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 - Exercises and Problems: 59

Answer

There are 8 wavelengths of the wave in this 2.00-meter section of the wire.

Work Step by Step

Since the frequency of a wave does not change when it goes from one medium to another medium, the frequency of this wave is 1500 Hz in both sections of the wire. We can find the speed in the left section of the wire as: $v_L = \sqrt{\frac{F_T}{\mu}}$ $v_L = \sqrt{\frac{2250~N}{0.00900~kg/m}}$ $v_L = 500~m/s$ We can find the wavelength in the left section of the wire as: $\lambda_L = \frac{v_L}{f}$ $\lambda_L = \frac{500~m/s}{1500~Hz}$ $\lambda_L = 0.333~m$ We can find the number of cycles $N_L$ in the left section of the wire as: $N_L = \frac{1.00~m}{\lambda_L}$ $N_L = \frac{1.00~m}{0.333~m}$ $N_L = 3~cycles$ We can find the speed in the right section of the wire as: $v_R = \sqrt{\frac{F_T}{\mu}}$ $v_R = \sqrt{\frac{2250~N}{0.0250~kg/m}}$ $v_R = 300~m/s$ We can find the wavelength in the right section of the wire as: $\lambda_R = \frac{v_R}{f}$ $\lambda_R = \frac{300~m/s}{1500~Hz}$ $\lambda_R = 0.20~m$ We can find the number of cycles $N_R$ in the right section of the wire as: $N_R = \frac{1.00~m}{\lambda_R}$ $N_R = \frac{1.00~m}{0.20~m}$ $N_R = 5~cycles$ The total number of complete cycles is 3 + 5 which is 8 complete cycles. There are 8 wavelengths of the wave in this 2.00-meter section of the wire.
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