A particle P of mass m=1 moves on the x-axis under the force field
F=36/(x^3) - 9/(x^2) (x>0).
Show that each motion of P consists of either (i) a periodic oscillation between two extreme points, or (ii) an unbounded motion with one extreme point, depending on the value of the total energy.
Regards
To analyze the motion of particle P under the given force field, we need to first express the force in terms of the position x. We can start by finding the derivative of the potential energy function U(x), since the force is the negative gradient of the potential energy.
Given that the force field is F(x) = 36/(x^3) - 9/(x^2), the potential energy function U(x) can be found by integrating the force function with respect to x.
U(x) = - ∫F(x) dx = - ∫(36/(x^3) - 9/(x^2)) dx = - (36 * (-1/x^2) + 9 * (1/x)) + C1,
where C1 is the constant of integration.
Simplifying the above expression, we get:
U(x) = 36/x^2 - 9ln|x| + C1.
Now, let's consider the total mechanical energy of the system, E, which is given by the sum of the kinetic energy (T) and the potential energy (U):
E = T + U.
The kinetic energy of the system is given by the expression:
T = (1/2) * m * (dx/dt)^2.
Since the mass m = 1, we can rewrite the expression for T as:
T = (1/2) * (dx/dt)^2.
Now, rearranging the above expression, we get:
(dx/dt)^2 = 2T.
Taking the square root of both sides, we have:
dx/dt = √(2T).
Next, we can express the total mechanical energy E in terms of the position x and velocity dx/dt:
E = T + U = (1/2) * (dx/dt)^2 + U(x).
Now, let's substitute dx/dt = √(2T) into the equation for E:
E = (1/2) * 2T + U(x) = T + U(x).
Replacing T with (dx/dt)^2, we have:
E = (dx/dt)^2 + U(x).
Substituting the expression for U(x) that we derived earlier:
E = (dx/dt)^2 + 36/x^2 - 9ln|x| + C1.
Rearranging the above equation, we get:
(dx/dt)^2 = E - 36/x^2 + 9ln|x| - C1.
Now, let's consider the different cases depending on the value of the total energy E:
Case 1: E > 0
If the total energy E is positive, then the right-hand side of the above equation (dx/dt)^2 will always be positive. Therefore, (dx/dt) must be either positive or negative, implying that the velocity dx/dt can have both positive and negative values. This means that particle P can have motion in both directions along the x-axis. Since (dx/dt) can be continuously increasing or decreasing, the motion will be unbounded and will not return to the starting point. In this case, there will be one extreme point but no periodic oscillation.
Case 2: E = 0
If the total energy E is zero, then the equation becomes:
(dx/dt)^2 = - 36/x^2 + 9ln|x| - C1.
In this case, the right-hand side of the equation can never be positive since all the terms on the right-hand side are negative or zero. Therefore, there will be no motion and no extreme points.
Case 3: E < 0
If the total energy E is negative, then the equation becomes:
(dx/dt)^2 = E - 36/x^2 + 9ln|x| - C1.
In this case, the right-hand side of the equation can be positive or negative depending on the values of x. When the right-hand side is positive, there will be motion, and when it is negative, there will be no motion. The motion will be bounded between two extreme points on the x-axis, and there will be periodic oscillations.
To summarize, each motion of particle P consists of either a periodic oscillation between two extreme points or an unbounded motion with one extreme point depending on the value of the total energy E.