Answer
$4.53mm$
Work Step by Step
The amplitude of vibration can be determined as follows:
$\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{2\times 800}{450}}$
$\implies \omega_n=1.8856rad/s$
We know that
$T=\frac{1}{f}$
$\implies T=\frac{2\pi}{\omega_{\circ}}$
$\implies 15Km/h\times \frac{1000m/Km}{3600s/h}=\frac{4\omega_{\circ}}{2\pi}$
This simplifies to:
$\omega_{\circ}=6.5rad/s$
Now $x=|\frac{\delta_{\circ}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}|$
We plug in the known values to obtain:
$x=4.53\times 10^{-6}m=4.53mm$
