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

$y=0.107sin(7t)+0.1cos(7t)$ $\phi=43.0^{\circ}$
We can determine the required equation and phase angle as follows: We know that $y=\frac{v_{\circ}}{\omega_n}sin\omega_n t+y_{\circ}cos\omega_n t$..eq(1) Now $K=\frac{mg}{dx}$ $\implies K=\frac{15\times 9.81}{0.2}=735.75N/m$ and $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{735.75}{15}}=7rad/s$ We plug in the known values into eq(1) to obtain: $y=\frac{0.75}{7}sin (7t)+0.1 cos(7t)$ $\implies y=0.107sin(7t)+0.1cos(7t)$ and $\phi=tan^{-1}(\frac{y_{\circ}\omega_n}{y_{\circ}})$ $\implies \phi=tan^{-1}(\frac{0.1\times 7}{0.75})$ $\implies \phi=43.0^{\circ}$