# Understanding Energy in standing waves

1. Jan 1, 2009

### shehri

Hi all,

I've been unable to understand how energy remains standing b/w nodes & antinodes.What's the behaviour of energy in nodes & antinodes.Plz. help.Thanks.

2. Jan 1, 2009

### Astronuc

Staff Emeritus
A vibrating spring or string is an example of harmonic motion (much like a pendulum), and assuming it is undamped (or free of dissipative forces), the motion continues. A vibrating string or spring is also an example of elastic behavior (Hooke's law).

See these -
http://hyperphysics.phy-astr.gsu.edu/hbase/permot2.html
http://hyperphysics.phy-astr.gsu.edu/hbase/permot.html
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/standw.html
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/funhar.htm [Broken]
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/rescon.html

When something is vibrating, there is a continuous transformation between mechanical potential energy and kinetic energy. When a spring or string is displaced from its equilibrium (rest) position, and then released, the elastic force (tension in string or tension or compression in spring) pulls (or pushes) the spring or string back toward equilibrium. But since there is momentum (and velocity) in the string or spring, the particles (mass) of the string or spring overshoot the rest position (also point of maximum speed/kinetic energy), and the particles are displaced in the opposite direction.

Nodes are simply those points that do not exhibit motion (translation), while the antinodes experience maximum displacement. Please refer to the discussions of standing waves (standw), fundamental harmonics (funhar), and resonance concepts (rescon).

Also, be aware of longitudinal and lateral motion/displacement, and the different media, e.g. solids, liquids, and gases.

Last edited by a moderator: May 3, 2017
3. Dec 1, 2010

### emailanmol

Sir,
I guess this is true only when the system is isolated and the energy can't be transmitted as in standing waves.However if the wave had been a travelling wave none of the potential and kinetic energy would hv changed into each other.At the peak both would have been minimum(KE would have been zero) and at zero displacement both KE and PE would have had max value.

So why is this difference between standing and travelling waves that at the peak the former has maximum potential energy while the latter has minimum PE

4. Dec 1, 2010

### sophiecentaur

Funny things, standing waves. They can make the brain hurt.
I think you can explain this by pointing out that the motion and displacement in a standing wave is the result of addition of many progressive waves ('sloshing' up and down many times before being dissipated by friction (or whatever). Just consider a single wave, incident on a reflector and what happens to the wave when reflected.

The KE and the PE of a travelling wave are π out of phase (that's supposed to be a pi) but they will each interfere with the reflected wave so as to produce maxes and mins in the same place. The position of maxs and mins depend just on geometry and not on the phase of a wave so both KE and PE will max in the same place. For a travelling wave, their positions are changing all the time, of course.

Despite this, there is still a π phase difference between motion and tension at any point on the standing wave. Maxes in the same place but at different times.

So, when you have a resonant system, like a tuned string, the ends of the string need to be the right distance apart so that all the left travelling waves are co-phased and so are the right travelling waves. i.e. the string has to be the right length for the particular excitation frequency.

btw, here's an excellent link with animation. Phasors are introduced as a concept but the animation sort of explains them as it goes along.
http://resonanceswavesandfields.blogspot.com/2008/02/complex-phasor-representation-of.html" [Broken]

Last edited by a moderator: May 5, 2017
5. Dec 1, 2010

### emailanmol

in a travelling wave KE and PE are in phase.Right?
As DK/Dt=Du/dt(note that its not - of as in a travelling wave a particle is not an isolated system.work is being done on it by neighbouring atoms.)*You can see the derivations here.Resnick haliday also has the same derivation)
http://galileo.phys.virginia.edu/cla...yzingWaves.htm [Broken]

Section Energy and Power in a Traveling Harmonic Wave

I hope am right as both are increasing at the same time and decreasing at the same time

Also if the PE maximum at zero displacement then this means that the string is most stretched at that point.So Why is this happening.?
How do we know its most stretched?(they showed in a site that its most stretched .but i dint find it convincing enough)

Also when we derive the potential energy like in the link given

why are they taking only the force exerted by left particle on right particle.
Why aren't we taking work done by the net force on a particle which is Fd(tanx)

Last edited by a moderator: May 5, 2017
6. Dec 1, 2010

7. Dec 1, 2010

### Studiot

emailanmol
Please look at post#9 in your own thread on this subject before misleading (unintentionally I'm sure) others here.

