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Wave amplitude question?

  1. Jun 9, 2005 #1
    I am interest in finding the amplitude and wavelength of an actual wave in nature that apply to the equation:
    wavelength = v/f where v = speed of light.
    This kind of wave would be example of light wave, electron wave, or radio wave, I assume. What I need is the exact measurement of it's amplitude and wavelength observe in nature. Can any please help with finding this kind of information.

    This information will help me find the vertical vibration velocity of a wave in nature. If amplitude = wavelength/4 then vertical vibration velocity is equal speed of light.
    If amplitude > wavelength/4 then vertical vibration velocity is greater then speed of light which should not be possible. But I cannot prove until the I know the actual wave behavior in nature.
  2. jcsd
  3. Jun 9, 2005 #2

    Claude Bile

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    For a medium with a linear restoring force, wavelength is independant of the amplitude.

    In any case, you can't compare amplitude to wavelength when talking about electromagnetic waves because these quantities have completely different units. It would be comparing apples to oranges.

  4. Jun 9, 2005 #3

    Doc Al

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    Don't confuse electromagnetic waves (like light and radio) or quantum mechanical "matter waves" (like electron waves) with mechanical waves (like sound or water waves). Only with mechanical waves is a material medium actually vibrating. And I assure you, the vibrating material comes nowhere near light speed.
  5. Jun 9, 2005 #4
    Thanks to the two post have answer some of my questions.
    But it only lead to more questions.

    True, electromagnetic wave may not have the vibration nature of the machanical wave. But if we choose to view
    it in this way what would that vibration speed be?
    Is the amplitude of an electromagnetic wave ever meausure or calculated before? Are we able to detect both the amplitude and the wavelength of an electromagnetic wave before in the lab or through an equation, what would that number be?
  6. Jun 9, 2005 #5
    Of course! Radio stations are assigned frequencies around which they broadcast their programs. These are approximately in the range from 500KHZ up to around 100MHz. If you divide the speed of these waves (3E8 m/s) by these frequencies you'll get the wave lengths. They range from a few meters (high frequencis) up to hundreds of meters (low frequencies).

    As to amplitude, that can also be measured. But your posts make it seem like you think this amplitude is a length. That isn't true. The amplitude of an electromagnetic wave is the strength of the electric and magnetic fields that are the wave.
  7. Jun 10, 2005 #6
    Correct I am trying to view the amplitude as length. Is it totally wrong or irrelevant to view it this way? Is the definition of wave totally different when it come to electormagnetic wave, then why is the equation to describe it the same as machanical wave.
    Is seem the amplitude of an electromagnetic wave is not an easy answer, cause it view of as intensity.
    Looks like I'll have to try and calculate the amplitude base on your idea of the
    radio wave, but it's probably not precise.
  8. Jun 10, 2005 #7
    If you consider, say, a wave in the sea and approximate it by a sine (or cosine) wave of given amplitude (height) moving at a constant velocity, it's a simple matter mathematically to work out the vertical velocity of a fixed point. Bear in mind that this vertical velocity isn't going to be constant.

    To quantify mathematically, assume the displacement of a wave at an arbiatary fixed point and variable time is given by the expression
    [tex]d(t)=A\sin{\omega t}[/tex]
    where A is the amplitude, and [itex]\omega[/itex] is the angular frequency given by [itex]\omega=2\pi f[/itex]. Remembering [itex]v=f\lambda[/itex] you can express the angular frequency in terms of velocity and wavelength if you want.

    So, to find the 'vertical velocity' you differentiate the expression for the displacement with respect to time, to give
    [tex]v_{vert}=\omega A\cos{\omega t}[/tex]

    This is clearly at a maximum when [itex]\cos{\omega t}=1[/itex], so the maximum vertical velocity is [itex]\omega A[/itex].
    Last edited: Jun 10, 2005
  9. Jun 10, 2005 #8
    Taking the example of a radio wave at 100Mhz, if the average amplitude = 3/4 meter or greater then there
    is a virtual speed (the vertical velocity) that is near or
    greater then the speed of light. I assume that this kind
    of wave have an average amplitude in the millimeter range
    no where near the meter range, so the vibration of the
    antenna(for imagine) is no where near the speed of light.
  10. Jun 10, 2005 #9
    The analogy to 'vertical velocity' in the wave you're describing is the rate of change of the field (either electric or magnetic). It has no physical meaning in terms of height.
  11. Jun 10, 2005 #10
    The analogy to 'vertical velocity' in the wave you're describing is the rate of change of the field (either electric or magnetic). It has no physical meaning in terms of height.

