect some of the light falling it. That's because there's a 'discontinuity' in the density as you go from air into glass.
The waves on the string are going along happily until they meet the 'flapping end' instead of another length of rope of similar construction. The energy in the wave has to go somewhere and it can't couple itself into the air very efficiently so it has to be reflected. In the same way, if the end of the rope is tied to a rigid support, only a minuscule amount of the energy will couple into the massive wall (or whatever) and the rest will be reflected.
When you do calculations about what will happen when a wave hits a discontinuity (or boundary), there are certain 'boundary conditions' that apply and will help in the calculation. For instance, the phase must be continuous at the boundary so that means the frequency of the waves leaving the boundary must be the same as the incident waves.

The only way to stop reflections is to 'terminate' the path of the wave with the appropriate amount of resistance / friction (the characteristic impedance) and that will absorb all the incident energy, behaving just like you hung an infinitely long line there. Then you get no standing wave at all.
Standing waves can be a real problem with transmission lines (e.g. UHF TV transmitter feeders) because you can get 'echos' which distort the signal and also 'over-voltage' at the unwanted antinodes.
You have chosen a huge topic to get interested in!!! There's no end to it.

The waves on the string are going along happily until they meet the 'flapping end' instead of another length of rope of similar construction. The energy in the wave has to go somewhere and it can't couple itself into the air very efficiently so it has to be reflected
two points:
1) why can't the energy dissapate. Since there is no medium to reflect the wave, surley it should NOT be reflected and it should be easier to dissapate energy

2) If we imagine a wave being reflected at a point, it will change phase by 180 degrees.
http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=Pic.jpg

Therefore, at each point on the wave, everything SHOULD be 180 degrees out of phase (anti-phase) and our standing wave should always have an amplitude of zero at each point. Therefore there should be NO antinodes BUT only nodes. This clearly is not the case. My question is, why not?

sophiecentaur
Gold Member
1. Wave your hands in the air and tell me if you can actually do any 'work' on the air. You will find that your hands just push against nothing (unless they travel at many m/s). That means that you are not coupling energy into the air. The flappy end of the rope just can't move enough air to dissipate the energy.
2. You make the wrong conclusion (this is not easy to visualise). The phase of the approaching wave is later and later with increasing distance from the reflection point, and the phase of the reflected wave is earlier and earlier. They will only be in complete antiphase again at a point λ/2 away - the next node. They will, of course, be in phase at a distance of λ/4 ( the antinode in between).

1. Wave your hands in the air and tell me if you can actually do any 'work' on the air. You will find that your hands just push against nothing (unless they travel at many m/s). That means that you are not coupling energy into the air. The flappy end of the rope just can't move enough air to dissipate the energy.
2. You make the wrong conclusion (this is not easy to visualise). The phase of the approaching wave is later and later with increasing distance from the reflection point, and the phase of the reflected wave is earlier and earlier. They will only be in complete antiphase again at a point λ/2 away - the next node. They will, of course, be in phase at a distance of λ/4 ( the antinode in between).
In response to point 1). Surely I am displacing air, and that in itself is a form of energy transfer?

for 2) what do you mean "The phase of the approaching wave is later and later with increasing distance from the reflection point". If it is a wave, surely it should remain constant - i..e: there should be no phase change of a specific wave other than a point on the wave being out of phase with another point at a later stage on the same wave.

This is not strictly true - despite what one may see when doing the experiment. Because there is a discontinuity at the loose end of the rope there must also be a reflection and a wave must travel back towards your hand.
I am usually in agreement with SC, but not this time. The above statement does not even make sense I'm afraid.

The loose end is in your hand. There is no discontinuity.
The node is at the point of fixity (by definition).
The antinode is at the driven end and since this follows exactly the driving 'waggle' .

Please note also that you have changed the carefully specified conditions for establishing a travelling wave in the rope.
By hanging the rope down you are inducing a longitudinal tension. The instant you do this you cannot apply the simple wave equation to the mechanics.
This was why I specified the rope was to be laid loosely and horizontally on the ground.

Of course you can generate standing waves if you apply tension to the rope, even more so if you hang a weight on the free end.

No jsmith, your original statements were in error as I outlined. I don't know what at level you are studying, but your description suggests you are confusing what are known as the steady state and transient solutions to equation of forced vibrations. Does this mean anything to you?

No jsmith, your original statements were in error as I outlined. I don't know what at level you are studying, but your description suggests you are confusing what are known as the steady state and transient solutions to equation of forced vibrations. Does this mean anything to you?

no it does not really! I am at AS Level Physics - Edexcel GCE from 2008

I am at AS Level Physics
Unfortunately too many half truths are promulgated as generalisations or simplifications at this level these days.
In the past such statements were more carefully worded.

