Sound Waves β€” Question 8

Section 17-1 Β· Physics pages 2646–2647

Question 8 A moving particle pattern can remain purely longitudinal A plane sound wave propagates along Λ†n = (3Λ†x + 4Λ†y)/5. Its particle displacement is ΞΎ = aΛ†n cos(kΛ†n Β· r βˆ’Ο‰t) with a = 5.0 Β΅m. x y Λ†n, ΞΎ Propagation and particle displacement share a direction; lengths are schematic. (a) Find the Cartesian displacement amplitudes and the displacement perpendicular to propagation. (b) Explain whether a graph of ΞΎy versus x would make this a transverse wave. Compute the phase gradient along the x axis.
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Question 8 - Solution A moving particle pattern can remain purely longitudinal x y Λ†n, ΞΎ (a) Method and calculation Project the displacement vector onto coordinate axes: ax = 3 5a = 3.0 Β΅m, ay = 4 5a = 4.0 Β΅m. A perpendicular unit vector is Λ†t = (βˆ’4Λ†x + 3Λ†y)/5, giving ΞΎ Β· Λ†t = a(βˆ’12 + 12) cos Ξ¦/25 = 0. (b) Interpretation and further calculation Transverse versus longitudinal is defined relative to propagation, not the plotted coordinate. Here ΞΎ βˆ₯Λ†n. Along the x axis, Ξ¦ = 3k 5 x βˆ’Ο‰t, βˆ‚Ξ¦ βˆ‚x = 3k 5 , Ξ»x = 5Ξ» 3 . The apparent spatial period along an oblique sampling line exceeds the wavelength measured along the propagation direction. Physical check The amplitudes satisfy a2 x + a2 y = a2. Rotating the coordinate axes cannot change wave polarization.

Question 8

A moving particle pattern can remain purely longitudinal

A plane sound wave propagates along 𝒏̂=(3𝒙̂+4π’šΜ‚)/5\hat{\mathbf n}=(3\hat{\mathbf x}+4\hat{\mathbf y})/5. Its particle displacement is 𝝃=a𝒏̂cos⁡(kπ’Μ‚β‹…π’“βˆ’Ο‰t)\boldsymbol\xi=a\hat{\mathbf n}\cos(k\hat{\mathbf n}\cdot\mathbf r-\omega t) with a=5.0ΞΌma=5.0\,\mu\mathrm m.

Diagram shown in the original SVG worksheet.

  1. Find the Cartesian displacement amplitudes and the displacement perpendicular to propagation.

  2. Explain whether a graph of ΞΎy\xi_y versus xx would make this a transverse wave. Compute the phase gradient along the xx axis.

Question 8 - Solution

A moving particle pattern can remain purely longitudinal

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Project the displacement vector onto coordinate axes:ax=35a=3.0ΞΌm,ay=45a=4.0ΞΌm.a_x=\frac35a=3.0\,\mu\mathrm m,\qquad a_y=\frac45a=4.0\,\mu\mathrm m.A perpendicular unit vector is 𝒕̂=(βˆ’4𝒙̂+3π’šΜ‚)/5\hat{\mathbf t}=(-4\hat{\mathbf x}+3\hat{\mathbf y})/5, giving 𝝃⋅𝒕̂=a(βˆ’12+12)cos⁡Ξ¦/25=0\boldsymbol\xi\cdot\hat{\mathbf t}=a(-12+12)\cos\Phi/25=0.

(b) Interpretation and further calculation

Transverse versus longitudinal is defined relative to propagation, not the plotted coordinate. Here 𝝃βˆ₯𝒏̂\boldsymbol\xi\parallel\hat{\mathbf n}. Along the xx axis,Ξ¦=3k5xβˆ’Ο‰t,βˆ‚Ξ¦βˆ‚x=3k5,Ξ»x=5Ξ»3.\Phi=\frac{3k}{5}x-\omega t,\quad \frac{\partial\Phi}{\partial x}=\frac{3k}{5},\quad\lambda_x=\frac{5\lambda}{3}.The apparent spatial period along an oblique sampling line exceeds the wavelength measured along the propagation direction.

Physical check

The amplitudes satisfy ax2+ay2=a2a_x^2+a_y^2=a^2. Rotating the coordinate axes cannot change wave polarization.