Answer
$0.0408ft$
Work Step by Step
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:
$x=|\frac{2.588/1800}{1-(\frac{20}{19.7})^2}|$
$\implies x=0.0408ft$
