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Can someone please give me some links to prove or disprove it? thanks

- Thread starter Steven2007
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- #1

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Can someone please give me some links to prove or disprove it? thanks

- #2

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Frequency will decrease when travelling away from a very large gravitational mass.

- #3

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thanks.

Frequency will decrease when travelling away from a very large gravitational mass.

but lets don't consider facters like gravitational or Doppler effects. Will atenuation make frequency change?

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Not to my knowledge.

- #5

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Thanks mateNot to my knowledge.

- #6

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If there is propagation attenuation in the media, there has to be force between the neighboring vibrating media points, the point near wave source pull/push the neighboring point faraway from wave source, so that the wave sustains.

To produce such force, there has to be phase difference between the two neighboring vibrating points. The accumulation of phase difference of every points will

make frequency lower when wave propagates.

- #7

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For a periodic wave ##y=e^{i \omega t}##, attenuated by ##e^{-kt}##,

##y'=e^{-kt} e^{i \omega t}##.

During the interval of attenuation the period is ##\frac{2 \pi i }{i \omega-k} ##.

##y'=e^{-kt} e^{i \omega t}##.

During the interval of attenuation the period is ##\frac{2 \pi i }{i \omega-k} ##.

Last edited:

- #8

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But in reality we deal with finite pulses of waves (either sound or EM or other) which are composed from many components of different frequencies.

Many times the attenuation depends on frequency. For example, in general, the high frequency ultrasound is attenuated more than low frequency, in many media.

So if we analyse the composition of the attenuated pulse, we will find that the frequency distribution is different than in the original pulse.

In this sense, the frequency is modified by the attenuation. But not a single specific frequency but rather something like average or mean frequency of the pulse.

In case of sound, the pitch as detected by ear, may change due to attenuation.

- #9

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By examination, say, a function ##sin(\omega t)## attenuated by ##-kx##, and ##k>0##, the wave peaks and nodes will occur at a period less than ##\omega##. This should coincide with the spectral peak frequency.For an ideal, monochromatic wave the attenuation does not change the frequency.

- #10

blue_leaf77

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First of all, the attenuation in common propagation in a medium is characterized by the decreasing amplitude withFor a periodic wave ##y=e^{i \omega t}##, attenuated by ##e^{-kt}##,

##y'=e^{-kt} e^{i \omega t}##.

During the interval of attenuation the period is ##\frac{2 \pi i }{i \omega-k} ##.

Ok following your example the attenuated wave would look like ##e^{-kx}\sin{\omega t}##, now imagine again you stand in a fixed ##x## and measure the oscillation over time. It's still a sinusoidal function with the same frequency, and hence same period.By examination, say, a function sin(ωt)sin(\omega t) attenuated by −kx-kx, and k>0k>0, the wave peaks and nodes will occur at a period less than ω\omega. This should coincide with the spectral peak frequency.

- #11

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Feel free to make the obvious corrections.First of all, the attenuation in common propagation in a medium is characterized by the decreasing amplitude withdistance, not time. For example, imagine you are standing inside an attenuating medium certain distance away from the boundary surface with the air where a plane wave is coming from. If somehow you can measure the E field oscillation in time, you will observe that the amplitude of the oscillation at your place is constant with time. In other words, you should observe the wave as being still monochromatic as it was before entering the medium. What you did to the wave when you multiply it with ##e^{-kt}## is that you have changed its spectrum, which is not the case in reality.

Ok following your example the attenuated wave would look like ##e^{-kx}\sin{\omega t}##, now imagine again you stand in a fixed ##x## and measure the oscillation over time. It's still a sinusoidal function with the same frequency, and hence same period.

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