Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 16 - Waves-I - Problems - Page 476: 71b

Answer

$12.27\;N$

Work Step by Step

let, the position of a point on the string after time $t$ makes the angle $\theta$ with the positive $x$ axis. The tension $(T)$ is along the string and its direction at that point also makes the angle $\theta$ with the positive $x$ axis. Therefore, the transverse component of the tension is: $T_{trans}=T\sin\theta$ The displacement of any point on the string follows the following equation $y(x,t)=y_m\sin(kx-\omega t)$ Now, $\tan\theta=\frac{\partial y}{\partial x} $ or, $\tan\theta=ky_m\cos(kx-\omega t)$ Therefore, $\tan\theta|_{max}=ky_m$ or, $\theta_{max}=\tan^{-1}(ky_m)$ According to the given values $2y_m=1\times10^{-2}\;m$ or, $y_m=5\times10^{-3}\;m$ $f=120\;Hz$ $\omega=2\pi f$ or, $\omega=240\pi\;rad/s$ $v=\sqrt {\frac{T}{\mu}}$ or, $v=\sqrt {\frac{90}{120\times10^{-3}}}\; m/s$ or, or, $v=5\sqrt {30}\; m/s$ We know $v=\frac{\omega}{k}$ or, $k=\frac{\omega}{v}$ or, $k=\frac{240\pi}{5\sqrt {30}}\;rad/m$ or, $k=27.53\;rad/m$ Substituting the above values, we obtain $\theta_{max}=\tan^{-1}(27.53\times5\times10^{-3})\;rad$ or, $\theta_{max}=0.13679\;rad$ Therefore, the maximum value of the transverse component of the tension is: $T^{max}_{trans}=T\sin\theta_{max}$ or, $T^{max}_{trans}=90\sin(0.13679)\;N$ or, $T^{max}_{trans}=12.27\;N$ Thus, the maximum value of the transverse component of the tension is $12.27\;N$
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