Energy-Standing and Travelling waves

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  • #1
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Please look at this

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

Energy in Travelling Wave

[The same information is in Resnick Haliday & Krane,so its not wrong.Whatever am typing is from an authentic source]



Now I have a doubt.
Question 1
When they derived the equation for potential energy they used the tension force acting on the particle by the particle to the left of it?
Why dint they use the net force which is F*d2y/dx2 ...ie Td(tan q) where q is the angle made by the force with the horizontal.
If we see a particle then its left particle is pulling it with a force F and its right particle is also pulling it with a force F.These forces are just a little tilted so they dont cancel out.


Question 2

We see that for a travelling wave dU/dt =dK/dT ie change in potential energy is equal to change in Kinetic energy.(Not -ve of it which is reasonable as energy is being transmitted.)
We find that at the peak the KE and PE is 0(or minimum) and at its equilibrium position its maximum.
The reason they give for PE being maximum at the equilibrium position is that the string is stretched maximum at that position.(as dy/dx is maximum then).So PE is maximum


However if we look at it for a standing wave the PE is maximum at the peak and minimum at the equilibrium position.

Now how do we derive that

{this is what I used-dint find it,so this can be wrong}

If we use the standing wave equation y=2Asin(kx)cos(wt)

& use dK=1/2*u*dx*(dy/dt)^2 {u being linear mass density so this is (1/2)*m*v^2)

and dU as 1/2*F*(dy/dx)^2 {which is the formula used in the link and Resnick Halliday)

we find dU as 2*u*w^2*dx*(a)^2*cos^2(kx)*cos^(wt) {cos^2 (kx) is cos(kx) whole square }

and dK as 2*u*w^2*dx*a^2*sin^2(kx)*sin^(wt)

which shows that for a standing wave the PE is 0 at peak as cos^2(kx) is 0


However Sources all over the net say that for a standing wave the PE is maximum at peak.

Also this equation shows that dU/dt is not equalt to -dK/dt

Question 3
So how do we actually prove mathematically that PE is maximum at the Peak and minimum at equilibrium positions.



Also why is this difference in standing and travelling waves coming into play ?


Please try to help guys.
Thanks
 
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Answers and Replies

  • #2
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I think I am beginning to see your difficulty.

In your question 2 you are referring to equations for two different things.

In your reference

equations 18.26 and 18.29 refer, quite clearly, to energy transport by a travelling wave.

whereas the equations you play with later refer to the energy on a particular particle of the wave.

Just as it is important to distinguish between the wave velocity and particle velocity, it is important to distinguish between wave energy and particle energy.

Hope this helps, but you ref is a deal to wade through.
 
  • #3
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Okay Okay,
Sir can you tell me the difference then.
Or provide an appropriate source?

Also if you check this link

http://cnx.org/content/m16027/latest/

only check the pe thing

then they say that Mechanical energy is not constant.
And they also say that this is different from SHM
So is that wrong.?
The same words are used in Resnick Halliday and Krane


And thanks a lot for your help.
 
  • #4
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What I am supposed to make of your latest link?

Wave represents distributed energy. Except for standing waves, the wave transports energy along with it from one point to another. In the context of our course, we shall focus our attention to the transverse harmonic wave along a string and investigate energy being transported by it.

He has told you just what your previous link told you.

Travelling or progressive waves transport energy along the horizontal axis. There is a rate of change of energy with time.

Stationary or standing wave do not transport energy along the horizontal axis.
This simplest explanation is that this is becuase there a fixed points where there is no oscillation. (= nodes). Clearly if nothing happens at nodes no energy passes.

However this new link seems to suggest you can have a travelling wave in a string under tension and then goes on to present the solution, applicable to a standing wave.

I don't wonder you are confused.
 
  • #5
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Okay but it also says that the string is sost stretched at mean position.
So doesnt that also mean that the particles have max potential energy at mean position (apart from the fact that they transmit more energy then) or is it that the particle energy is independent of string stretch?

Also when they say that the mechanical energy is not constant,what do they actually mean?


