Can someone solve this in order to provide an example for solving central force problems?

Question

Let $F(x)=x$ be a function that describes the magnitude and direction of a force that varies with distance from the origin. I understand that $m \frac {d^2} {dt^2} p(t) = F(p(t))$ is used to derive the motion of a point particle $P$ placed within this force field, but could someone please show me how to do this for this simple case? I just need an example of how to do this before I move on to more complex fields.

Answer 1

Well firstly $p$ might not be the best letter for the position of a particle since $p$ is commonly used for linear momentum and is used in Newtons laws, $F=\frac{dp}{dt}$.

Anyway, suppose a particle whose position I will denote $x(t)$ feels a force directly proportional to it's displacement from the origin of our co-ordinate system $0$, is given by $F(x)=x$, then it's motion is given by solving the resultant differential equation:

$$x=F(x)=\frac{dp}{dt}=m\frac{dv}{dt}=m\frac{d^2 x}{dt^2}$$

We'll firstly take this as a $1d$ equation for simplicity.

If $F(x)=-x$ it would be a restoring force like a spring, however, this force pushes away from the origin, and the force is bigger the further you are away.

$$\frac{d^2 x}{dt^2}=\frac{1}{m}x(t) \text{ is solved by } x(t)=Ae^{\,t/m^{1/2}}+Be^{-t/m^{1/2}}$$

This can be solved either by checking, or in principle by power series or some method for constant coefficient odes. Note mass $m$ is a positive quantity so there's no problem with square roots.

Now to solve this fully we need initial conditions. You could specify $x(t_1)=x_1$ and $x(t_2)=x_2$ which would be Dirichilet conditions, $v(t_1)=v_1$ and $v(t_2)=v_2$, which would be Neumann, or mixed or so on.

Either way we need two conditions to fully specify the motion, and solve for the two undetermined constants $A$ and $B$.

If you would like to analyse motions, consider the energy of the system, $E=\frac{1}{2}mv^2+U(x)$, take $U(0)=0$ and $U(x)=-\int_0^x x' dx'=-x^2/2$, so the energy $E=-2AB$, check by $v(t)=\frac{dx}{dt}$ and sub in.

As an over view of the motions, if the energy $E=0$, see green line, the particle can exist and stay at the origin iff $A=B=0$, i.e., it starts there and has no velocity. This is an unstable equilibrium point. If it does not have a velocity $=0$ at $x=0$ it will move off to infinity in the direction it crosses $0$.

Otherwise regardless of where else the particle starts it will feel a force away from the origin, and will move off to either $\pm \infty$. See the red curve. If it moves in from the left, it slows down and stops where it meets the red line, then turns and moves back to $-\infty$, or if it approaches from the right of $0$, it again slows and stops at the red line and moves away to $+\infty$.

A final remark on more than one dimension.

$\vec{x}=\vec{F}=m\frac{d^2\vec{x}}{dt^2}$ is three in three dimensions. You can solve each component separately, and try a similar analysis in that case.