University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 15 - Mechanical Waves - Problems - Exercises - Page 502: 15.67


(a) $m = 2.22~grams$ (b) $a_{max} = 22400~m/s^2$

Work Step by Step

(a) The expression for the maximum transverse speed at the anti-node is $\omega A$. We can find the angular frequency. $\omega A = v_{max}$ $\omega = \frac{v_{max}}{A}$ $\omega = \frac{28.0~m/s}{0.0350~m}$ $\omega = 800~rad/s$ We can find the frequency of the first overtone, which is the second harmonic. $f_2 = \frac{\omega}{2\pi}$ $f_2 = \frac{800~rad/s}{2\pi}$ $f_2 = 127.32~Hz$ We can find the speed of the wave on the string. $f_2 = \frac{2v}{2L}$ $v = f_2~L$ $v = (127.32~Hz)(2.50~m)$ $v = 318.3~m/s$ We can find the mass of the string. $v = \sqrt{\frac{F}{\mu}}$ $\mu = \frac{F}{v^2}$ $\frac{m}{L} = \frac{F}{v^2}$ $m = \frac{F~L}{v^2}$ $m = \frac{(90.0~N)(2.50~m)}{(318.3~m/s)^2}$ $m = 0.00222~kg = 2.22~grams$ (b) The expression for the maximum transverse acceleration at the anti-node is $\omega^2A$. We can find the maximum acceleration. $a_{max} = \omega^2 A$ $a_{max} = (800~rad/s)^2(0.0350~m)$ $a_{max} = 22400~m/s^2$
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