Fundamentals of Physics Extended (10th Edition)

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

Chapter 15 - Oscillations - Problems - Page 436: 20

Answer

$\phi = 1.03 \,\text{rad}$

Work Step by Step

The velocity of SHM is given by \begin{align*} v_s =- \omega x_{m} \sin (\omega t+\phi) \end{align*} At time $t$ = 0 the displacment is $x_s$ = 0.02 m, so get the displacement using the next equation \begin{gather*} x_s = x_{m} \cos (\omega (0)+\phi) \\ 0.02 \,\text{m} = x_{m} \cos \phi \\ x_m = \dfrac{ 0.02 \,\text{m} }{\cos \phi } \end{gather*} For the velocity, $t$ = 0 and $v_s$ = 0.04 m/s will be \begin{gather*} v_s = - \omega x_{m} \sin (\omega t+\phi) \\ -0.04 \,\text{m/s} = - \omega x_{m} \sin \phi \\ x_m = \dfrac{ 0.04 \,\text{m/s} }{ \omega \sin \phi } \end{gather*} From both these equations we get the next \begin{gather*} \dfrac{ 0.02 \,\text{m} }{\cos \phi } = \dfrac{ 0.04 \,\text{m/s} }{ \omega \sin \phi } \\ \dfrac{ \sin \phi }{\cos \phi } = \dfrac{ 0.04 \,\text{m/s} }{ \omega (0.02 \,\text{m} ) } \\ \tan \phi = \dfrac{ 2 }{ \omega } \\ \phi = \tan^-1 \left( \dfrac{ 1 }{ \omega } \right) \end{gather*} Substitute by $\omega$ to get $\phi$ \begin{align*} \phi = \tan^-1 \left( \dfrac{ 2}{ \omega } \right) = \tan^-1 \left( \dfrac{ 2}{ 1.2 \,\text{rad/s}} \right) = 59.03^o \end{align*} In rad, the phase constant is $$\boxed{\phi = 1.03 \,\text{rad}}$$
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