How Close is a Copper Bar to Breaking Under High Tensile Stress and Sound Waves?

AI Thread Summary
The tensile stress in a thick copper bar is at 99.5% of its elastic breaking point, which is 13.0 X 10^10 N/m^2. To determine the displacement amplitude that would cause the bar to break, the maximum pressure change is calculated as 6.5 X 10^8 N/m^2, using the difference between the breaking point and the current stress. The equation for maximum pressure change can be applied, incorporating density and angular frequency. For maximum particle speed, the relationship v_max = ωA is used, where A is the amplitude. The bulk modulus B can be determined using the formula v = √(B/ρ).
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The tensile stress in a thick copper bar is 99.5% of its elastic breaking point of 13.0 X 10^10 N/m^2. A 500 Hz sound wave is transmitted through the material. (a) What displacement amplitude will cause the bar to break? (b) What is the maximum speed of the particle at this moment?

Comments:

(b) is easy because one we know the maximum displacement (or amplitude), we may use v_{max} = \omega A

For (a), though, it seems like the frequency isn't related to this part of the question. If the copper is stretched so far, beyond it's elastic breaking point, then it will break, but how do I determine this? Thank you.
 
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Well, we know that \Delta P = 13.0 \times 10^{10} - .995 \times 13.0 \times 10^{10} = 6.5 \times 10^8
and we can assume that this is \Delta P_{max} and we may use the equation \Delta P_{max} = \rho v \omega s_{max} where rho and omega are easily determined.

What about v? The book says v = \sqrt{\frac{B}{\rho}} so how would I determine B?
 
B is the bulk's modulus
 
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