197. A weight supported by a spiral spring is raised to a distance $b$ and then released. It begins to fall, with its acceleration determined by the equation $g=-p^{2} s$, where $p$ is a constant, and $s$ is the distance from the equilibrium position. Find the equation of motion.
Solution. Let's write down the differential equation

$$
s^{\prime \prime}=-p s, \text { or } s^{\prime \prime}+p^{2} s=0 .
$$

We form the characteristic equation and find its roots: $k^{2}+p=0, k= \pm p i, a=0, b=p$. Therefore,

$$
s=e^{0}(A \sin p t+B \cos p t), \text { i.e. } s=A \sin p t+B \cos p t
$$

To determine the arbitrary constants $A$ and $B$, we use the initial conditions. At $t=0$ we have $s=b$; therefore,

$$
b=A \sin 0+B \cos 0=A \cdot 0+B \cdot 1, \text { i.e. } b=B
$$

We find the velocity at this moment:

$$
v=s^{\prime}=A p \cos p t-B p \sin p t
$$

and use the fact that $v=0$ at $t=0$. We have

$$
0=A p \cos 0+B p \sin 0 ; 0=A p+0 ; A p=0
$$

since $p \neq 0$, then $A=0$.

Thus, the equation of motion is $s=b \cos \rho t$. The function $\cos p t$ is periodic, so $s$ is also a periodic function; the function $s$ will periodically increase and decrease, i.e., the weight will perform oscillatory motion, moving up and down.