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 20 - Traveling Waves - Conceptual Questions: 2

Answer

(a) The speed of the wave is 283 cm/s (b) The speed of the wave is 100 cm/s (c) The speed of the wave is 400 cm/s (d) The speed of the wave is 200 cm/s

Work Step by Step

We can write an expression for the original speed of the wave which is 200 cm/s: $v = \sqrt{\frac{F_T}{\mu}}$ $v = \sqrt{\frac{F_T}{(m/L)}}$ $v = \sqrt{\frac{F_T~L}{m}}$ (a) We can find the new speed of the wave: $v' = \sqrt{\frac{2F_T~L}{m}}$ $v' = \sqrt{2}~\sqrt{\frac{F_T~L}{m}}$ $v' = \sqrt{2}~(200~cm/s)$ $v' = 283~cm/s$ (b) We can find the new speed of the wave: $v' = \sqrt{\frac{F_T~L}{4m}}$ $v' = \sqrt{\frac{1}{4}}~\sqrt{\frac{F_T~L}{m}}$ $v' = \frac{1}{2}~\sqrt{\frac{F_T~L}{m}}$ $v' = \frac{1}{2}~(200~cm/s)$ $v' = 100~cm/s$ (c) We can find the new speed of the wave: $v' = \sqrt{\frac{F_T~(4L)}{m}}$ $v' = \sqrt{4}~\sqrt{\frac{F_T~L}{m}}$ $v' = 2~\sqrt{\frac{F_T~L}{m}}$ $v' = (2)~(200~cm/s)$ $v' = 400~cm/s$ (d) We can find the speed of the wave: $v' = \sqrt{\frac{F_T~(4L)}{4m}}$ $v' = \sqrt{\frac{F_T~L}{m}}$ $v' = 200~cm/s$
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