## Engineering Mechanics: Statics & Dynamics (14th Edition)

$0.0408ft$
The required amplitude can be determined as follows: $K=\frac{W}{\delta}$ $\implies K=\frac{150}{1in\times \frac{1}{12}ft/in}=1800lb/ft$ The angular velocity is given as $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{Kg}{W}}$ $\implies \omega_n=\sqrt{\frac{1800\times 32.2}{150}}$ $\implies \omega_n=19.7rad/s$ The periodic force is given as $F_{\circ}=mr\omega_{\circ}^2$ $F_{\circ}=\frac{W}{g}r\omega_{\circ}^2$ We plug in the known values to obtain: $F_{\circ}=\frac{0.25}{32.2}\times \frac{10in}{12in/ft}=2.588lb$ Now $x=|\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}|$ We plug in the known values to obtain: $x=|\frac{2.588/1800}{1-(\frac{20}{19.7})^2}|$ $\implies x=0.0408ft$