College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 11 - Problems - Page 430: 47

Answer

From speaker #2, this point is a distance of $2.61~m$, $3.27~m$, or $3.93~m$.

Work Step by Step

We can find the wavelength: $\lambda~f = v$ $\lambda = \frac{v}{f}$ $\lambda = \frac{343~m/s}{523~Hz}$ $\lambda = 0.66~m$ Note that $\frac{\lambda}{2} = \frac{0.66~m}{2} = 0.33~m$ In order for destructive interference to occur, the path difference between the two speakers must have the form $(n~\lambda +\frac{\lambda}{2})$ for some integer $n$. Therefore, the possible distances from speaker #2 must have the form $2.28~m + (n~\lambda+\frac{\lambda}{2})$. We can find the possible distances from speaker #2: $d_0 = 2.28~m+(0)(0.66~m)+0.33~m = 2.61~m$ $d_1 = 2.28~m+(1)(0.66~m)+0.33~m = 3.27~m$ $d_2 = 2.28~m+(2)(0.66~m)+0.33~m = 3.93~m$ From speaker #2, this point is a distance of $2.61~m$, $3.27~m$, or $3.93~m$.
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