Never mind, keep enquiring and plugging away.

At GCSE you consider waves as a repetitive function of amplitude as time progresses.
At A/AS level you will encounter simple harmonic motion. All the waves you will deal with will be of this type. You will consider more properties than just amplitude.

At GCSE you consider waves as a repetitive function of amplitude as time progresses.
At A/AS level you will encounter simple harmonic motion. All the waves you will deal with will be of this type. You will consider more properties than just amplitude.
I am still, nonetheless, confused slightly about the standing wave (more confused than when I asked the inital question - because I reaslied something important).

Sophie said:
"The phase of the approaching wave is later and later with increasing distance from the reflection point, and the phase of the reflected wave is earlier and earlier. They will only be in complete antiphase again at a point λ/2 away - the next node. They will, of course, be in phase at a distance of λ/4 ( the antinode in between)."
I am slighlty confused by what this means, and why, from what i understand of what she is saying, why it occurs.

(I think she is saying that the approaching wave takes longer and longer to reach the boundary, whereas the reflected wave leaves earlier and earlier. If the approaching wave takes longer to reach the boundary, then the leaving wave should also take longer)

cont....

Actually, watching this video, i think i understand:

this diagram
http://s359.photobucket.com/albums/oo40/jsmith613/?action=view&current=Pic.jpg
ONLY SHOWS a single point on the wave. This entire wave speciemen, so to speak, is not relfected at once, rather in parts, so the first part of that wave is reflected and propagates along, followed by the second..third.... etc.
This may seem unclear, but essentially is this what is happening?
(each individual point reflects RATHER THAN 1/3 of a wavelength reflecting at one point.

This is why I despise motionless pictures (pictures!) in physics. They don't always show the whole story, rather they show a generallised picture to make things 'easier' to understand.

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Click on second video from left in bottom row (Creating standing wave) after the travelling vave simulation finishes in:

This demonstrates zero, one, two, and three nodes (not counting end).

Bob S

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sophiecentaur
Gold Member
I am usually in agreement with SC, but not this time. The above statement does not even make sense I'm afraid.

The loose end is in your hand. There is no discontinuity.
The node is at the point of fixity (by definition).
The antinode is at the driven end and since this follows exactly the driving 'waggle' .

Please note also that you have changed the carefully specified conditions for establishing a travelling wave in the rope.
By hanging the rope down you are inducing a longitudinal tension. The instant you do this you cannot apply the simple wave equation to the mechanics.
This was why I specified the rope was to be laid loosely and horizontally on the ground.

Of course you can generate standing waves if you apply tension to the rope, even more so if you hang a weight on the free end.

No jsmith, your original statements were in error as I outlined. I don't know what at level you are studying, but your description suggests you are confusing what are known as the steady state and transient solutions to equation of forced vibrations. Does this mean anything to you?
Pistols at dawn is it now? :)

This has gone along a few tangents. Take the rope somewhere with a very small amount of gravity and let it hang - just to remove the friction with the floor. The weight force is then negligible and so would be the external friction. Better still - have a slightly rigid rope and have it under zero gravity - but in air - it will then maintain its original span (minus a bit to account for the wiggles). This is as 'ideal' as I can think of. The end of the rope can either be attached or unattached and the model still works. I was actually addressing the 'free end' option.
Your hand, the driving point, is a high impedance source i.e. the force is high and the displacement is whatever your hand wants it to be. In fact, once the wave has settled down and you have found the right frequency, you need only move your hand a very little for a large amplitude standing wave to exist. The hand (driver) is almost at a node position. Here there is a big discontinuity because the driving impedance of your hand is nothing like the characteristic impedance of the wave on the rope. This will cause a further reflection of waves travelling to the hand and increase the 'standing wave ratio'. Your hand merely feeds more power in to make up for the power lost.
The unattached end (which was the case I referred to) is a low impedance because the force on the rope end is small. The end will definitely not remain still - it will always waggle about - more or less at an antinode.

@jsmith: Compared with the energy in the rope wave, very little energy will transfer to the air. The air will be disturbed a bit but that is all. Waving your hands about can transfer very little energy (just a waft of air) - if it could transfer more then you could swim through the air using your hands - you can't. That's the point I am making. If virtually no energy is lost then the energy must travel back along the rope.

I would suggest that the simplest analysis would treat the system in a steady state. In electrical terms, you have a transmission line with imperfect terminations but they approximate to a closed circuit (a loop at the end) one end and a current source at the other (hand end). Energy losses are due to friction in the fibres (the resistance of the line) and power lost to the air (radiating power from the loop at the end). The standing wave you see is a result of multiple reflections and the standard transmission line equations will tell you about the amplitudes of maxes and mins.

