Answer
$12.2rad/s$
$7.07rad/s$
Work Step by Step
We can determine the two possible values of $\omega$ as follows:
$\omega_n=\frac{K}{m}$
$\omega_n=\frac{2\times 2500}{50}=10rad/s$
We know that
$y_{max}=\frac{\delta_{\circ}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$
We plug in the known values to obtain:
$\pm 0.4=\frac{0.2}{1-(\frac{\omega_{\circ}}{10})^2}$
$\implies \pm 0.4\times (1-(\frac{\omega_{\circ}}{10})^2)=0.2$
This simplifies to:
$\omega_{\circ}=10\sqrt{1\pm 0.5}$
This simplifies for two values:
$\implies \omega_{\circ}=5\sqrt{6}=12.2rad/s$
and $\omega_{\circ}=5\sqrt{2}=7.07rad/s$
