The Speed of Sound — Question 6

Question 6

Hard material bounds produce asymmetric speed bounds

A fluid has independent hard bounds $1.8\le B\le2.2\,\mathrm{GPa}$ and $950\le\rho\le1050\,\mathrm{kg/m^3}$. Use $c=\sqrt{B/\rho}$.

1. Find the tight guaranteed speed interval.

2. Compare the interval with the first-order worst-case estimate about $B_0=2.0\,\mathrm{GPa}$ and $\rho_0=1000\,\mathrm{kg/m^3}$.

Question 6 - Solution

Hard material bounds produce asymmetric speed bounds

(a) Method and calculation

Speed increases with $B$ and decreases with $\rho$. Independent allowed extremes give$$c_{\min}=\sqrt{1.8\times10^9/1050}=1309.3\,\mathrm{m/s},$$$$c_{\max}=\sqrt{2.2\times10^9/950}=1521.8\,\mathrm{m/s}.$$Both endpoints are attainable within the rectangular parameter bounds.

(b) Interpretation and further calculation

The central speed is $c_0=1414.2\,\mathrm{m/s}$. Logarithmic differentiation gives$$\frac{dc}{c}=\frac12\frac{dB}{B}-\frac12\frac{d\rho}{\rho},\qquad\frac{|\delta c|}{c_0}\lesssim\tfrac12(0.10+0.05)=0.075.$$The approximate interval is $1308.1$ to $1520.3\,\mathrm{m/s}$. It is close but not an exact guarantee; hard bounds are not combined in quadrature.

Physical check

The asymmetry arises from square-root and reciprocal curvature. Correlations between parameters would change the attainable extrema.