I agree that the visualisation of the wave, which varies both in position and time, is hard but, as we know that standing waves occur, it might be a good idea to see a way how it can happen by considering the phases of the wave in different places rather than trying to make them fit your 'model of disbelief'. We can't demand animations for every phenomenon - our brains sometimes have to do the work! :)

Just bear in mind that the phase of the wave, at different positions, are all different - steadily advancing as the wave progresses (i.e. the phase here is earlier than the phase of the wave just before it arrives here). This applies after reflection too and full constructive and destructive interference only occurs at the appropriate places.

BTW I am a bloke - my boat is a girl. With hindsight It may have been better to choose a better moniker but it's too late now.

I agree that the visualisation of the wave, which varies both in position and time, is hard but, as we know that standing waves occur, it might be a good idea to see a way how it can happen by considering the phases of the wave in different places
Please amplify. What do you mean by phase of a standing wave? Is it any different from phase as applied to a travelling wave?

sophiecentaur
Gold Member
will continue with this when time permits

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sophiecentaur
Gold Member
Please amplify. What do you mean by phase of a standing wave? Is it any different from phase as applied to a travelling wave?
The phase I am talking about is the phase of the travelling wave (as it goes forwards and after reflection) at different points along its path. The phase of the more distant parts of the wave will be advanced wrt the parts nearer the source (which were generated later). The resultant (stationary) wave is produced by the algebraic sum of the forward and reflected wave. In the case of a resonant system, there will be many forward and reflected waves involved and the path length needs to be right so that they all add up appropriately.
There is a difference between the phase situation in a single, progressive wave and that in a standing wave. The resultant of (all) the forward and reflected waves will vary in the same phase (except for a change of sign) at all points although the phases of all the individual progressive waves varies according to
A =A0 cos(wt-kx) (standard notation).

Yes, this is all rather more accurate.

The term phase by itself is pretty vague and almost meaningless.

You can have a phase difference between two different waves.

Within a single wave you can only consider phase difference between two participating particles of the vibrating medium.

The big difference between a travelling wave and a stationary wave is that the for a travelling wave, the phase difference between two such particles depends upon their relative positions on the horizontal axis, whereas for a stationary wave it is independent of this.

All the particles in a stationary wave vibrate in phase.
Only particles that are a whole multiple of wavelengths apart are in phase in a travelling wave.

I do not have time now for diagrams, but will try to post some tonight. I am sorry I can't offer video.

so what we are saying about reflection is that we don't need a medium for reflection to occur? reflecttion is simply a 'measure' of the ease at which a process will occur - meaning:
If a wave reaches the end of a moving string that is loose, it is easier for it to 'reflect itself' and travel back along the wave at a 180 degrees phase change than it is for the energy to totally dissapate?

sophiecentaur
Gold Member
so what we are saying about reflection is that we don't need a medium for reflection to occur? reflecttion is simply a 'measure' of the ease at which a process will occur - meaning:
I can't see what you mean here.
You need a discontinuity in your medium for reflection to occur. The proportion of energy that is reflected will depend upon the amount by which the two media differ.

If a wave reaches the end of a moving string that is loose, it is easier for it to 'reflect itself' and travel back along the wave at a 180 degrees phase change than it is for the energy to totally dissapate?
Remember, at a 'loose' end, there is an antinode so the phase of the reflection will be unchanged - a maximum is formed. At a fixed end, the 180degree change occurs because there is zero net displacement. In either case, very little / zero energy will / can pass out into the surroundings so, because it HAS to go somewhere, it goes back the way it came.
Like I said, the only way to avoid energy being reflected would be to dissipate the energy at the end. You could, at least partially, do this by letting the end drag through a viscous liquid or by having a fan-like structure which could move plenty of air without being too heavy.

Salter 'Ducks' do a good job of taking the energy out of water waves so the forward wave and the reflected wave are very reduced versions of the incident wave.

Pure wave motion, as defined by the wave equation, contains no dissipative term. This means that such an (ideal) wave does not dissipate energy.

It should be noted that the equation posted by SC in post#21 shows this behaviour.

A good real world example of this would be electromagnetic waves travelling in vacuo.

Paradoxically, an ideal (ie completely flexible, non dissipative) string cannot support transverse wave motion unless it is also subject to longitudinal tension.
Real world strings behave in the manner described by SC because they possess a measure of transverse stiffness and because they incorporate dissipative mechanisms in their structure.

jsmith, it would be helpful to know if you have yet studied simple harmonic motion? This is defined as an oscillation of a system subject to a restoring force proportional to the displacement and leads to the type of wave motion described in post#21.