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: 47

Answer

$L_1 = 2.34~m$ $L_2 = 1.66~m$

Work Step by Step

We can write an expression for the speed along string 1 as: $v_1 = \sqrt{\frac{F_T}{\mu_1}}$ We can find the time $t_1$ for the pulse to reach the end of string 1. $t_1 = \frac{L_1}{v_1}$ $t_1 = \frac{L_1}{\sqrt{\frac{F_T}{\mu_1}}}$ $t_1 = L_1~\sqrt{\frac{\mu_1}{F_T}}$ Note that $\mu_2 = 2\mu_1$. We can write an expression for the speed along string 2. $v_2 = \sqrt{\frac{F_T}{\mu_2}}$ $v_2 = \sqrt{\frac{F_T}{2\mu_1}}$ We can find the time $t_2$ for the pulse to reach the end of string 2. $t_2 = \frac{L_2}{v_2}$ $t_2 = \frac{L_2}{\sqrt{\frac{F_T}{2\mu_1}}}$ $t_2 = \sqrt{2}~L_2~\sqrt{\frac{\mu_1}{F_T}}$ If the pulses reach the end at the same time, then $t_1=t_2$. $L_1~\sqrt{\frac{\mu_1}{F_T}} = \sqrt{2}~L_2~\sqrt{\frac{\mu_1}{F_T}}$ $L_1 = \sqrt{2}~L_2$ We also know that the sum of the two lengths is $4.0~m$: $L_1+L_2 = 4.0~m$ $\sqrt{2}~L_2+L_2 = 4.0~m$ $L_2 = \frac{4.0~m}{\sqrt{2}+1}$ $L_2 = 1.66~m$ We can find $L_1$: $L_1+L_2 = 4.0~m$ $L_1 = 4.0~m - L_2$ $L_1 = 4.0~m - 1.66~m$ $L_1 = 2.34~m$
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