8. Dec 1, 2010

### sophiecentaur

This appears to be a bit mixed up.
The energy in a travelling wave is shared between KE and PE - going from one to the other and back, both in space and in time. How can they be in phase?
When the resonating string is going past its 'straight' position, it is going fastest (KE max and the string is at its normal tension (PE zero)) - surely??? - When the string is at it biggest displacement, then it is most stretched (PE max) and stationary (KE zero). That means they are π out of phase. This is an entirely separate issue from the spatial position of nodes and antinodes - which coincide for KE and PE. The energies are at maxima but the phases different.

Last edited by a moderator: May 5, 2017
9. Dec 1, 2010

### emailanmol

No if you just have a look at this

https://www.physicsforums.com/showthread.php?t=212620 or

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/powstr.html

even in Resnick Halliday Krane

{I am not trying to confuse you guys.This is genuine.Sorry for the inconvinience causes)

The change of Ke into PE is valid in simple harmonic motion.
But in a travelling wave neighbouring elements are doing work on each other.So the Ke doesnt change into PE.Thats how it transfers energy.Right?
Otherwise noo energy would be transmitted

Also you can look in the two links.
Or maybe you can search for the power transmitted by a travelling wave.From there you can see that the dU/dt = dK/dt

i can post the derivations if that can help.

10. Dec 1, 2010

### emailanmol

11. Dec 1, 2010

### sophiecentaur

"But in a travelling wave neighbouring elements are doing work on each other.So the Ke doesnt change into PE.Thats how it transfers energy.Right?"

I can't see the diagrams or the proper layout on that last link you gave. But, to me, it is agreeing with what I am saying. Also the previous posts are agreeing. They are not saying that the instantaneous KE and the PE are equal. They are saying that the integrals over a cycle are equal.

In a stationary wave in a resonant system the same thing is happening as with a progressive wave; elements are acting on their neighbours. But a stationary wave is the superposition of a whole set of progressive waves in a space of appropriate size where the phases and distances happen to coincide and enhance the standing wave pattern.
You can have a stationary wave with just one incident wave and one reflected wave (perfectly terminated source /sink of waves at the other end of some transmission path). There is superposition of two waves - travelling in opposite directions and obeying the rules of waves. The fastest moving elements are at the 'mean' position and the stationary elements are at the maximum displacement at all points on the stationary wave. Can this be wrong? Is that not what that link of mine clearly demonstrates so nicely?

12. Dec 1, 2010

### emailanmol

http://jpkc.fudan.edu.cn/picture/ar...e965/0b6e686f-6716-4d52-ba65-79ae2a87d8fa.ppt

Try this link
Its a ppt file.Around 6mb.

Even the instantaneous KE and PE are equal.
You can see that they clearly state that the mechanical energy is not constant.
The same is stated in Resnick
have a go through it

I was also very confused at first sight.now i am clear with this that the mechanical energy is not constant.
But now the next problem comes why is it then that the standing wave has its mechanical energy as constant.
If we use the same trick they used mathematically we still cant prove it

Thanks for your help btw

Last edited: Dec 1, 2010
13. Dec 1, 2010

### sophiecentaur

I can't see that powerpoint properly on my old Office or in Openoffice - especially the Maths symbols (a pity). People should use non proprietory formats for publication. PDF is fine for most people and it is designed to be Portable.
What you are saying seems to imply that where the displacement is maximum then so is the velocity.
A longitudinal wave carried by a series of masses joined by springs has some masses stationary - where the adjacent springs are compressed of stretched and some masses moving quickly where the springs are slack. That shows that the energy of the elements fluctuates from PE to Ke and back. But that powerpoint is discussing the actual Flow of energy - which may not be the same thing that I have just described, which is the energy of each element. Energy is flowing through the medium and in a steady state, at least, the elements may 'possess' energy which doesn't actually flow regularly along the chain but is passed to the adjacent elements in an irregular way. I think that's the nub of the discussion.

