Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 22 - Vibrations - Section 22.1 - Undamped Free Vibration - Problems - Page 652: 9


$\omega_n=8.165rad/s$ $x=-0.05cos (8.16t)m$ $C=50mm$

Work Step by Step

We can determine the required equation, natural frequency and the amplitude as follows: $x=\frac{v_{\circ}}{\omega_n}sin(\omega_n t)+y_{\circ}cos(\omega_n t)$.eq(1) We know that $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{200}{3}}=8.165rad/s$ We plug in the known values in eq(1) to obtain: $x=\frac{0}{8.165}sin(8.165t)-0.05cos(8.165t)$ This simplifies to: $x=-0.05cos (8.16t)m$ Now the amplitude can be determined as $C=\sqrt{(\frac{v_{\circ}}{\omega_n})^2+(x_{\circ})^2}$ $\implies C=x_{\circ}=0.05m=50mm$
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