Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 21 - Superposition - Exercises and Problems - Page 624: 61

Answer

$7.15\;\rm cm$

Work Step by Step

We need the sound waves reflected from the mesh to interfere destructively with the sound waves reflected from the wall so we can minimize the annoying hum from the machine. Assuming that the separation distance between the wall and the mesh is $d$. This means that the reflected waves from the wall will move an extra distance than that reflected off the mesh equals $\Delta x=2d$ since it penetrates the mesh and moves to the wall and move back toward the mesh and penetrates again. We know, for perfect destructive interference, that $$\Delta \phi=\dfrac{2\pi \Delta x}{\lambda}+\Delta \phi_0=\left[m+\frac{1}{2}\right]2\pi$$ where $m=0,1,2,3,...$ Recalling that $\lambda=v/f$; $$ \dfrac{2\pi f\;{\overbrace{\Delta x}^{=2d}}}{v}+\overbrace{\Delta \phi_0}^{=0}=\left[m+\frac{1}{2}\right]2\pi$$ $$ \dfrac{4 \color{red}{\bf\not} \pi fd}{v} =\left[m+\frac{1}{2}\right] \color{red}{\bf\not} \pi$$ The minimum distance is given st $m=0$ $$ \dfrac{4 fd}{v} =1$$ Hence, $$d=\dfrac{v}{4f}$$ Plugging the known; $$d=\dfrac{343}{4(1200)}=\bf 0.07145\;\rm m$$ $$d=\color{red}{\bf 7.15}\;\rm cm$$
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