Find the law of motion of a point mass m attracted toward a…

Find the law of motion of a point mass $m$ attracted toward a fixed center $O$ by a force whose magnitude is proportional to the distance $r$ from $O$: $F=-kr$, $k>0$.

Step 1: Equations of motion

In Cartesian coordinates with $O$ at the origin, Newton's second law gives independent equations along each axis, all of the same form; it suffices to treat one coordinate, say $x$:

\[ m\ddot x = -kx, \qquad m\ddot y=-ky, \qquad m\ddot z=-kz \ \implies\ m\ddot x+kx=0. \]

Step 2: Standard form and characteristic equation

Introduce $\omega^2=k/m$: \[\ddot x+\omega^2 x=0, \qquad \lambda^2+\omega^2=0 \implies \lambda=\pm i\omega.\]

Step 3: General solution

\[ x(t) = C_1\cos\omega t + C_2\sin\omega t = A\cos(\omega t-\alpha), \]

with $A=\sqrt{C_1^2+C_2^2}$ and $\alpha=\arctan(C_2/C_1)$.

Step 4: Initial conditions

With $x(0)=x_0$, $\dot x(0)=v_0$: \[C_1=x_0, \qquad C_2=\frac{v_0}{\omega}, \qquad A=\sqrt{x_0^2+\frac{v_0^2}{\omega^2}}, \qquad \alpha=\arctan\!\left(\frac{v_0}{\omega x_0}\right).\]

The particular solution satisfying these conditions: $x(t) = x_0\cos\omega t + \frac{v_0}{\omega}\sin\omega t.$

Step 5: Motion in the plane: the trajectory is an ellipse

If the point moves in the $(x,y)$-plane, the same equation holds along each axis with its own amplitude and phase: $x=A_x\cos(\omega t-\alpha_x), \qquad y=A_y\cos(\omega t-\alpha_y).$

Eliminating $t$ (expand $y=A_y\cos(\omega t-\alpha_x-\delta)$ with $\delta=\alpha_y-\alpha_x$, then combine with $x/A_x=\cos(\omega t-\alpha_x)$) gives the trajectory equation:

\[ \frac{x^2}{A_x^2} - \frac{2xy\cos\delta}{A_xA_y} + \frac{y^2}{A_y^2} = \sin^2\delta, \qquad \delta=\alpha_y-\alpha_x, \]

an ellipse centered at $O$. Special cases: $\delta=0$ degenerates to a straight line segment through $O$; $\delta=\pi/2$ with $A_x=A_y=A$ gives a circle of radius $A$; a general $\delta$ gives an ellipse.

Step 6: Period and energy

\[ T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{k}}, \]

independent of amplitude (isochronism). Mechanical energy is conserved: \[E=\frac{1}{2}m\dot x^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2=\mathrm{const}, \qquad \text{or in the plane } E=\frac{1}{2}k\left(A_x^2+A_y^2\right)=\mathrm{const}.\]

Step 7: Verify

With $x(t)=A\cos(\omega t-\alpha)$: $\dot x=-A\omega\sin(\omega t-\alpha)$, $\ddot x=-A\omega^2\cos(\omega t-\alpha)=-\omega^2 x$, so $\ddot x+\omega^2 x=0\ \checkmark$.

Step 8: Law of motion

The solution is recorded in the result below.