Where are the pressure nodes on a standing acoustic wave in water?

In summary, the transducer will supply energy to allow the standing wave to build up over many cycles, until the standing wave has sufficient amplitude that the losses in the resonator absorb all the transducer power.
  • #1
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I have calculated the wave length of a 36 kHz acoustic wave in 20 °C water to be around 41.16mm.
Suppose I have a transducer that produces a 36 kHz acoustic wave and a small water container with a length of 41.6 mm. How will the standing acoustic wave look like, which is produced by the reflection of the incident ultrasound wave upon hitting the hard boundary (plexiglas wall) of the container?

I am not sure but as far as I have understood, the resulting standing wave has to have a pressure anti-node (displacement node) at the rigid boundary, which is the container wall, as the wall will not be able to oscillate. Does this mean that I draw one wave length backwards from the container wall and get to the transducer, which will result in another pressure anti-node on the transducer side?
Is the incident wave being emitted from the transducer completely ignored and only the standing acoustic wave plays a role in the container?
Also, am I understanding this correctly? On a pressure anti- node, the pressure changes with time , oscillating +- around the hydrostatic pressure and on a pressure node, the pressure remains contant (only hydrostatic pressure)?

I have included a simple sketch of this constellation.

1621937672406.jpeg
Thank you for your help.
 
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  • #2
I believe that is a useful and correct description (the definition of dynamic and static pressure is always fraught). You understand I think that the Rayleigh waves at the surface are somewhat exotic creatures. But the pressure must support the water above.
 
  • #3
You will not have a displacement node at the precise position of the transducer. It'll be close, though.
 
  • #4
The transducer will initially supply energy to allow the standing wave to build up over many cycles, until the standing wave has sufficient amplitude that the losses in the resonator absorb all the transducer power.
 
  • #5
Mister T said:
You will not have a displacement node at the precise position of the transducer. It'll be close, though.
This is true for any resonator. In order for Energy to enter the system there needs to be a real (resistive) component in the input and that implies a phase shift at reflection by the input port that's not Zero or 180 degrees - so there can't be a precise Node or Antinode exactly there.
 
  • #6
My understanding is that the displacement and pressure standing waves are for the most part in time quadrature, having 90 degrees phase difference. However, as we move across a node, the phase rapidly changes by 180 degrees in a very short distance. Within this region the two are in phase, and we are seeing the resistive component of the resonator. So a transducer in this position will see a resistive load, and will also apply a resistive load to the resonator.
 
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  • #7
tech99 said:
the displacement and pressure standing waves are for the most part in time quadrature, having 90 degrees phase difference.
The higher the Q, the nearer they are to quadrature. It's all just like the electrical equivalent but the effective location of the transducer is hard to specify because of its impedance.
hutchphd said:
Rayleigh waves at the surface are somewhat exotic
Too 'ard guv, afaiac. I understood that Rayleigh Waves are more a phenomenon in solids but there is bound to be more displacement of water near the surface and then you could get coupling into 'gravity waves' which are many orders of magnitude slower. My only experience of ultrasonic waves in water is in a cleaning bath and I seem to remember standing waves appearing on the surface - but they could be due to some other effect.
 
  • #8
tech99 said:
My understanding is that the displacement and pressure standing waves are for the most part in time quadrature,

That is true. For the waves near the interface (using Airy's solution) the pressure manifests as an additional transverse (upward) displacement of the surface which is in phase with the "pressure wave". Motion of the surface molecule of water is therefore circular (elliptical) as the wave (or waves) pass. For the standing waves I believe the net displacement at the surface to be mostly vertical .
Its very pretty math.
 
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Related to Where are the pressure nodes on a standing acoustic wave in water?

1. Where are the pressure nodes located on a standing acoustic wave in water?

The pressure nodes on a standing acoustic wave in water are located at points where the pressure remains constant. These points are typically located at the peaks and troughs of the wave.

2. How are pressure nodes formed on a standing acoustic wave in water?

Pressure nodes are formed on a standing acoustic wave in water due to the interference of two waves traveling in opposite directions. At these points, the waves cancel each other out, resulting in a constant pressure.

3. Are the pressure nodes evenly spaced on a standing acoustic wave in water?

No, the pressure nodes on a standing acoustic wave in water are not evenly spaced. The spacing between nodes depends on the wavelength of the wave, with longer wavelengths resulting in greater spacing between nodes.

4. Do the pressure nodes on a standing acoustic wave in water always remain in the same location?

Yes, the pressure nodes on a standing acoustic wave in water will always remain in the same location as long as the conditions (such as water temperature and pressure) remain constant. Any changes in these conditions can cause the nodes to shift.

5. Can the pressure nodes on a standing acoustic wave in water be visualized?

Yes, the pressure nodes on a standing acoustic wave in water can be visualized using techniques such as Schlieren imaging or shadowgraphy. These techniques allow us to see the changes in pressure and density caused by the standing wave, including the locations of the pressure nodes.

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