The Speed of Sound — Question 1

Section 17-2 · Physics pages 2652–2653

Question 1 Elastic compression and inertia produce the acoustic speed law A uniform fluid has equilibrium density ρ0 and bulk modulus B. In one dimension, small displacement ξ produces pressure perturbation p = −B∂xξ. Consider a thin slab of area A and thickness dx. pA (p + dp)A dx Thin fluid slab; pressure forces act on opposite faces. (a) Derive the displacement wave equation and sound speed. (b) Evaluate the speed for B = 2.25 × 109 Pa and ρ0 = 1000 kg/m3. Explain why a larger density alone lowers speed.
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Question 1 - Solution Elastic compression and inertia produce the acoustic speed law pA (p + dp)A dx (a) Method and calculation The force difference is A[p(x) −p(x + dx)] ≃−Apxdx. Newton's law gives ρ0Adx ξtt = −Apxdx = BAdx ξxx. ξtt = B ρ0 ξxx, c = √ B ρ0 . The area and slab thickness cancel; this is a local material speed. (b) Interpretation and further calculation Substitution yields c = √ 2.25 × 109 1000 = 1500 m/s. At fixed B, doubling density gives c →c/ √ 2: restoring force per deformation is unchanged while inertia increases. Comparing different materials requires both B and ρ0, so density alone cannot rank their speeds. Physical check B/ρ0 has units m2/s2. Compression must provide a restoring, not destabilizing, acceleration.

Question 1

Elastic compression and inertia produce the acoustic speed law

A uniform fluid has equilibrium density ρ0\rho_0 and bulk modulus BB. In one dimension, small displacement ξ\xi produces pressure perturbation p=−B∂xξp=-B\partial_x\xi. Consider a thin slab of area AA and thickness dxdx.

Diagram shown in the original SVG worksheet.

  1. Derive the displacement wave equation and sound speed.

  2. Evaluate the speed for B=2.25×109PaB=2.25\times10^9\,\mathrm{Pa} and ρ0=1000kg/m3\rho_0=1000\,\mathrm{kg/m^3}. Explain why a larger density alone lowers speed.

Question 1 - Solution

Elastic compression and inertia produce the acoustic speed law

Diagram shown in the original SVG worksheet.

(a) Method and calculation

The force difference is A[p(x)−p(x+dx)]≃−ApxdxA[p(x)-p(x+dx)]\simeq-Ap_xdx. Newton's law givesρ0Adxξtt=−Apxdx=BAdxξxx.\rho_0 A dx\,\xi_{tt}=-A p_xdx=B A dx\,\xi_{xx}.ξtt=Bρ0ξxx,c=Bρ0.\xi_{tt}=\frac B{\rho_0}\xi_{xx},\qquad c=\sqrt{\frac B{\rho_0}}.The area and slab thickness cancel; this is a local material speed.

(b) Interpretation and further calculation

Substitution yieldsc=2.25×1091000=1500m/s.c=\sqrt{\frac{2.25\times10^9}{1000}}=1500\,\mathrm{m/s}.At fixed BB, doubling density gives c→c/2c\to c/\sqrt2: restoring force per deformation is unchanged while inertia increases. Comparing different materials requires both BB and ρ0\rho_0, so density alone cannot rank their speeds.

Physical check

B/ρ0B/\rho_0 has units m2/s2\mathrm{m^2/s^2}. Compression must provide a restoring, not destabilizing, acceleration.