The free-vibration solution of a two-degree-of-freedom system can be determined by solving the equations
$$[m] \ddot{\vec{x}}+[k] \vec{x}=\overrightarrow{0}$$
with $\vec{x}=\left\{\begin{array}{l}x_{1}(t) \\ x_{2}(t)\end{array}\right\},$ using the initial conditions
$$\vec{x}(t=0)=\vec{x}_{0}=\left\{\begin{array}{l}
x_{10} \\
x_{20}
\end{array}\right\} \quad \text { and } \quad \dot{\vec{x}}(t=0)=\dot{\vec{x}}=\left\{\begin{array}{l}
\dot{x}_{10} \\
\dot{x}_{20}
\end{array}\right\} .$$
If $\omega_{1}$ and $\omega_{2}$ are the natural frequencies and $\vec{u}_{1}$ and $\vec{u}_{2}$ are the mode shapes of the system obtained from the solution of the characteristic equation
$$\left[[m] s^{2}+[k]\right] \vec{u}=\overrightarrow{0}$$
with $s=\pm \omega_{1}, \pm \omega_{2}$ (characteristic roots), the solution of Eq. (E.1), $\vec{x}(t),$ can be found as a linear combination of different solutions as:
$$\vec{x}(t): \quad C_{1} \vec{u}_{1} e^{-i \omega_{1} t}+C_{2} \vec{u}_{1} e^{+i \omega_{1} t}+C_{3} \vec{u}_{2} e^{-i \omega_{2} t}+C_{4} \vec{u}_{2} e^{+i \omega_{2} t}$$
where $C_{i}, i=1,2,3,4,$ are constants. Show that the solution in Eq. (E.3) can be expressed, in equivalent form, as
$$\vec{x}(t)=A_{1} \sin \left(\omega_{1} t+\phi_{1}\right) \vec{u}_{1}+A_{2} \sin \left(\omega_{2} t+\phi_{2}\right) \vec{u}_{2}$$
where $A_{1}, A_{2}, \phi_{1},$ and $\phi_{2}$ are constants.