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Energy of a Wave

  1. Mar 15, 2007 #1
    I just read the section in my physics book that deals with the energy transported by a wave and I have a couple of questions: A wave travels through a medium and given the right conditions, the medium will move in SHM.

    1) Is the energy transported by a wave the energy of a single point particle of the medium moving in SHM or is it the energy of the whole system (say, a rope, the area of rippling water or the volume of vibrating air)?

    2) Say I move a piece of rope up and down twice such that two traveling crests are produced on the rope. I would imagine that the energy of the wave is proportional to the two crests somehow. If I were to move the rope to create more moving crests, I would have added more energy to the traveling wave right?
    Last edited: Mar 15, 2007
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  3. Mar 17, 2007 #2
    I was thinking: Say I move a rope up and down to produce a wave one wavelength long. Each portion of the rope where the wave passes oscillates once up and down in SHM right? But what is a portion exactly? I figure its easier to just say that the particles of the rope oscillate up and down once in SHM. But then, how do I determine the energy of the wave? Do I take the sum of the energies of all the particles that are undergoing SHM?
  4. Mar 17, 2007 #3


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    1)I think it's best to consider the energy of a wave in a medium as the energy of the system of particles through which the wave moves.
    (This will not apply for E&M waves, since they have no medium) The energy is transfered from one particle to the next as they oscillate and thus the wave moves forward.

    2)Yes, the energy of a wave is proportional to its frequency, in your words, the number of crests on the rope. Think of light, x-rays are light, but they have so much energy that they could pass right through you, thus X-rays have a higher frequency than visible light.
    Last edited: Mar 17, 2007
  5. Mar 17, 2007 #4
    It seems I was making things more complicated for myself than necessary. Consider an infinitesimal segment of a wave on a string of length dx. The energy dE of this segment is

    [tex]2 \pi^2 f^2 D^2 dm[/tex]

    where [itex]dm = \mu dx[/itex], [itex]\mu[/itex] being the mass per unit length of the string. So, the energy E(x) of a segment of the wave of length x is:

    [tex]E(x) = 2 \pi^2 f^2 D^2 \mu x[/tex]

    If the wave measures l wavelengths, then the total energy of the wave is E(l). Now, if the wave grows, the length grows in proportion to the velocity at which it moves. In other words, l = vt, v being the velocity of the wave so E(l) becomes

    [tex]E(l) = 2 \pi^2 f^2 D^2 \mu v t[/tex]

    Last edited: Mar 17, 2007
  6. Mar 19, 2007 #5
    Now, for a circular wave (such as the wave the forms after dropping a stone in a puddle of water), the energy transported by the wave is given by:

    [tex]E(A) = \int_A 2 \pi^2 f^2 D^2 dm = 2 \pi^3 f^2 D^2 \rho_A h (2r + h)[/tex]

    where [itex]A = \pi ((r + h)^2 - r^2) = \pi h (2r + h)[/itex] is the area of the circular wave and [itex]\rho_A[/itex] is the area density of the medium. Now, since the energy of the wave is constant and the area of the wave is increasing in size, that means either the frequency or the amplitude must decrease over time. In other words, E(A) = E, A = A(t), r = r(t), f = f(t) and D = D(t). Let v be the velocity of the wave such that r(t) = vt. Then,

    [tex]E = 2 \pi^3 f(t)^2 D(t)^2 \rho_A h (2vt + h)[/tex]

    Last edited: Mar 19, 2007
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