Positions of nodes on a resonating wine glass & symmetry

[Happiness]

When immersed in a high-frequency sound, the rim of a wine glass oscillates as shown in the picture below.

If the sound has equal intensity in all directions from the center of the rim, then where are the nodes of the stationary waves (in the rim)?

By symmetry, every point on the rim is the same, subject to the same environmental conditions. Then there is no reason why one particular point is the node while another point isn't. By symmetry, I conclude that either all points on the rim are nodes or all points are not. Then, it's impossible to have a stationary wave. How can this contradiction be resolved?

 

[kuruman]

By symmetry, I conclude that either ...

Indeed the glass and its contents are symmetric, but I don't think that the exciting sound field is. Sound fields are normally produced by transducers/loudspeakers and normally have a direction of propagation along an axis. Some points on the rim are on or near the axis and other points are to the left or right of it which excites normal modes other than the symmetric "breathing" mode that has no nodes.

 

[olivermsun]

When immersed in a high-frequency sound, the rim of a wine glass oscillates as shown in the picture below.

View attachment 106885

If the sound has equal intensity in all directions from the center of the rim, then where are the nodes of the stationary waves (in the rim)?

By symmetry, every point on the rim is the same, subject to the same environmental conditions. Then there is no reason why one particular point is the node while another point isn't. By symmetry, I conclude that either all points on the rim are nodes or all points are not. Then, it's impossible to have a stationary wave. How can this contradiction be resolved?

You set up the standing wave pattern by disturbing the glass in some way, e.g., projecting a sound wave onto the glass from some direction or rubbing along the rim starting from a certain point and going a certain direction. To would introduce an asymmetry into the problem even if the glass itself were perfectly axisymmetric.

 

[Happiness]

Indeed the glass and its contents are symmetric, but I don't think that the exciting sound field is. Sound fields are normally produced by transducers/loudspeakers and normally have a direction of propagation along an axis. Some points on the rim are on or near the axis and other points are to the left or right of it which excites normal modes other than the symmetric "breathing" mode that has no nodes.

You set up the standing wave pattern by disturbing the glass in some way, e.g., projecting a sound wave onto the glass from some direction or rubbing along the rim starting from a certain point and going a certain direction. To would introduce an asymmetry into the problem even if the glass itself were perfectly axisymmetric.

There is a way to maintain symmetry of the sound wave. Try figuring it out yourself before looking at the answer below.

Spoiler

Place the source (the loudspeaker) at the center of the rim or at any point on the vertical axis passing through it.

Spoiler

The answer above is incomplete, because some sources, such as a tuning fork, are inherently asymmetrical. So even if a tuning fork is placed at the center of the rim, the symmetry is broken due to the vibration of its prongs in some particular direction. Could you think of any symmetrical sound sources before reading the next spoiler?

Spoiler

Hitting a drum right in the center perpendicularly maintains the symmetry.

And so the contradiction is still unresolved.

 

[olivermsun]

There is a way to maintain symmetry of the sound wave. Try figuring it out yourself before looking at the answer below.
[...]

And so the contradiction is still unresolved.

In theory, yes.

In practice, I think the excitation will likely not be perfectly axisymmetric, and hence you will excite particular "phases" of the mode differently, with different "nodes" belonging to each alignment. These could be thought of as analogous to the superposed string that is vibrating in, e.g., both its second and third modes.

If the excitation is near-enough to perfectly symmetric, then perhaps the zero mode (no nodes at all) would be the only one that is significantly excited. (That would also be a stationary case.)

 

[Merlin3189]

I don't know anything about this (at all!) but I had two thoughts when I saw this:
First, complete symmetry is an ideal rather than a real or practical state and any tiny (maybe transient, say due to thermal vibrations) asymmetry could provide the start of an asymmetric mode
Second, even once the glass started vibrating in the fundamental symmetric mode, wouldn't its energy gradually become partitioned equally among all available modes? (This is just a vague notion I had from the molecules in a gas, via random collisions, distributing their energy equally among all available <options> - translation, rotations, vibrations.)

 

[Happiness]

If the excitation is near-enough to perfectly symmetric, then perhaps the zero mode (no nodes at all) would be the only one that is significantly excited. (That would also be a stationary case.)

The zero-node mode is stationary in the sense that energy is localized, but it fails to be stationary in the sense that it is not made up of two progressive waves traveling in opposite directions, I believe.

And by taking the wavelength ##\lambda## as twice the distance between adjacent nodes, we have ##\lambda=\infty##. Then, ##v=f\lambda=\infty##, which is unrealistic.

 

[olivermsun]

Given the definition of wavelength when there are no nodes at all, why do you find the (necessary) result ##v = \infty## to be unrealistic?

 

[Khashishi]

The mallet's position breaks the symmetry. If the mallet somehow hit every point at once, then it would not excite this sound mode in the glass, but perhaps some other mode.

 

[Happiness]

Given the definition of wavelength when there are no nodes at all, why do you find the (necessary) result ##v = \infty## to be unrealistic?

Because we will then have a wave that travels faster than the speed of light.

 

[olivermsun]

That isn't a problem. It's only the phase speed that would be moving faster than the speed of light.