Like could you be more descriptive in explaining the major difference between transport of energy and energy of the particle itself using both physics and mathematics.Like more in detail.



Also in the transport of energy why are we taking the work done by tension as energy transported.
Shouldnt it be the work done by impulsive force?

Please clear this
And also why arent we the taking the net force on particle for energy transmitted.Why are we just taking the tension force exerted ny left hand particle on right hand particle?
Why are we taking into account force from all sides which is T*d[sin (theta)]?



Also what happens to the energy after reaching the nodes.
As points before it were transmitting the energy.So where exactly the energy being transmitted upto now get lost?
Is it lost?


What I am supposed to make of your latest link?
.

However this new link seems to suggest you can have a travelling wave in a string under tension and then goes on to present the solution, applicable to a standing wave.

I don't wonder you are confused.



I am damn confused.
And which was the place where they made this mistake.I could find it.Is it the last section where they evaluate the work done by tension?


Please help me by answering all the questions I posted.
Thanks Studiot.
 
  • #6
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Since your references are particularly long it would be a great help if you would quote page number and/or equation number in your questions as I have done in my comments.

I will answer some questions tonight, but you do need to grasp the basics before asking looking at the more involved stuuf.

Firstly energy.

The total energy at any point of a travelling simple wave is constant. There is a continual interchange between KE and PE. We say that it is a conservative system.

A simple standing wave is also a conservative system, but the total energy varies from point to point. I suspect this was what was meant by the statement about mechanical energy (reference as above please). The total energy in a standing wave can only be found by summing (integrating) betwen nodes. The energy at a node is, of course, zero. Since the energy is not travelling there is no question of it disappearing at a node.

Since you are keen on the vibrating string (actually quite a difficult example) I have sketched out the first stage of its analysis in the attachment.

The string is assumed fixed at each end and pretensioned by a tension T. Since the string is totally flexible the tension acts along its length and is constant along its length. It is therfore tangential to the string at all points. So no work is done by the tension. All the energy of oscillation is in one direction - the y direction and come from the differential of the component of the tension in the y direction alone.

Is there anything you are unhappy with about this?
Once you have understood this we can proceed to the next bit of analysis of the string.
 

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  • #7
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Okay...if u say the pic http://cnx.org/content/m16027/latest/

the figure 1,they say the string is most stretched at mean position.
So doesnt that also mean that the particles have max potential energy at mean position (apart from the fact that they transmit more energy then) or is it that the particle energy is independent of string stretch?



In the same link Mechanics of energy transmission section


Also in the transport of energy why are we taking the work done by tension as energy transported.
Shouldnt it be the work done by impulsive force?
Please clear this
And also why arent we the taking the net force on particle for energy transmitted.Why are we just taking the tension force exerted ny left hand particle on right hand particle?
Why are we taking into account force from all sides which is T*d[sin (theta)]?


In my first post i derived the two eqtns
If we use the standing wave equation y=2Asin(kx)cos(wt)

& use dK=1/2*u*dx*(dy/dt)^2 {u being linear mass density so this is (1/2)*m*v^2)

and dU as 1/2*F*(dy/dx)^2 {which is the formula used in the link and Resnick Halliday)

we find dU as 2*u*w^2*dx*(a)^2*cos^2(kx)*cos^(wt) {cos^2 (kx) is cos(kx) whole square }

and dK as 2*u*w^2*dx*a^2*sin^2(kx)*sin^(wt)

which shows that for a standing wave the PE is 0 at peak as cos^2(kx) is 0

So here if these equations are for not energy of particles bt energy transmitted then they indicate that energy is travelling till the node.Right?
Also what happens to the energy after reaching the nodes.
As points before it were transmitting the energy.So where exactly the energy being transmitted upto now get lost?
Is it lost?


In the thumbnail u posted have we taken the d[Tsin(phi)]dx as cause the both tensions are aligned a bit?
please explain it
 
  • #8
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I think we should deal with travelling waves on your string first (and separately).

Since it was first inflicted on me many years ago at Uni I have never been that impressed by R&H.

Like many others he says that the displacements are small, then draws these huge sine waves.

