Wave frequency and propagation attenuation

[Steven2007]

In my memory, the wave frequency will decrease because the attenuation in wave propagation, is that true?
Can someone please give me some links to prove or disprove it? thanks

 

[Dr. Courtney]

It depends on what the propagation is toward or away from.

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

 

[Steven2007]

It depends on what the propagation is toward or away from.

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

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

 

[Dr. Courtney]

Not to my knowledge.

 

[Steven2007]

Not to my knowledge.

Thanks mate

 

[Steven2007]

I think it is true.

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.

 

[stedwards]

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 $\frac{2 \pi i }{i \omega-k}$.

 

[nasu]

For an ideal, monochromatic wave the attenuation does not change the frequency.
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.

 

[stedwards]

For an ideal, monochromatic wave the attenuation does not change the frequency.

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.

 

[blue_leaf77]

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 $\frac{2 \pi i }{i \omega-k}$.

First of all, the attenuation in common propagation in a medium is characterized by the decreasing amplitude with distance, 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.

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.

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.

 

[stedwards]

First of all, the attenuation in common propagation in a medium is characterized by the decreasing amplitude with distance, 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.

Feel free to make the obvious corrections.