Suppose that a mass weighing $10 \mathrm{lb}$ stretches a spring $2 \mathrm{in}$. If the mass is displaced an additional 2 in. and is then set in motion with an initial upward velocity of $1 \mathrm{ft} / \mathrm{s}$, by determining the position of the mass at any later time, calculate the phase of the motion.
The spring constant is $k=10 \mathrm{lb} / 2 \mathrm{in} .=60 \mathrm{lb} / \mathrm{ft}$, and the mass is $m=w / g=10 / 32 \mathrm{lb} \cdot \mathrm{s}^2 / \mathrm{ft}$. Hence the equation of motion reduces to
$$
u^{\prime \prime}+192 u=0
$$
and the general solution is
$$
u=A \cos (8 \sqrt{3} t)+B \sin (8 \sqrt{3} t)
$$
The solution satisfying the initial conditions $u(0)=1 / 6 \mathrm{ft}$ and $u^{\prime}(0)=-1 \mathrm{ft} / \mathrm{s}$ is
$$
u=\frac{1}{6} \cos (8 \sqrt{3} t)-\frac{1}{8 \sqrt{3}} \sin (8 \sqrt{3} t)
$$
The natural frequency is $\omega_0=\sqrt{192} \cong 13.856 \mathrm{rad} / \mathrm{s}$, so the period is $T=2 \pi / \omega_0 \cong 0.45345 \mathrm{~s}$. The amplitude $R$ and phase $\delta$ are found from Eqs. (17). We have
$$
R^2=\frac{1}{36}+\frac{1}{192}=\frac{19}{576}, \quad \text { so } \quad R \cong 0.18162 \mathrm{ft}
$$
The second of Eqs. (17) yields $\tan \delta=-\sqrt{3} / 4$. There are two solutions of this equation, one in the second quadrant and one in the fourth. In the present problem $\cos \delta>0$ and $\sin \delta<0$, so $\delta$ is in the fourth quadrant, namely,
$$
\delta=-\arctan (\sqrt{3} / 4) \cong-0.40864 \mathrm{rad}
$$
