# Resonating wine glass

1. Jun 26, 2011

### pkc111

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

2. Jun 26, 2011

### danR

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

3. Jun 27, 2011

### Fewmet

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" [Broken](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.

Last edited by a moderator: May 5, 2017
4. Jun 27, 2011

### danR

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.

Last edited by a moderator: May 5, 2017
5. Jun 27, 2011

### HallsofIvy

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.