Interference — Question 8

Section 17-4 · Physics pages 2706–2707

Question 8 A fixed cancellation delay has a finite useful bandwidth Two paths deliver equal pressure amplitudes P. Their relative transfer phase is 2πfτ + ϕ0, with fixed delay τ = 2.0 ms. Choose ϕ0 so the paths cancel at f0 = 500 Hz. S2 S1 P r2 r1 Two unequal-delay acoustic paths; a fixed phase adjustment cancels one chosen frequency. (a) Find the residual pressure amplitude at f = f0 + ∆f. (b) Find the largest |∆f| in the band around f0 for which the residual amplitude is at most 0.10P.
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Question 8 - Solution A fixed cancellation delay has a finite useful bandwidth S2 S1 P r2 r1 (a) Method and calculation The chosen phase obeys 2πf0τ + ϕ0 = π modulo 2π. At the offset frequency, δ = π + 2π∆fτ, Pres = 2P| cos(δ/2)| = 2P| sin(π∆fτ)|. (b) Interpretation and further calculation In the central band the sine magnitude increases monotonically. Therefore | sin(π∆fτ)| ≤0.05, |∆f| ≤sin−1(0.05) π(0.002) = 7.961 Hz. A longer mismatch delay narrows the useful band. The approximate bound 0.05/(πτ) differs very little here. Physical check At ∆f = 0 the residual vanishes. Additional remote cancellation bands exist, but the requested bound is the contiguous band around f0.

Question 8

A fixed cancellation delay has a finite useful bandwidth

Two paths deliver equal pressure amplitudes PP. Their relative transfer phase is 2πfτ+ϕ02\pi f\tau+\phi_0, with fixed delay τ=2.0ms\tau=2.0\,\mathrm{ms}. Choose ϕ0\phi_0 so the paths cancel at f0=500Hzf_0=500\,\mathrm{Hz}.

Diagram shown in the original SVG worksheet.

  1. Find the residual pressure amplitude at f=f0+Δff=f_0+\Delta f.

  2. Find the largest |Δf||\Delta f| in the band around f0f_0 for which the residual amplitude is at most 0.10P0.10P.

Question 8 - Solution

A fixed cancellation delay has a finite useful bandwidth

Diagram shown in the original SVG worksheet.

(a) Method and calculation

The chosen phase obeys 2πf0τ+ϕ0=π2\pi f_0\tau+\phi_0=\pi modulo 2π2\pi. At the offset frequency,δ=π+2πΔfτ,Pres=2P|cos⁡(δ/2)|=2P|sin⁡(πΔfτ)|.\delta=\pi+2\pi\Delta f\tau,\quad P_{\mathrm{res}}=2P|\cos(\delta/2)|=2P|\sin(\pi\Delta f\tau)|.

(b) Interpretation and further calculation

In the central band the sine magnitude increases monotonically. Therefore|sin⁡(πΔfτ)|≤0.05,|Δf|≤sin⁡−1(0.05)π(0.002)=7.961Hz.|\sin(\pi\Delta f\tau)|\le0.05,\quad |\Delta f|\le\frac{\sin^{-1}(0.05)}{\pi(0.002)}=7.961\,\mathrm{Hz}.A longer mismatch delay narrows the useful band. The approximate bound 0.05/(πτ)0.05/(\pi\tau) differs very little here.

Physical check

At Δf=0\Delta f=0 the residual vanishes. Additional remote cancellation bands exist, but the requested bound is the contiguous band around f0f_0.