The Speed of Sound — Question 5

Section 17-2 · Physics pages 2660–2661

Question 5 An acoustic measurement determines compressibility rather than density alone A liquid has measured sound speed c = 1200 m/s and density ρ = 800 kg/m3. Its small uniform compression obeys ∆V/V = −∆p/B, with acoustic bulk modulus B = ρc2. ∆p liquid Compression experiment; piston displacement is schematic. (a) Infer B and the fractional compression under ∆p = 2.0 MPa. (b) If the speed calibration is high by 1% while density is exact, find the exact fractional bias in the inferred modulus.
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Question 5 - Solution An acoustic measurement determines compressibility rather than density alone ∆p liquid (a) Method and calculation First infer stiffness, then use the small-compression law: B = 800(1200)2 = 1.152 × 109 Pa, ∆V V = −2.0 × 106 1.152 × 109 = −1.736 × 10−3. The volume decreases by about 0.174% under the stated model. (b) Interpretation and further calculation Because the inverse inference squares the speed, Bmeas Btrue = (1.01)2 = 1.0201. The modulus is high by 2.01%, slightly above the differential estimate dB/B ≃2 dc/c = 2%. The same constitutive response must apply to the compression process for the inferred acoustic modulus to be usable. Physical check The small fractional compression supports linearization; acoustic and slow isothermal moduli need not coincide in a general material.

Question 5

An acoustic measurement determines compressibility rather than density alone

A liquid has measured sound speed c=1200m/sc=1200\,\mathrm{m/s} and density ρ=800kg/m3\rho=800\,\mathrm{kg/m^3}. Its small uniform compression obeys ΔV/V=−Δp/B\Delta V/V=-\Delta p/B, with acoustic bulk modulus B=ρc2B=\rho c^2.

Diagram shown in the original SVG worksheet.

  1. Infer BB and the fractional compression under Δp=2.0MPa\Delta p=2.0\,\mathrm{MPa}.

  2. If the speed calibration is high by 1%1\% while density is exact, find the exact fractional bias in the inferred modulus.

Question 5 - Solution

An acoustic measurement determines compressibility rather than density alone

Diagram shown in the original SVG worksheet.

(a) Method and calculation

First infer stiffness, then use the small-compression law:B=800(1200)2=1.152×109Pa,B=800(1200)^2=1.152\times10^9\,\mathrm{Pa},ΔVV=−2.0×1061.152×109=−1.736×10−3.\frac{\Delta V}{V}=-\frac{2.0\times10^6}{1.152\times10^9}=-1.736\times10^{-3}.The volume decreases by about 0.174%0.174\% under the stated model.

(b) Interpretation and further calculation

Because the inverse inference squares the speed,BmeasBtrue=(1.01)2=1.0201.\frac{B_{\mathrm{meas}}}{B_{\mathrm{true}}}=(1.01)^2=1.0201.The modulus is high by 2.01%2.01\%, slightly above the differential estimate dB/B≃2dc/c=2%dB/B\simeq2\,dc/c=2\%. The same constitutive response must apply to the compression process for the inferred acoustic modulus to be usable.

Physical check

The small fractional compression supports linearization; acoustic and slow isothermal moduli need not coincide in a general material.