Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 16 - 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. $v_L = \sqrt{\frac{F_T}{\mu}}$ $v_L = \sqrt{\frac{2250~N}{0.00900~kg/m}}$ $v_L = 500~m/s$ We can find the wavelength in the left section of the wire. $\lambda_L = \frac{v_L}{f}$ $\lambda_L = \frac{500~m/s}{1500~Hz}$ $\lambda_L = 0.333~m$ We can find the number of cycles $N_L$ in the left section of the wire. $N_L = \frac{1.00~m}{\lambda_L}$ $N_L = \frac{1.00~m}{0.333~m}$ $N_L = 3~cycles$ We can find the speed in the right section of the wire. $v_R = \sqrt{\frac{F_T}{\mu}}$ $v_R = \sqrt{\frac{2250~N}{0.0250~kg/m}}$ $v_R = 300~m/s$ We can find the wavelength in the right section of the wire. $\lambda_R = \frac{v_R}{f}$ $\lambda_R = \frac{300~m/s}{1500~Hz}$ $\lambda_R = 0.20~m$ We can find the number of cycles $N_R$ in the right section of the wire. $N_R = \frac{1.00~m}{\lambda_R}$ $N_R = \frac{1.00~m}{0.20~m}$ $N_R = 5~cycles$ 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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