Speed of a wave and its particles

[sophiecentaur]

?
The definition of a period is the length of time it takes for that to happen. The 'exactness' is from the definition.
What else could happen? If you find it difficult to answer that question then that could suggest an answer to your question.

 

[BruceW]

I'm not sure what is meant by "gets formed" ? Is that supposed to mean something about how the system goes from 'no wave' to 'wave' ?

Or does it mean in the case when the wave is always there, and how each particle goes through its oscillation?

 

[sophiecentaur]

I just can't seem to visualise how a complete wave gets formed in exactly
One period ... How is it that all wavelengths get formed in exactly one
Period...

I just re-read your post No26. You seemed to have got things well sorted out in that post. I don't see why you as the above question except that you may not have grasped the following point. When one particle is moving, it is the result of the 'previous particle' moving and its motion lags slightly behind the previous particle. This is because of its mass and the spring constant (or equivalent) associated with the forces that couple them together. Similarly, the 'next particle' will move, but slightly lagging with respect to the middle particles. Every particle will move in the same way but just in a different phase. The lower the masses and the stiffer the coupling, the less lag there will be - so the quicker the forces will be transmitted and the longer the wavelength will be. For massive particles and soggy springs, the disturbance will take longer to travel - so the wavelength will be shorter.
This follows the wave equation c=λf. Once you have established the value of c, f and λ are bound together. (f being 1/the period of oscillation)

 

[xAxis]

I think I know what is the problem. You don't understand why while the first particle, tho source, makes one oscilation, the wave has traveled exactly one wavewlent "down the line". I.e. why not half wavelents, or 1.23 wavelents or any other number.
Is that what you can't visualise?

 

[Celluhh]

I think I know what is the problem. You don't understand why while the first particle, tho source, makes one oscilation, the wave has traveled exactly one wavewlent "down the line". I.e. why not half wavelents, or 1.23 wavelents or any other number.
Is that what you can't visualise?

yep,bingo.

 

[BruceW]

mathematically speaking, it is because we have specified that the wave has only one frequency. And as sophiecentaur said, this means the wavelength is then fixed.

 

[sophiecentaur]

yep,bingo.

If you are having this problem then I think it's because you are putting the cart before the horse.
The fact is that there must be a delay in the propagation of the disturbance down the line. For a uniform system, after a while, the situation will establish a steady state, with waves traveling from the loudspeaker / driver / whatever. The distance that a particular amount of displaacement (say a maximum) happens to get from the source before the source is again at a maximum, is called the length of the wave. This only applies to a repeated wave - usually a sine wave - of course. If you send a single pulse down the line then there is not an identifiable 'wavelength'. If you think in terms of a propagation delay causing a physical spacing of points along the wave then wave'length' follows. Start with speed and frequency and the idea of a λ then comes out of it.

 

[xAxis]

Well, consider the simplest form of wave where exactly one particle is excited to vibrate, end let it be the first one on the left in some elastic strip . Now think that wave is transfer of energy, so as the first particle goes up from equilibrium position, the next one to the right follows, but it follows in exactly the same manner. But now think what happenes to the 3rd particle. It follows exactly the same motion of the 2.nd particle. So does the 4.th particle and so on. (we are assuming only one particle streight line as a medium) So you can see how as particle goes up, the wave travels to the right. But don't forget that that right movement of the wave is nothing but the transfer of energy, which is the transfer of motion of the first particle.This happens almost instantly between two particles, but there is a small lag. So the next particle will always be little behind the previous, but it's motion (oscilation) will be identical.
Now imagine the moment when the first particle's just finished the first oscillation. Because the lag in transfering energy is so small, the second particle has almost finnished it's first oscillation and so on. Looking in the direction of the wave propagation, you will see "backward chronological copy" of the movement of the first particle, because all the particles' motions are exact copies of the motion of the preceding particle.

 

[Celluhh]

@xaxis, and so your point is ?

 

[Celluhh]

The distance that a particular amount of displaacement (say a maximum) happens to get from the source before the source is again at a maximum, is called the length of the wave. This only applies to a repeated wave - usually a sine wave - of course. it.

I don't understand what you mean...

 

[sophiecentaur]

If you take a rope and just give it a jerk then there is just an impulse passing along it. There is no 'wavelength'. If you set up a continuous side to side motion then a continuous progressive wave is set up. Each piece of the rope will oscillate continuously. The distance between two pieces is a wavelength (or a whole number of wavelengths) when their motions are in step. We are discussing continuous waves, aren't we?

 

[xAxis]

@xaxis, and so your point is ?

