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Question

Calculus Problem

ATTRACTION VERSUS REPULSION

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

Variables are x (position), Ka (attraction constant), Kr (repulsion constant), m (mass of particle), v (velocity), a (acceleration), t (time) and initial conditions of x = xo and v = vo at t=0. From the description Fa = -Ka/(x+1) and Fr=Kr/(X-1). Note that the sum of the two forces can result in motion where a = ((Ka+Kr)x+Ka-Kr)/(x²-1)/m. Note that if Ka=Kr, a=(Ka+Kr)x/(x²-1)/m. Note further that if at t=0, x=0, a=0, vo=0 then no motion occurs. Note that in the case where vo=0 and xo=(Kr-Ka)/(Ka+Kr) at t=0, no motion occurs as the sum of the forces =0. Note that in the case where vo<>0 and/or xo<>(Kr-Ka)/(Kr+Ka), oscillary motion continues forever as this system has no losses (friction) oscillating about (Kr-Ka)/(Kr+Ka) according to the equation below:

x3/6-xlnx+x=(Ka+Kr)t2/2m+Ct+C1 C and C1 are determined by xo, vo which is found by setting a=d²x/dt² and solving the resulting separable differential equation.