# Properties of an ondulatory movement

1. Sep 15, 2011

### jaumzaum

Hi, I'm finished studying ondulatory and I have maany questions I still don't understand. I would be thankful if someone could help me in that, Because I'm pretty lost. I'm going to ask them in letters (A, B, C...), each one is a different doubt.

A man has photographed (with a fixed camera in a fixed place) the movemment of a group of particles that describes an ondulatory movemment. He has taken 6 pictures at all, with an interval of 0.1s from one to another.

[PLAIN]http://img88.imageshack.us/img88/2383/unled1hb.jpg [Broken] [Broken]

[PLAIN]http://img40.imageshack.us/img40/5232/36055600.jpg [Broken]

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[PLAIN]http://img3.imageshack.us/img3/7914/54590617.jpg [Broken]

Suppose the particles were photons

A) We can see that the particles has a pure ondulatory movement (POM) + a horizontal movement. What do I mean by a pure ondulatory movement?

This is a pure ondulatory movement

[PLAIN]http://img88.imageshack.us/img88/2383/unled1hb.jpg [Broken] [Broken]

[PLAIN]http://img580.imageshack.us/img580/9205/94982373.jpg [Broken]

[PLAIN]http://img51.imageshack.us/img51/6260/97579754.jpg [Broken]

[PLAIN]http://img30.imageshack.us/img30/2192/45941021.jpg [Broken]

[PLAIN]http://img855.imageshack.us/img855/7759/21054272.jpg [Broken]

[PLAIN]http://img830.imageshack.us/img830/3765/90509177.jpg [Broken]

Note that in a this movement given a fixed position, like x = 3, we have a constant y.
In a POM the equation is like that y = F(x) = A.cos(z.x)

Where A and z are constants.

In a variable ondulatory movement (VOM) we have y = F(X, t) = A.cos(z.x - w.t)
Where A, z, and w are constants, x and t variables.

First question, the sound wave motion is a VOM or a POM?
* Note that I really don't know the names of this type of movement in english, so I've tried to traduce from portuguese. I think that acronyms don't even exist. But let's assume it's like that.

B) We can see that we have 2 lambda (wave lenght) for the VOM. The first is if we compare that the reference is the first particle in relation to the last one.

The second is if you compare the movement of one particle only (by the whole movement). In the case was the first.

The frequency does not change, so what's the lambda that we calculate from
$v = \lambda.f$ ?

I mean, if sound is VOM, the 340m/s represents lambda 1 or lambda 2?

And here it's another question. The VOM can be decomposed in pure ondulatory movement + horizontal extra moviment. Let's say the velocity of the e pure ondulatory movement in Vp and the velocity of the extra horizontal is Vh. Vpx is the Vp in x axe and Vpy in the y axe.

So if ann exercise gives you the velocity of a wave, is it the Vpx + Vh, VH only or Vpx only?

C) What types of waves are VOM/ POM?

I have more questions, but I will have to understand all theese before I post the others.

[]'s
João

Last edited by a moderator: May 5, 2017
2. Sep 15, 2011

### Philip Wood

Your second set of diagrams seems to show the positions of particles in a medium through which a progressive wave is passing. [Wave = undulation] In your terminology this is a VOM, judging by your equation:

"In a variable ondulatory movement (VOM) we have y = F(X) = A.cos(z.x - w.t)
Where A, z, and w are constants, x and t variables."

I don't understand your comment: "Note that in a this movement given a fixed position, like x = 3, we have a constant y. In a POM the equation is like that y = F(x) = A.cos(z.x)" But the particle at x = 3 does change its y value, in the second set of diagrams.

Is this of any help? I know I haven't addressed much of your question so far.

3. Sep 15, 2011

### jaumzaum

I was not clear : )

Sorry
I mean that in the second diagram, if you take 3 cm after the left border of the picture (x = 3), the particle in the correspondent x will be at an height y at that time, and any time after it, the correspondent particle in that x (which i s not the first) will be at the same y too. I don't mean that the third particle (for example) will be always in the same height (because it will not). I mean that If you take the position of the third particle in the time 0.3s for example, it will correspond to the exact same position of the second particle at the 0.4s and the exact same position of thee first at 0.5s.

[]'s

João

4. Sep 15, 2011

### Philip Wood

I think the fault is partly mine. I didn't spot that in your second set of diagrams, each particle has a steady movement to the right, as well as the up-and-down movement. So the second set of diagrams is like the first set, but with a smaller horizontal movement. [I take it that the left hand particle (for example) in each diagram in the time sequence is always the same particle.] I have to confess that I'm confused by the combination of oscillatory up and down motion, and steady horizontal motion. Can we get rid of this horizontal motion? How important is it to your understanding?

5. Sep 15, 2011

### jaumzaum

Pretty much, all the questions are about it

6. Sep 15, 2011

### Philip Wood

OK. So, if it's the same 6 particles of the medium in each of the six time-sequence diagrams, what you've got is a transverse wave, combined with a steady horizontal velocity shared by all the particles. This is like a wave passing along a piece of string, with the string along the length of a carriage in a moving train. The (stationary) camera is taking a picture of the string through the window of the carriage. [Six marks have been made on the string with red ink.]

This seems a bit complicated, which is why I asked you how important the horizontal movement was.

In B, I agree with your $\lambda$/2 label on the upper diagram (though the distance shown actually looks a bit less than $\lambda$/2.

In the lower diagram in B, the distance you've shown isn't $\lambda$/2. If you divide the distance you've shown (the horizontal distance gone by the first particle) by the time interval between the first and last pictures in the sequence you get the horizontal velocity of the particles, which is additional to the transverse wave; in my example, you get the velocity of the train in which the string is situated!

7. Sep 15, 2011

### jaumzaum

Thanks Phillip

It's what I thought too, but

If it were like that, $v = \lambda.f$ would give the Vpx velocity of a VOM.

And many sound waves (I think all of them) are represented by y = A.cos(zx - w.t), which is a VOM equation. So there has to be a Vh too. And the real sound velocity of the sound listened by a stopped fixed person would ber Vpx + Vh which would be more than 340m/s

So the 340 m/s represents what?

8. Sep 16, 2011

### Philip Wood

Once we've allowed for the additional horizontal velocity of the particles, we're left with a (progressive) wave. This wave is itself moving from left to right (the x-direction) in your diagrams. Do you agree? But the oscillatory motion of the particles is in the y-direction. So what you're showing is a transverse wave (with the particles of the medium having an additional fixed velocity to the right). It can't be a sound waves, because a sound wave is longitudinal: the particles of the medium oscillate parallel to the direction in which the wave is travelling. Do you agree?

Last edited: Sep 16, 2011
9. Sep 18, 2011

### Philip Wood

I'm sorry not to have helped much here. My advice would be to make sure that you understand progressive waves thoroughly before you start to consider additional superimposed particle velocities.