Traveling Sound Waves — Question 4

Section 17-3 · Physics pages 2678–2679

Question 4 A proposed pressure profile must satisfy acoustic dispersion A uniform fluid has c = 300 m/s. Consider p(x, t) = P cos(4x −1000t) with SI arguments. Free plane sound must satisfy ptt = c2pxx. x p A sinusoidal appearance alone does not establish dynamical consistency. (a) Test whether this profile is a free sound wave in the given fluid. (b) If wave number remains 4 m−1, find the required frequency. If angular frequency remains 1000 s−1, find the required wavelength.
Show solutionHide solution
Question 4 - Solution A proposed pressure profile must satisfy acoustic dispersion x p (a) Method and calculation Differentiate rather than identifying a drawn waveform as a physical solution: ptt = −106p, c2pxx = −(300)2(4)2p = −1.44 × 106p. The residual is ptt −c2pxx = 4.4 × 105p, so the proposed nonzero field fails the free-wave equation. (b) Interpretation and further calculation At fixed k, ω = ck = 1200 s−1, f = 1200/(2π) = 190.99 Hz. At fixed ω, k = 1000/300 = 3.333 m−1 and λ = 2π/k = 1.885 m. The original phase speed 1000/4 = 250 m/s does not match the material speed. Physical check The derivative test is equivalent to ω2 = c2k2 for a nonzero sinusoid. External forcing would require a different governing equation.

Question 4

A proposed pressure profile must satisfy acoustic dispersion

A uniform fluid has c=300m/sc=300\,\mathrm{m/s}. Consider p(x,t)=Pcos⁡(4x−1000t)p(x,t)=P\cos(4x-1000t) with SI arguments. Free plane sound must satisfy ptt=c2pxxp_{tt}=c^2p_{xx}.

Diagram shown in the original SVG worksheet.

  1. Test whether this profile is a free sound wave in the given fluid.

  2. If wave number remains 4m−14\,\mathrm{m^{-1}}, find the required frequency. If angular frequency remains 1000s−11000\,\mathrm{s^{-1}}, find the required wavelength.

Question 4 - Solution

A proposed pressure profile must satisfy acoustic dispersion

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Differentiate rather than identifying a drawn waveform as a physical solution:ptt=−106p,c2pxx=−(300)2(4)2p=−1.44×106p.p_{tt}=-10^6p,\quad c^2p_{xx}=-(300)^2(4)^2p=-1.44\times10^6p.The residual is ptt−c2pxx=4.4×105pp_{tt}-c^2p_{xx}=4.4\times10^5p, so the proposed nonzero field fails the free-wave equation.

(b) Interpretation and further calculation

At fixed kk,ω=ck=1200s−1,f=1200/(2π)=190.99Hz.\omega=ck=1200\,\mathrm{s^{-1}},\qquad f=1200/(2\pi)=190.99\,\mathrm{Hz}.At fixed ω\omega, k=1000/300=3.333m−1k=1000/300=3.333\,\mathrm{m^{-1}} and λ=2π/k=1.885m\lambda=2\pi/k=1.885\,\mathrm m. The original phase speed 1000/4=250m/s1000/4=250\,\mathrm{m/s} does not match the material speed.

Physical check

The derivative test is equivalent to ω2=c2k2\omega^2=c^2k^2 for a nonzero sinusoid. External forcing would require a different governing equation.