Answer
$19.7rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
$K=\frac{W}{\delta }$
$\implies K=\frac{150}{1in\times \frac{1}{12}ft/in}=1800lb/ft$
Now $\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{Kg}{W}}$
We plug in the known values to obtain:
$\omega_n=\sqrt{\frac{1800\times 32.2}{150}}$
$\omega_n=19.7rad/s$
