College Physics (4th Edition)

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

Chapter 11 - Problems - Page 430: 46


From speaker #2, this point is a distance of $2.28~m$, $2.94~m$, or $3.60~m$.

Work Step by Step

We can find the wavelength: $\lambda~f = v$ $\lambda = \frac{v}{f}$ $\lambda = \frac{343~m/s}{523~Hz}$ $\lambda = 0.66~m$ In order for constructive interference to occur, the path difference between the two speakers must have an integer number of wavelengths. Therefore, the possible distances from the speaker #2 must have the form $2.28~m \pm n~\lambda$. We can find the possible distances from speaker #2: $d_0 = 2.28~m+(0)(0.66~m) = 2.28~m$ $d_1 = 2.28~m+(1)(0.66~m) = 2.94~m$ $d_2 = 2.28~m+(2)(0.66~m) = 3.60~m$ From speaker #2, this point is a distance of $2.28~m$, $2.94~m$, or $3.60~m$.
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