It's interesting to note that 'the books' clearly show (and derive with yer actual Maths) that E and H fields in an em progressive wave, in vacuo, are in phase. The energy comes in dollops, twice per cycle - not smoothly over the cycle. This contrasts nicely with the fact that Voltage and Current variations in an antenna (and, hence, the local E and H fields) are in quadrature and confused me until I went through the derivation. The antenna definitely carries a standing wave.

I think the same sort of thing is happening here, too. No just my confusion, I mean! - it may be the interpretation of what is being said.

As for your final comment in the last post, any established standing wave that you can actually observe will be losing power at a rate equal to that supplied. Simulations all show an established wave but never make this clear. With no loss, the wave would just grow and grow. The loss, in an antenna, is due to the radiated energy and this shows up as a 'radiation resistance' component in the impedance of the antenna at the feeder.The situation in a lossless resonant system would be a steadily increase in stored energy and you would have a steady build up of energy in each element plus half cycle dollops of energy as each new cycle of the incoming wave arrives.
In a real system you will either get a decay (as with a plucked string) or a steady state situation in which the power is injected with an oscillator, somehow, by a mechanical connection to some point along the string. This source impedance is also a path through which energy from the high amplitude displacements can leak out. It is a mechanism that actually limits the Q. There will be a real and imaginary part to the motion and displacement vectors. The imaginary part corresponds to the true standing wave bit and the real part corresponds to the energy loss.

The variations in the transfer rate of mechanical energy over the cycle should be very easy to measure with strain gauges and accelerometers. I wonder whether it has been done.
I'm glad you wouldn't let this lie. I think I am learning something useful here! Thanks.

14. Dec 1, 2010

### emailanmol

Yes,stretching would be the right term i.e dy/dx.Meaning that where dy/dt is maximum so is dy/dx so the potential and kinetic energy.

(Using stretching is appropriate as displacement can confuse people with vertical displacement)
I mean to say that at the equilibrium position the wire is undergoing the maximum amount of stretching(it may appear strange but thats the truth as dy/dx is maximum there)

You mean to say that they mean the energy transfered across the particle whereas we mean the energy of the particle. I had thought of this but then ruled it out as i felt that
the energy acquired by a particle is given to the next particle at the next instance.Thats how energy is being transmitted.So energy transmitted is nothing but the extra energy the particle had an instance earlier.so energy transmitted is same to it !right?

Also the still the string is stretched maximum at equilibrium position right. as dy/dx is maximum then.(I hope this is the right way of deciding the stretching in the string)
But this is what the links,resnick and all mention so be it.
And here where my question comes in that then why are they sayig its not same for standing waves.

Maybe I am wrong here.Correct me please

I dint fully get this part.Maybe I am still small for this.In 12th std now.
But i guess you meant that if E and H fields are in phase then how voltage and current are in quadrature right.?
I dint get the antennae has a standing wave part.
Hmmm....Okay maybe this can be a similar case

Okay i dint get all the lines of the last paragraph.The dollops and all.
But i understand the part why standing waves cant transmit energy.
But the doubt also is that for a standing wave as well the wire is more stretched at equilubrium position(dy/dx)
So it should have more PE then.

So now my question still remains that why a standing wave has maximum potential energy at Peak position
Why is it different for travelling waves?

here they give the derivation.hope you (or anyone else) can crack the logic behing all this.
http://cnx.org/content/m16027/latest/

Also Here they clearly mention the difference between SHM and travelling waves elastic potentiall energy.And they also say that there is no difference in energy transferred and energy of the particle

Cheers :D
Thanks

Last edited: Dec 1, 2010
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