    What about the intensity, can it be view as having different height? What is the closest imagination for this kind of wave? Is it close to droping a rock in the pond generated wave or something different that I don't have the imagination for cause I don't understand electromagnetic wave?
  12. Jun 10, 2005 #11
    just bear in mind that the speed of an electromagnetic wave must be constant. only its amplitude, frequency and wavelenght can be changed. however, the amplitude of a wave is not effected by its wavelenght or frenquency, while the wavelenght is inverse proportional to the frequency.
  13. Jun 10, 2005 #12
    do u mean the intensity of the wave? if yes, it's suppose to be the intensity of the flux.
  14. Jun 10, 2005 #13


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    Yes. You would never ask what is the height of your bank account, i.e. mixing dollars and metres. Similarly, it makes no sense to ask the height of an electromagnetic wave; the amplitude is an electric field. Also, the intensity is the square of the amplitude.
  15. Jun 10, 2005 #14
    That's a very good question, and I'm not sure anyone knows the answer. It's just how nature is!

    However, the equation is a very simple one: the second time derivative is proportional to the second space derivative. So maybe it wasn't so unlikely that it would show up in more than one place. But that's really philosophy, not physics.
  16. Jun 11, 2005 #15
    Hmm, I don't know if I would go that far jdavel. If you start from an analysis of origin centred circular motion in the cartesian plane you have the expression

    [tex]y=r\cos{\omega t}[/tex]

    We know that

    [tex]\omega = \frac{2\pi}{T} = 2\pi f[/tex]

    and that

    [tex]\lambda=2\pi r[/tex] (circumference of circle)


    [tex]2\pi = \frac{\omega}{f}[/tex]

    Which gives

    [tex]\lambda =\frac{\omega r}{f}[/tex]


    [tex]f\lambda =\omega r = v[/tex]

    So that's for a 'physical' wave. We solve Maxwell's equations in free space and we get a nice planar sinusoidal wave as a solution, so the same relation holds.
  17. Jun 11, 2005 #16
    James Jackson,

    But you can't really do that and then say that lambda is also the wave length of a periodic wave. Because up to that point, you haven't even shown that there is a wave.

    To show that a mechanical wave can exist on, say, a stretched string, you have to show that the string (under no force except its tension) can assume a shape of the form f(kx - wt) where k and w (omega) are constants. The function f doesn't have to be sinusoidal (or even periodic for that matter). In fact f can be anything (as long as it's continuous etc.).

    The fact that this condition is approximately valid for lots of mechanical systems (stretched strings, air, the surface of water etc.) is very interesting. The fact that it is exactly valid for classical electric and magnetic fields is, I think, astounding!
    Last edited: Jun 11, 2005
  18. Jun 11, 2005 #17
    Oh totally, I was just generalising to an easily analysed example... It's easy enough to also show that a sinusoidal wave is the simplest solution to the wave equation for a tensioned string, much like it's easy to show a sinusoidal wave (well, two of them) are the simplest solution to Maxwell's field equations in free space.

    At the level you're talking at, this all comes down to the 'model vs reality' discussion.
  19. Jun 13, 2005 #18

    Claude Bile

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    Krab - Irradiance is proportional to the square of the amplitude (I'm being picky, I know), in SI units. Not sure about other units though.

    [tex] I=\frac{1}{2}c\epsilon_0E_0^2 [/tex]

    jdavel - The reason why the equations are similar is because the wave equation describes propagating phenomena! The solutions to the wave-equation are waveforms moving at a constant velocity.

    The fact that one disturbance is electromagnetic and one mechanical is irrelevant, the important similarity is that both scenarios involve a propagating disturbance, and it should therefore come as no surprise that all wave phenomena can be explained by the same form of equation.

    tmwong - The speed of an electromagnetic wave can change when propagating through media (i.e. not in a vacuum). In this circumstance the velocity will reduce and wavelength is reduced by the same factor. Frequency is the only true invariant quantity. In a vacuum however, velocity and wavelength are indeed constant.

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