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Transverse Wave properties

  1. Mar 27, 2012 #1
    Ok, so the wavelength is the length between two crests. Is it also the length of a complete wave? (sorry if I sound really stupid) The period is the time taken for one particle that makes up the wave to complete its oscillation, and it is also the time taken to make one complete wave. But I don't understand why this is so. I may be complicating matters, but I can't seem to stop thinking about it until I get it right. There is energy being transferred from particle to particle in a wave. When the first particle gets its kinetic energy from the source of vibration, it starts to move up and down, and transfers its energy to anther particle or it to start moving up and down, and this continues for all consecutive particles which make up the eventual wave. Since there is time lost while energy is being transferred from one particle to
    Another, doesn't this
    Mean that the first particle will complete it's oscillation before the other particles. How then, will they all manage to make one complete wave in a particular period?

    Here is the question that got me thinking:(it's not supposed to be tough, but again , I like to think a lot. )
    The speed of a transverse wave on a string is 450m/s. while the wavelength is 0.18m. The amplitude of the wave is 2.0mm. How much time is required for a particle for the string to move through a total distance of 1.0 km? I can deduce from this that a particle on the string covers 8.0mm in one oscillation, but I don't see how how the wavelength applies when we are talking about the particles. If anyone can enlighten me on these questions whih have been bugging me since a few days ago, pls help . Thank you !!
     
  2. jcsd
  3. Mar 27, 2012 #2
    you are over complicating matters. the wave length is the distance between two crests or troughs in the wave, you are correct. but to think of a wave as some kind of 'dominoes' effect, is confusing. as long as the wave length remains constant, so will your time period. and if you want to consider all the particles moving in a wave, then yes, each particle is in a slightly different position within the wave, but this does not mean that they have a slower time period.
     
  4. Mar 27, 2012 #3
    But how can all the particles in a wave gain kinetic energy at the same time, if that is so it won't be a wave anymore..... Ok maybe I don't make sense.
     
  5. Mar 27, 2012 #4
    consider a single photon travelling in a straight line. it has energy, and it is oscillating through a wave form.

    the particles in the string don't get given the energy all at the same instant, which is why parts of the string are in different parts of the wave. a wave travels through a medium, like you said, by passing on it's energy.

    when i say 'time period' i am referring to the time it takes for the particle to move through 2pi radians, not the time at which that particle was given energy.

    if you drop two identical balls onto a surface, at different times, they won't hit the ground at the same time, nor will they reach their highest point at the same time, but the time it takes for them both to do this, will be the same.
     
  6. Mar 27, 2012 #5
    Exactly! Because the time period is the same, the dirt particle will finish its oscillation faster than others, won't it ??
     
  7. Mar 27, 2012 #6
    yes. so the wave moves on through the medium.
     
  8. Mar 27, 2012 #7
    Yeah so how can one complete wave be made by the respective particles in the same time period??
     
  9. Mar 27, 2012 #8
    because if i start counting to ten at t=0 and you start counting to ten at t=1, i will reach ten before you do, but it will still take us both ten seconds to complete the count.

    this would be easier if i could draw some diagrams for you.
     
  10. Mar 27, 2012 #9
    If we are supposed to add up to twenty, at the end of ten seconds, we would only be nineteen!
     
  11. Mar 27, 2012 #10
    you misunderstand. if i am saying the time period of a certain event is ten seconds, it takes ten seconds for that event to complete.

    it makes no difference when another event starts, yours takes ten seconds too.

    your event doesn't stop simply because i finish before you.

    i think it would be useful for you to find some short and clearly defined definitions of wavelength, frequency, and time period.
     
  12. Mar 27, 2012 #11
    The period is the time taken to complete one wave. But at the end of the period, due to all the particles needing the same time to complete one oscillation, which is something like both of us needing 10s to complete the race, there is no way a complete wave would be formed once the period is up, and some particles re still halfway into their oscillation.

    Can I ask u something to see if we are actually visualising the same thing?? To you, when a particle on the wave comPletes one oscillation, does that mean a complete wave cycle is formed?
     
  13. Mar 27, 2012 #12
    yes, the particle starts at 0 radians, it then moves to pi/2 radians and so on, completing a whole 2pi radian rotation.

    2pi radians = 360 degrees.

    there are 2 pi radians in one oscillation on a sin curve
     
  14. Mar 27, 2012 #13

    sophiecentaur

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    If we are dealing with a wave on a string is REALLY can't be helpful to introduce Photons into the situation. Who ever said that photons are "oscillating through a wave form", anyway? Can we please stick to the classical wave stuff until we get it properly sorted?

    @Celluhh
    Remember that a wave (to be a wave) varies in distance and time. Each particle on your string (or whatever) goes through a complete cycle of motion in the same time as every other particle - there is just a slight delay for each successive particle on the wave path.
    A freeze frame picture of the wave as it travels will show you the spatial form / wavelength and watching just one point to see how the displacement varies in time will tell you the frequency.

    If you draw a wavy line on a long piece of paper and move it in front of you, the line goes up and down at the frequency of the wave and the space between identical parts of the wave is the wavelength. The speed that the wave is traveling will be the frequency times the wavelength - (how many waves pass in a second times their length).
     
  15. Mar 27, 2012 #14
    i was trying to create an example where there is one object oscillating without considering what happens to other particles.

    in hindsight - that was a bad choice
     
  16. Mar 27, 2012 #15
    @ intz I knew what you were trying to get at so it's ok. @sophiecentaur if I understand you correctly you are saying we can consider the freeze frame picture of the wave to be as if the complete wave is formed in the given period although there is a time lag between the movement of particles?
     
  17. Mar 27, 2012 #16
    To be perfectly frank, I know how to use the terms to find myself the answer. But what I want to know is not how to find the answer based on formulas, but truly understanding
    How the answer is derived. which is why I think too much most of the time . And why I can't understand the simple school question I posted in my first post.
     
  18. Mar 27, 2012 #17
    It is perhaps useful to first understand how energy is transferred between kinetic and potential energy in the classic pendulum (force = F = mg), then with a mass on a coil spring with a F= -kx restoration force, then finally with the tension on a string creating a restoration force and a traveling wave.
     
  19. Mar 27, 2012 #18
    is it safe to assume that all the waves we are considering are isochronous?

    to me it looks like you're moving into SHM, but is this a decent explanation of how a wave behaves?
     
  20. Mar 27, 2012 #19

    sophiecentaur

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    Yes. Particularly because there is no particular evidence that a 'photon' actually oscillates.
     
  21. Mar 27, 2012 #20

    sophiecentaur

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    At one instant of time, all the particles are in different places - along the profile of the wave (Think of a photo of water waves). Each one is oscillating a bit earlier in phase than its downstream neighbour and a bit later than its upstream neighbour.
    The definitions of wavelength and frequency support this idea.
     
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