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Energy/momentum relationship

  1. Feb 1, 2016 #1
    ω1. The problem statement, all variables and given/known data
    The following function represents a propagating wave function in a medium
    f(x,t) = Asin(3π/ λ(x-ct) + a)

    where A is the amplitude, λ is the wavelength, c is the speed of light in free space, and a is the initial phase.
    (a) Plot the energy-momentum relation of the wave in the medium
    (b) Derive an expression for the phase velocity of the wave.
    2. Relevant equations


    3. The attempt at a solution
    (a) if I plot energy on one axis and momentum on the other wouldn't graph look like an X= y kind of graph because energy and momentum are proportional?

    (b) I have the formula vp = ω/k = (dx/dt)φ
    Where φ is equal to everything after sin. So, do I just take the derivative of (3π/ λ(x-ct) + a)?
     
  2. jcsd
  3. Feb 1, 2016 #2

    Simon Bridge

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    Just checking some details:
    Do you mean: ##f(x,t) = A\sin\big(\frac{3\pi}{\lambda}(x-ct)+a\big)## ?

    So with ##E\propto p## a graph of ##E## vs ##p## would get a slope of 1, is that what you mean?

    ... what is the definition of phase velocity?
    You have ##v_p = \omega/k## ... in your equation, what is ##k## and what is ##\omega##.
     
  4. Feb 1, 2016 #3
    Yes that is what I mean.

    A) yes that's what I mean. A slope of 1.

    B) isn't what I posted the definition of phase velocity?
     
  5. Feb 1, 2016 #4

    Simon Bridge

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    OK. Then there are some issues to clear up:
    The 3 in front of the pi is a bit challenging ... what happens to f when x goes from ##x=0## to ##x=\lambda##?
    (may be easier to see what I am getting at if a=0)

    A) is it not possible to have other proportionalities? i.e. by Newton: force is proportional to acceleration, but if you plot F vs a the slope is m isn't it, not 1?
    B) You posted a mathematical consequence of the definition - it may not apply here. It is important not to use equations unless understand them.
    The "phase velocity" is a physical phenomena. If you were to measure the phase velocity of a physical wave in front of you, what would you have to do?
    What about the other question: what is k? etc.
     
  6. Feb 1, 2016 #5
    A) I see what you're saying. I just couldn't find any other relationships regarding the question in the text. We haven't even started covering mediums so I think that's only making me more confused.

    B) omega is angular frequency. k is the wave number.
    Im not following you with the "wave in front of me" question. I supposed I try to measure it's length (λ) and use that to solve for k? Not sure how I'd solve for ω without knowing the period or frequency.
     
  7. Feb 1, 2016 #6

    Simon Bridge

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    A) The slope of the graph will be the constant of proportionality. If you have only been given ##E\propto p## without the constant, then maybe the point of part A is to get you to work out what the constant is? For the equation you have been given, it won't be 1.
    Have you not been told how to find the energy and momentum of the wave from the wave function?

    B) Yes - but where is omega and k in the equation you have been given? How are these things related to the variables in the equation?

    When talking about "a wave in front of you", I am saying: what is the physical property of the wave that is called the "phase velocity".
    It is clear that you don't know - so look it up.

    This from wikipedia:
    "The phase velocity of a wave is the rate at which the phase of the wave propagates in space. This is the velocity at which the phase of any one frequency component of the wave travels.​
    ... look at the point of equal phase in the wave, like each of the crests, and time how long they take to travel a certain distance.
    Take a look at the equation - what is the speed that all points on the wave travel at?

    But there is still an issue with the 3 in front of the pi in the equation isn't there?
    It means the equation is wrong, that should be a 2. Either that or ##\lambda## is not the wavelength.
     
  8. Feb 2, 2016 #7
    I spoke to the instructor today and he basically quoted what you just said.

    He did say that the 3 was not a typo though.

    I'll give this problem another crack when I get home in a couple hours. Thanks again for your help!
     
  9. Feb 11, 2016 #8

    Simon Bridge

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    OK, I'll quickly walk you through the core maths.
    You need to understand the definition of a wavelength though, I give the pertinant bits below.

    A distortion of some quantity distributed over space may be written as y = f(x)
    If that distortion travels with speed v in the +x direction, then y(x,t) = f(x-vt), and f(x) would be the state of the distortion at time t=0. (This is just what the word "travelling" means.)
    Here the f is just the function describing the distortion.

    Example.
    If the distortion is a sine wave of amplitude A and wavelength L, then f is often written
    ##f(x) = A\sin kx## where ##k=2\pi /L##
    The reason it is ##2\pi## is so that f will have the same value and slope at x=a and at x=a+L (a is an arbitrary position in x)... from the definition of the wavelength. This means that ##L=2\pi /k##

    If the above sine wave travelled in the +x direction with speed v, then the equation becomes,
    ##y(x,t) = A\sin k(x-vt)## ... which has the form of the wave equation you are given.

    If there are no typos, then it seems you are being told that ##k=3\pi /\lambda##
    Because of the definition of "wavelength", this means that ##L= 2\pi/k = (2/3)\lambda##
    In other words: the wavelength is not ##\lambda##...

    This finding, from the physics, appears to contradict the information in post #1
    Since it is not a typo, then the wave is not a pure sine wave.

    However, there may be a trig fiddle you can do...
    Is there an identity for ##\sin a\theta =## some sum of sines and cosines.
    This would be a fourier decomposition... or you may have, in your notes, a general for for a dispursive wave.
     
    Last edited: Feb 11, 2016
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