What actually happens if you pluck a stretched string somewhere?
You get a standing wave. But we will deal with these later. Then I will show (mathematically) what happens to the energy.



Remember Tension is a vector. Vectors have magnitude and direction.

In a stretched string the energy input to create the wave only changes the direction, it does not change the magnitude of T.

Think about this carefully.
T is aligned parallel to the string and has no component transverse to it.
So variation in magnitude cannot produce or affect event (vibrations) tranverse to it (ie perpendicular to T).
The y direction force required to move the element of string comes from the fact that angle [tex]\psi[/tex] is different at each end of our differential element of string.

This is equivalent to a rotation.

One way to look at this is to think of the string as a chain with links. The links rotate about each junction, and in the limit we shrink the size of the link to zero.

I have no more time now, till later tonight, when I will come back to the maths.
 
  • #9
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With reference to slide 29ff , section 18.6 of your powerpoint link.

They do not say that the wave on the string is transverse or longitudinal (for a good reason). R&H section 17-7 has a similar skimpy presentation.

The point to ask is: How can stretching of the string be counted as a transverse wave, not a longitudinal one?

I suppose one way to look at things is to say that there is not one wave, but two in progress.

This is the approach taken for instance in geophysics where seismologists refer to the P (longitudinal) and S (transverse) waves from an event.

In fact this is an inevitable consequence of applying the wave equation to elastic solids and why I said that vibrating strings are more difficult.

Looking at things this way you can see that the elastic and kinetic energies are not being exchanged and the two waves are really quite separate. It just so happens that in the case of the string the waves are coincident so the max and min of the energy transport occur at the same times.

Talking of energy transport, the system is not conservative.
You can only have travelling waves if you continue to input energy. But that, of course, is what energy transport is all about.

Standing waves, on the other hand, are conservative (if we ignore frictional dissipation).
 
  • #10
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I got what you meant Sir in the 2nd last reply.

Except this.I dint get why Tension is constant.Could u explain again?

n a stretched string the energy input to create the wave only changes the direction, it does not change the magnitude of T.


Is it cause the Tension is equally divided and stretched at every part so Tension per unit length us constant.

Also

With reference to slide 29ff , section 18.6 of your powerpoint link.

They do not say that the wave on the string is transverse or longitudinal (for a good reason). R&H section 17-7 has a similar skimpy presentation.

No i think they mean it is transverse as they are discussing transverse waves.The only place where we can use the y(vertical displacement) is transverse waves.Right?


I suppose one way to look at things is to say that there is not one wave, but two in progress.

In fact this is an inevitable consequence of applying the wave equation to elastic solids and why I said that vibrating strings are more difficult.

Looking at things this way you can see that the elastic and kinetic energies are not being exchanged and the two waves are really quite separate. It just so happens that in the case of the string the waves are coincident so the max and min of the energy transport occur at the same times.

Dint understand this.

Can u tell a time(with the time zone) when we both can be onl9 nd we can discuss this out.This mite save your time Sir.Thanks for providing such personal interest.
 
  • #11
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It seems someone else has been asking about waves on a string recently so have a look at this thread; it discusses some of theis stuff.

https://www.physicsforums.com/showthread.php?t=454029&highlight=wave

As regards the tension.

1) The tension is always directed along to the string which is the same as saying tangential to it.

2) Any stretching due to the tension must therefore take place along the line of the string.

3) But this is, by definition, a longitudinal effect.

4) So any wave that alternately extends and relaxes a segment or element of string is necessarily a longitudinal wave.

5) So how (and why) do we talk about a transverse wave?

6) Further, if the string is originally horizontal the tension has no vertical component.

7) So how does the transverse force necessary to move an element vertically arise?

8) If the element does not move but rotates in place then the tension still points along the string, but it now has both vertical and horizontal components.

9) The greater the rotation, the steeper the slope and the greater the vertical component.

10) A sine wave has a maximum slope of 1 at x=0.

11) So the max vertical component of tension applied to the element occurs at x=0

My time zone is GMT - I am in the South West of the United Kingdom.
 

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