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 - Page 588: 52

Answer

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Work Step by Step

a) The wave is moving in the $x$-direction since it is $D_{(x,t)}$. And it is moving in the negative direction of the $x$-axis since the sign in front of $t$ is positive. b) Knowing that $$D_{(x,t)}=A\sin\left( kx-\omega t+\phi_0\right)=A\sin\left( \frac{2\pi}{\lambda}x-\frac{2\pi }{T} t+\phi_0\right)$$ Hence, $$A\sin\left( \frac{2\pi}{\lambda}x-2\pi ft+\phi_0\right)=3.0\times 10^{-2}\sin\left( \frac{2\pi}{2.4}x+ \frac{2\pi }{0.2}t+2\pi \right)$$ Thus the frequency is given by $$\frac{2\pi}{0.2}=2\pi f$$ $$f= \color{red}{\bf 5}\;\rm Hz$$ The wavelength $$\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{2.4}$$ so $$\lambda =\color {blue}{\bf 2.4}\;\rm m$$ The wave speed is given by $$v=\lambda f=(2.4)(5)=\color {red}{\bf 12}\;\rm m/s$$ The wave number $$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{2.4} =\color {red}{\bf 2.62}\;\rm rad/m$$ c) $$D_{(0.2,0.5)}=3.0 \sin\left( \frac{2\pi}{2.4}(0.2)+ \frac{2\pi }{0.2}(0.5)+2\pi\right)=\color{red}{\bf -1.5}\;\rm cm$$
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