A particle moves under a force directed away from the Oz…
A particle moves under a force directed away from the $Oz$ axis, proportional to its distance $\rho=\sqrt{x^2+y^2}$ from that axis, giving Newton's equations: $$ \ddot x = kx,\qquad \ddot y = ky,\qquad \ddot z = 0 $$
Explanation of Method
The $z$-motion is free (uniform), while the $x$- and $y$-equations are identical exponential-growth ODEs; eliminating the time parameter between the exponential solutions for $x$ and $y$ produces a conic-section relation, showing the projected path is a hyperbola.
- Solve $\ddot z=0$ for uniform motion along the axis.
- Solve $\ddot x=kx$ and $\ddot y=ky$ using $\omega=\sqrt k$.
- Eliminate the exponential parameter $u=e^{\omega t}$ between $x(t)$ and $y(t)$.
- Interpret the resulting relation as a conic section (a hyperbola) in the $xy$-plane.
Step 1 · Motion along $Oz$
$$ z(t)=\alpha t+\beta. $$
Step 2 · Motion in $x$ and $y$
With $\omega=\sqrt k$,
$$ x(t)=A_1e^{\omega t}+A_2e^{-\omega t},\qquad y(t)=B_1e^{\omega t}+B_2e^{-\omega t}. $$
Step 3 · Eliminate the time parameter
Setting $u=e^{\omega t}$ and solving the two linear equations for $u$ and $1/u$, then using $u\cdot(1/u)=1$, gives a quadratic relation between $x$ and $y$:
$$ B_1B_2\,x^2-(A_1B_2+A_2B_1)\,xy+A_1A_2\,y^2+(A_1B_2-A_2B_1)^2=0. $$
Step 4 · Classify the conic
The discriminant of this quadratic form equals $(A_1B_2-A_2B_1)^2\geq0$, so the projected path onto the $xy$-plane is a hyperbola (or a pair of lines in the degenerate case).
Step 5 · Verification
Differentiating $x(t)$ and $y(t)$ twice reproduces $\ddot x=kx$ and $\ddot y=ky$ exactly, and $\ddot z=0$ holds trivially. ✓
Summary Table
| Quantity | Value |
|---|---|
| Axial motion | uniform: $z=\alpha t+\beta$ |
| Radial motion | exponential: $x,y\sim A e^{\omega t}+Be^{-\omega t}$ |
| Projected trajectory | hyperbola in the $xy$-plane (discriminant $\geq0$) |
Additional Notes
- The full three-dimensional path is a "hyperbolic helix": a hyperbola in cross-section combined with uniform drift along $Oz$.
- Special cases include a straight-line trajectory (if $A_2=B_2=0$) or motion confined to a fixed plane $z=\text{const}$ (if $\alpha=0$).
- Verified by direct double differentiation of $x(t),y(t),z(t)$ against the original equations of motion.
Result
The trajectory, projected and reparametrised in terms of $z$, is given by: