196. A material point of mass $m$ is attracted to a fixed point $O$ with a force proportional to the mass $m$ and the distance $x$ from point $O$; the coefficient of proportionality is $\omega^{2}$. Find the law of motion.
Solution. $1^{0}$. From mechanics, it is known that if a material point of mass $m$ moves in a straight line under the action of a force $F$, then $F=m a$, i.e., $F=m x^{\prime \prime}$, where $t$ is time, and $x^{\prime \prime}$ is the acceleration caused by the force $F$.

Considering that the attractive force $F$ is proportional to $m$ and $x$, and that this force is directed towards the decrease of $x$ when $x$ is positive, we find

$$
F=-\omega^{2} m x
$$

By substituting the value of $F$ in this equation with $F=m x^{\prime \prime}$, we obtain the equation of motion:

$$
m x^{\prime \prime}=-\omega^{2} m x, \text { or } x^{\prime \prime}+\omega^{2} x=0 .
$$

$2^{0}$. Let's solve the obtained linear homogeneous second-order differential equation with constant coefficients. The characteristic equation $k^{2}+\omega^{2}=0$ has imaginary roots: $k_{1}=\omega i, k_{2}=$ $=-\omega i$. The general solution is the function

$$
x=C_{1} \cos \omega t+C_{2} \sin \omega t
$$

This dependence can be simplified by assuming $C_{1}=$ $=R \sin \alpha, C_{2}=R \cos \alpha$, where $R$ and $\alpha$ are constant values. We have

$$
x=R(\sin \omega t \cos \alpha+\cos \omega t \sin \alpha)
$$

In the parentheses, the expanded expression of the sine of the sum of two angles $\omega t$ and $\alpha$ is written. As a result, we obtain the relation

$$
x=R \sin (\omega t+\alpha)
$$

expressing the law of simple harmonic oscillatory motion.