Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 19 - Think and Solve: 36

Answer

The skipper first finds the frequency from the observed period. $f = \frac{1}{T} = \frac{1}{5 s}= 0.2 Hz$. The distance between wave crests is the wavelength. Finally, the wave speed is frequency multiplied by wavelength. $v = f \lambda = (0.2 Hz)(15 m) = 3 \frac{m}{s}$. Alternatively, 15 meters of wave passes in 5 seconds. The speed of the wave must be $\frac{15 m}{5 s} = 3 \frac{m}{s}$.
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