Can a Wineglass Shatter from Two Notes?

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A wineglass can shatter when exposed to a loud enough pure high C, which matches its resonance frequency. If both a high C and an F note are played simultaneously at the same volume, the combined sound waves may create additive effects that could exceed the glass's failure strength. Some participants express uncertainty about whether the stress distribution within the glass could mitigate the overall impact of the combined frequencies. The discussion touches on the physics of wave superposition and resonance, suggesting that the glass's failure could occur rapidly due to the nature of sound wave interactions. Ultimately, the consensus leans towards the likelihood that the glass would shatter under these conditions.
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pkc111 Says: 10:09 PM Y



If a wineglass can be made that shatters when a loud enough pure high C is played (because the resonace frequency of the winegalss is a high C).If a pure high C (just loud enough to shatter the glass) and an F (same volume) were played at the same time, would the wineglass shatter ?

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pkc111 said:
pkc111 Says: 10:09 PM Y



If a wineglass can be made that shatters when a loud enough pure high C is played (because the resonace frequency of the winegalss is a high C).If a pure high C (just loud enough to shatter the glass) and an F (same volume) were played at the same time, would the wineglass shatter ?

Thanks

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I will call it 'yes'. There are going to be additive peaks and troughs that will drive the glass past the failure strength. IMO.
 
danR said:
I will call it 'yes'. There are going to be additive peaks and troughs that will drive the glass past the failure strength. IMO.

That seems plausible, but I am wrestling with whether any points on the glass would shift to reduce overall stress despite the greater amplitude experiences by some parts of the glass. http://zonalandeducation.com/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html" (if you click on the first and third boxes to display the equivalent of F and then C) makes it look like that won't happen on a d 2D string. I am not confident of my ability to mentally extend this to three dimensions.

It seems like you could craft a neat max-min calculus problem out of this situation.
 
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Fewmet said:
That seems plausible, but I am wrestling with whether any points on the glass would shift to reduce overall stress despite the greater amplitude experiences by some parts of the glass. http://zonalandeducation.com/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html" (if you click on the first and third boxes to display the equivalent of F and then C) makes it look like that won't happen on a d 2D string. I am not confident of my ability to mentally extend this to three dimensions.

It seems like you could craft a neat max-min calculus problem out of this situation.

My gut reaction is that if the original sine wave was driving the system just to the point of failure, then the velocity of a peak formation would be even faster than the glass' ability to shift the stress in the first instance, and when glass fails, it's incredibly fast.
 
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You are talking about solving a d.e. of the form
\frac{d^2x}{dt^2}+ a^2x= sin(at)+ sin(bt)
with the 'sin(ax)' being the "high C" and sin(bx) the "F"

The general solution would be
x(t)= Ccos(at)+ Bsin(at)- \frac{1}{2a}t cos(at)+ \frac{1}{b^2+ a^2}sin(at)

The multiplied "t" in the third terms is the "resonance" that causes the crystal to break. It is the "superposition" property- the solutions from the two separate waves add. Since the solution for the "high C" alone causes x to be unbounded, so will the superposition of the two solutions.
 
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