I tried to help you visualise. So just imagine the picture I propose. there is a streep which is say one molecule thin, like a line of molecules.
Now imagine the moment when the first particle's just finished the first oscillation. Because the lag in transfering energy is so small, the second particle has almost finnished it's first oscillation. So if you can see that (together with "backwards time copy") the wave started traveling to the right almost the same instant that the first particle started first oscilation, it shouldn't be difficult to see that now, as it have finished, the wave has traveled down the line some distance. But remember the main point of wave? The motion of it is because the motion of the source (up down oscillation) is comunicated to the next particle. Now you should be able to see that that distance cannot be more than one wavelent because it would mean that some particles have already finnished one full osculation and started the second, but it cannot be cause their motion is just the copy of the motion of previous particle, and none of them had finnished full oscilation. Does this help?

 

[Celluhh]

@xaxis yup ok so for this to happen all particles are traveling at the same speed right ?

 

[sophiecentaur]

If you look at the Maths, it is described perfectly for sinusoidal wave:
d = A sin(ωt - 2πx/λ)
where d is the displacement at time t and distance x, ω is the angular frequency and λ is the wavelength.
A particle at distance x is vibrating at the rate ω/2π times per second and the phase of its vibration lags by x/λ.
'Why' the wave repeats itself every λ is not an answerable question. The fact is that it does repeat itself (it has to return to a mean position every cycle and the pattern will be regular) and the distances involved must be regular. The wavelength is DEFINED by this spacing and the spacing is determined by the stiffness and density (or some equivalent pair of quantities) of the medium that carries the waves.
How do the particles "know" where to be at any time? They don't; it just takes time for the forces to propagate through the medium and to move them into the appropriate places at the appropriate speeds.

 

[xAxis]

@xaxis yup ok so for this to happen all particles are traveling at the same speed right ?

Particles don't travell in the direction of the wave. Their speed is zero (so yes, they're all traveling at the same speed :) ).
Oscilating particle has only vertical speed (in this ideal case). That speed changes according to v(t) = awsin(wt). This speed is maximal when particle is in equilibrium position. So that the first particle starts oscillation up for instance with its maximum speed. It decelerates and reaches zero at maximum distance 'a' which we call amplitud of the wave. But this motion is vertical. Particles never travel along the wave. It only comunicate it's motion to the next particle and this takes some time. And this comunication is the actual 'speed of wave'. All the particles in the medium are motionless in the wave direction.

 

[Celluhh]

Which is why th speed of the wave remains constant right ? Oh yeah I meant their vertical speed so at the same phases thy travel at the same speed right ? Ok thank you !

 

[davenn]

Particles don't travel in the direction of the wave. Their speed is zero (so yes, they're all traveling at the same speed :) )...
Particles never travel along the wave.

Not true !
In a compressional wave like a sound wave and a seismic P wave the particle motion (oscillation) is back and forwards along the line of the wave motion

Their motion is NOT perpendicular to the wave direction of travel as with some other waves
eg a seismic S wave

Dave

 

[sophiecentaur]

For a very good reason, we use the term "displacement" to describe the temporary motion of the particles (or even the fields) which are involved in a Wave. This can correspond to movement in ANY DIRECTION and the actual direction is irrelevant in nearly every discussion of the wave itself and the direction of the energy flow. Half of the misunderstandings on this thread have been because people haven't used the right terms.

 

[xAxis]

... Oh yeah I meant their vertical speed so at the same phases thy travel at the same speed right ?

Yes, at the same phase they have equal speeds. Note that in this example, during the first period no two particles are at the same phase, and that only the first particle has finnished one full oscillation.

 

[BruceW]

Yes, at the same phase they have equal speeds. Note that in this example, during the first period no two particles are at the same phase, and that only the first particle has finnished one full oscillation.

Agreed :)

 

[Celluhh]

Omg
Ok thank you all I understand
Now !

 

[Celluhh]

I tried to help you visualise. So just imagine the picture I propose. there is a streep which is say one molecule thin, like a line of molecules.
Now imagine the moment when the first particle's just finished the first oscillation. Because the lag in transfering energy is so small, the second particle has almost finnished it's first oscillation. So if you can see that (together with "backwards time copy") the wave started traveling to the right almost the same instant that the first particle started first oscilation, it shouldn't be difficult to see that now, as it have finished, the wave has traveled down the line some distance. But remember the main point of wave? The motion of it is because the motion of the source (up down oscillation) is comunicated to the next particle. Now you should be able to see that that distance cannot be more than one wavelent because it would mean that some particles have already finnished one full osculation and started the second, but it cannot be cause their motion is just the copy of the motion of previous particle, and none of them had finnished full oscilation. Does this help?

Then again, why can't it be half a wavelength then?

 

[sophiecentaur]

Half a wavelength away the particles will be traveling in opposite directions. One going up and the other going down - they are in antiphase.

 

[Celluhh]

Ok wait um a wavelength is decided only after a stable and periodic wave motion had been set up right? It cannot be deduce after only one wave right?

 

[sophiecentaur]

That's right. To set up a wave that you can recognise as such requires some time to switch it on (and off). If you just try a, 'up/down' jerk of a string, the wave will tend to disperse into an odd shape because it will in fact consist of a host of different waves at different frequencies.