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There are just two requirements, 1) correct frequency, and 2) sufficient amplitude. The correct frequency is, the resonant frequency of the glass cup (pane, cube, etc.). You will know you have sufficient amplitude, when the glass breaks! Both requirements will vary, depending on the material, shape, dimensions of the object, and other variables. If you ...


1

Any structure that leads to a high Q system (the glass) will work and the trick is precisely matching the resonant (natural frequency ). By mounting the glass in a clamp that dissipates energy at a lesser rate than the sound energy that feeds it, the glass is doomed regardless of thickness or lack of imperfections. If the rate of energy input exceeds the ...


5

First: what frequency should you hit? There are many, many different factors at play in determining the natural frequency of an object I know from experience. These are (not limited to): Thickness, density, elasticity modulus (you'll need two of those, e.g. Young's Modulus and Poisson Ratio), and of course shape. I'm not aware of any papers publishing a ...


0

As you said $y_{cm}=x_{cm}=0$ by symmetry, now $$z_{cm}=\frac{\int \rho z dV}{\int \rho dV}=\frac{\int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{r_1}^{r_2} \rho z r^2 \sin\theta dr d\theta d\phi }{\int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{r_1}^{r_2} \rho r^2 \sin\theta dr d\theta d\phi}. $$ Next recall in the spherical ccordinates $z=r\cos\theta$, integrate to get ...


4

A reasonable sub woofer at sound power level of 130 dB would produce pressure fluctuations of 60 Pa. Compare this to the ambient pressure of 100'000 Pa and you will see that related temperature fluctuations would be negligible. It extinguishes fire because it pushes the air back and forth. For the small fire in this video you could take a small air blower ...



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