## Engineering Mechanics: Statics & Dynamics (14th Edition)

$y=-0.1265sin(\sqrt{10}t)-0.09cos(\sqrt{10}t)$ $0.1552m$
We can determine the required equation and the maximum upward displacement as follows: $y=\frac{v_{\circ}}{\omega_n}sin(\omega_n t)+y_{\circ} cos(\omega_n)t$eq(1) We know that $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{80}{8}}=\sqrt{10}rad/s$ We plug in the known values in eq(1) to obtain: $y=-\frac{0.4}{\sqrt{10}}sin(\sqrt{10}t)-0.09cos(\sqrt{10}t)$ $\implies y=-0.1265sin(\sqrt{10}t)-0.09cos(\sqrt{10}t)$ Now we find the maximum upward displacement $C=\sqrt{(\frac{v_{\circ}}{\omega_n})^2+(y_{\circ})^2}$ We plug in the known values to obtain: $C=\sqrt{(\frac{0.4}{\sqrt{10}})^2+(0.09)^2}$ This simplifies to: $C=0.1552m$