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: 43

Answer

The speed will be $\frac{v_0}{2}$

Work Step by Step

Let $m_1$ be the mass of the first string; $m_1 = \rho~V_1$ $m_1 = \rho~(L~\pi~R^2)$ We can find an expression for the mass of the new string as: $m_2 = \rho~V_1$ $m_2 = \rho~L~\pi~(2R)^2$ $m_2 = 4~\rho~(L~\pi~R^2)$ $m_2 = 4~m_1$ We can find an expression for the original speed as: $v_0 = \sqrt{\frac{F_T}{\mu}}$ $v_0 = \sqrt{\frac{F_T}{m_1/L}}$ $v_0 = \sqrt{\frac{F_T~L}{m_1}}$ We can find an expression for the new speed $v_2$ as: $v_2 = \sqrt{\frac{F_T}{\mu}}$ $v_2 = \sqrt{\frac{F_T}{m_2/L}}$ $v_2 = \sqrt{\frac{F_T~L}{m_2}}$ $v_2 = \sqrt{\frac{F_T~L}{4~m_1}}$ $v_2 = \frac{1}{2}\sqrt{\frac{F_T~L}{m_1}}$ $v_2 = \frac{1}{2}~v_0$ The speed will be $\frac{v_0}{2}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.