Answer
$\omega_n=8.165rad/s$
$x=-0.05cos (8.16t)m$
$C=50mm$
Work Step by Step
We can determine the required equation, natural frequency and the amplitude as follows:
$x=\frac{v_{\circ}}{\omega_n}sin(\omega_n t)+y_{\circ}cos(\omega_n t)$.eq(1)
We know that
$\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{200}{3}}=8.165rad/s$
We plug in the known values in eq(1) to obtain:
$x=\frac{0}{8.165}sin(8.165t)-0.05cos(8.165t)$
This simplifies to:
$x=-0.05cos (8.16t)m$
Now the amplitude can be determined as
$C=\sqrt{(\frac{v_{\circ}}{\omega_n})^2+(x_{\circ})^2}$
$\implies C=x_{\circ}=0.05m=50mm$
