Sound Waves — Question 3

Section 17-1 · Physics pages 2636–2637

Question 3 An acoustic packet outruns the motion of its material particles A 400 Hz tone is emitted for 0.050 s into still air at c = 340 m/s. Inside the packet, particle displacement amplitude is a = 3.0 µm. Neglect turn-on and turn-offtransients and acoustic streaming. x c ℓ A compression pulse travels through the medium. (a) Find the number of cycles, wavelength and packet length. (b) Find maximum particle speed and compare it with packet speed. Does a tagged particle travel the packet length?
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Question 3 - Solution An acoustic packet outruns the motion of its material particles x c ℓ (a) Method and calculation The duration counts temporal cycles, while each cycle occupies one wavelength: N = f∆t = 20, λ = c f = 0.850 m, ℓ= c∆t = Nλ = 17.0 m. (b) Interpretation and further calculation For ξ = a cos ωt, u = −ωa sin ωt, umax = 2π(400)(3 × 10−6) = 7.54 × 10−3 m/s. umax c = 2.22 × 10−5. A tagged particle oscillates over a peak-to-peak distance 2a = 6 µm. Over the ideal twenty complete cycles its net displacement is zero; the packet advances through successive particles. Physical check Information and energy can move over macroscopic distances while material displacement remains microscopic. The no-drift conclusion is within the stated linear model.

Question 3

An acoustic packet outruns the motion of its material particles

A 400Hz400\,\mathrm{Hz} tone is emitted for 0.050s0.050\,\mathrm s into still air at c=340m/sc=340\,\mathrm{m/s}. Inside the packet, particle displacement amplitude is a=3.0μma=3.0\,\mu\mathrm m. Neglect turn-on and turn-off transients and acoustic streaming.

Diagram shown in the original SVG worksheet.

  1. Find the number of cycles, wavelength and packet length.

  2. Find maximum particle speed and compare it with packet speed. Does a tagged particle travel the packet length?

Question 3 - Solution

An acoustic packet outruns the motion of its material particles

Diagram shown in the original SVG worksheet.

(a) Method and calculation

The duration counts temporal cycles, while each cycle occupies one wavelength:N=fΔt=20,λ=cf=0.850m,ℓ=cΔt=Nλ=17.0m.N=f\Delta t=20,\quad\lambda=\frac cf=0.850\,\mathrm m,\quad\ell=c\Delta t=N\lambda=17.0\,\mathrm m.

(b) Interpretation and further calculation

For ξ=acos⁡ωt\xi=a\cos\omega t,u=−ωasin⁡ωt,umax=2π(400)(3×10−6)=7.54×10−3m/s.u=-\omega a\sin\omega t,\quad u_{\max}=2\pi(400)(3\times10^{-6})=7.54\times10^{-3}\,\mathrm{m/s}.umaxc=2.22×10−5.\frac{u_{\max}}c=2.22\times10^{-5}.A tagged particle oscillates over a peak-to-peak distance 2a=6μm2a=6\,\mu\mathrm m. Over the ideal twenty complete cycles its net displacement is zero; the packet advances through successive particles.

Physical check

Information and energy can move over macroscopic distances while material displacement remains microscopic. The no-drift conclusion is within the stated linear model.