Answer
$35.5mm$
Work Step by Step
We can find the required amplitude as follows:
$K=\frac{mg}{\delta}$
$K=\frac{25\times 9.81}{0.05}=4905N/m$
$\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{4905}{25}}=14 rad/s$
and $F_{\circ}=mr\omega_{\circ}^2$
$\implies F_{\circ}=3.5\times 0.1\times (18)^2=113.4N$
Now $x=\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$
We plug in the known values to obtain:
$x=\frac{35/4905}{1-(\frac{18}{14})^2}$
$\implies x=0.0355m=35.5mm$
