Attraction vs Repulsion Calculus Problem

Question

Consider the motion of a particle in one

dimension (x-axis) under the simultaneous influence of two forces. One force repels the particle from the location x=1, while the second one attracts the particle towards the location x = -1. In both cases, the force is inversely proportional to the distance from x = 1 and x = -1 respectively, but the proportionality constants may be different. Describe the trajectory of the particle over time, for different values of the two proportionality constants. Does the motion of the particle change qualitatively as you vary the proportionality constants?

• Define variables that capture the important components of the situation.

• Set up equations to describe the desired relationship between the variables you have defined.

• Describe how to solve the equations you set up.

• Interpret the solutions you obtained in the context of the original situation.

Answer 1

Here is the trajectory in words.

There are three zones of interest:

Zone 1: x₀>1

Zone 2: -1<x₀<1

Zone 3: x₀<-1

Different forces apply in each zone, and x=1 and x=-1 are asymptotes. x₀ is the starting point for the particle and applying Newton’s Law F=ma, so a=F/m.

In Zone 1, there is a repulsive force only which causes the particle to accelerate away from x=1. We can write the acceleration a=k₁/(x-1) where k₁ is a constant which incorporates the mass of the particle. The acceleration decreases as the particle is pushed further and further away from its starting point x₀. The asymptote shields the particle from the attractive force.

In Zone 2, two forces act on the particle and the acceleration is -k₁/(1-x)-k₂/(1+x), where k₂ is the constant relating to the attractive force. The net force drives the particle towards x=-1.

In Zone 3, the attractive force is -k₂/(x+1). Since x₀<-1, x₀+1<0 which makes the acceleration increase as the particle moves towards x=-1. The asymptote shields the particle from the repulsive force.

The acceleration becomes infinite x→-1 or x→1 from either side, hence the three zones and the asymptotes.

Now the mathematical model. Let v=dx/dt. then acceleration a=dv/dt=d²x/dt² and dx/dv.dv/dt=dx/dt=v, dx/dv=v/(d²x/dt²). So (d²x/dt²)dx=vdv. This is "integration ready" form.

Applying this to Zone 1, (k₁/(x-1))dx=vdv. Integrating we get: k₁ln(x-1)=v²/2+C, where C is a constant of integration. When x=x₀ let v=0, meaning that initially the particle is at rest when x=x₀.

C=k₁ln(x₀-1) and k₁ln(x-1)=v²/2+k₁ln(x₀-1), (x-1)/(x₀-1)=e^(v²/(2k₁)).

Alternatively, v²=2k₁ln[(x-1)/(x₀-1)]. Both x and x₀ are greater than 1 for this equation to apply.

Applying the equation to Zone 3, (-k₂/(x+1))dx=vdv. Integrating we get: -k₂ln|x+1|=v²/2+C, where C is a constant of integration. When x=x₀ let v=0, meaning that initially the particle is at rest when x=x₀. Remember that x₀<-1 for this zone.

C=-k₂ln|x₀+1| and -k₂ln|x+1|=v²/2-k₂ln|x₀+1|, (x₀+1)/(x+1)=e^(v²/(2k₂)). (x₀+1 and x+1 are both negative.)

v²=2k₂ln[(x₀+1)/(x+1)].

Now to apply the equation to Zone 2:

First, simplify the formula -k₁/(1-x)-k₂/(1+x)=-(k₁+k₁x+k₂-k₂x)/(1-x²)=(k₁+k₂+k₁x-k₂x)/(1-x²).

(d²x/dt²)dx=vdv becomes ((k₁+k₂+k₁x-k₂x)/(1-x²))dx=-vdv.  

This simplifies:

[k₁/(1-x)+k₂/(1+x)]dx=-vdv and integrating:

-k₁ln|1-x|+k₂ln|1+x|=C-v²/2, (1+x)^k₂/(1-x)^k₁=e^(C-v²/2)=(e^C)(e^-(v²/2)).

If v=0 when x=x₀, e^C=(1+x₀)^k₂/(1-x₀)^k₁, [(1+x)^k₂/(1-x)^k₁][(1-x₀)^k₁/(1+x₀)^k₂]=e^-(v²/2).