Interference two waves

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Rabbilo

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Joined 2022
#1
@Galu
@AllenB
@newvirus2008 u
@Andrew Eckhardt
Good evening guys here I am then, this new thread is true because the project has changed, I will use 2 lower frequencies (5khz) to be able to overcome the impedance. Their interference will produce a 10khz wave, which is not ultrasound, but I'll start from there. However, for now I believe that the important question is the mathematical formulas, such as how does the frequency and amplitude vary based on the materials they pass through? I hope to make it with your help
 
#3
@Galu hi galu 😄😄. no you don't understand, I use 5 khz because they are more penetrating then when they interfere at the point (x,y) they will produce an addition frequency of 10 khz, in this way I overcome the impedance problem. However, where can I find the formulas relating to attenuation based on frequency and power? because if we don't talk about math I can't do anything
 
#4
Rabbilo said:
if we don't talk about math I can't do anything
Click to expand...

As an example of the math, here's one equation used in ultrasound imaging:

Attenuation (in dB) = µ × f × x

Where:

µ is is the attenuation coefficient (in dB/cm at a frequency of 1 MHz)
f is the ultrasound frequency (in MHz)
x is the absorbing material thickness (in cm)

You don't understand that?

Me neither! 🤓
 
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#6
Galu said:
As an example of the math, here's one equation used in ultrasound imaging:

Attenuation (in dB) = µ × f × x

Where:

µ is is the attenuation coefficient (in dB/cm at a frequency of 1 MHz)
f is the ultrasound frequency (in MHz)
x is the absorbing material thickness (in cm)

You don't understand that?

Me neither! 🤓
Click to expand...
exactly my friends, on Wikipedia it gives me a formula derived from a power law, but on which the coefficient of the medium to be crossed does not depend, and therefore it is impossible
 
#8
@Galu
@AllenB
guys so for now let's leave the question of formulas aside because I can't find them either. therefore I would like to know how to focus as best as possible a sound beam with a frequency of 5khz which is equivalent to approximately 6 cm of wavelength. they told me to use a parabolic reflector, I want a 6 cm "wide" beam that can remain 6 cm wide for at least 40 cm of distance in the air, how can I do it? I'm talking about directives
 
#10
AllenB said:
Measure at the surface and at the wanted depth.
Click to expand...

We can't quite seem to get across to Rabbilo the obvious requirement to measure the sound intensity at the surface and at the required 4 cm depth in the soil.

He should consider how the measurement may be undertaken - what measuring equipment would be required for the task?

P.S. For the baffled or simply curious:

Rabbilo's goal, as I understand it, is to investigate the use of ultrasound to break down pesticides that have been absorbed by the roots of a plant.
 
#11
Rabbilo said:
@Galu
@AllenB
guys so for now let's leave the question of formulas aside because I can't find them either. therefore I would like to know how to focus as best as possible a sound beam with a frequency of 5khz which is equivalent to approximately 6 cm of wavelength. they told me to use a parabolic reflector, I want a 6 cm "wide" beam that can remain 6 cm wide for at least 40 cm of distance in the air, how can I do it? I'm talking about directives
Click to expand...
Dirt Song.jpg
A parabolic reflector can concentrate a wider beam source(s) into a planar wave, delivering more SPL to the soil surface in a concentrated area.
Using multiple piezo drivers focused using either a parabolic or hyperbolid projector like Curt Graber's HyperSpike, SPL (sound pressure level) at 5kHz-30kHz could reach the 160dB range at the soil surface. in the set up above.

An estimate of the penetration and propagation level of the VHF into the soil is impossible without knowing the soil composition.

The effect of the airborne sound on the plant compared to the tiny fraction of that energy that would reach the roots at 3 to 4 centimeters below the soil surface won't be known.
Using ultrasonic transducers coupled to the soil on one control group and airborne transducers for another group would be useful to document.
My opinion would be ultrasonic transducers coupled to the soil will be impractical, even if they prove more effective than airborne response.

The effect of a VHF "chirp" on plant stoma has been documented by Dan Carlson/ Sonic Bloom, small arrays of piezo tweeters can be effective at over 70meters distance, the SPL at that distance would be well under 90dB.
https://dancarlsonsonicbloom.com/faq
Sonic Bloom.png


Anyway, you have a "tough row to hoe" ahead 😉

Art
 
#12
Rabbilo said:
I would like to know how to focus as best as possible a sound beam with a frequency of 5khz which is equivalent to approximately 6 cm of wavelength. they told me to use a parabolic reflector
Click to expand...

It appears to be a rule of thumb that the parabolic refector should have a diameter (D) twice that of the wavelength (λ) of the lowest desired frequency, i.e., D/λ is suggested to be 2.

That means a diameter of 12 cm in the case of your 5 kHz sound wave.

The emitter, of course, must be placed at the focal point of the reflector, and pointing in at the reflector.

1702255729005.png



This would create, in effect, a flat emitter the size of the reflector's diameter - 12 cm in your case.

P.S. Using this calculator, the wavelength is 6.86 cm using the speed of sound in air at 20 degrees C:

https://www.omnicalculator.com/physics/sound-wavelength
 
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#13
Galu said:
We can't quite seem to get across to Rabbilo the obvious requirement to measure
Click to expand...
I though we were making progress in #7?

I think Rabillo is getting ahead of himself with the reflector. Since the single source is assumed to be too low in power, why not measure how much. Multiple sources could then produce a plane wave by arraying them.
 
#15
Hmm, seems like whether the 'cart's before the horse' or not, what's needed is a WG to focus it over a narrower polar response and then if still not strong enough it will be easy to figure:

Horn mouth polar frequency: c = 10^6/(a*b)

a = wall angle in deg

b = mouth width (or height) in inches

c = -6 dB frequency

metric = (2.54*10^6)/(a*b) in cm
 
#16
Galu said:
The emitter, of course, must be placed at the focal point of the reflector, and pointing in at the reflector.
This would create, in effect, a flat emitter the size of the reflector's diameter - 12 cm in your case.
Click to expand...
One (major) problem is most VHF "emitters" have very narrow dispersion, so the emitter shadows a good portion of the greatest output, and the outer portion of the reflector is not "iluminated".
In actual practice, the VHF output of the parabolic reflector using conventional transducers is more of a "hazy torus ring" than a planar wave.
At the short distance the OP suggests, the losses in the reflection would probably be greater than the limited planar wave benefit.
 
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