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 there is a single equilibrium position for P and that it is stable. Find the period of small oscillations about this point.
To determine the equilibrium position(s) for particle P under the given force field, we need to find the point(s) where the net force experienced by the particle is zero.
First, let's find the net force by setting F equal to zero:
36/(x^3) - 9/(x^2) = 0
To solve this equation, let's find the common denominator and simplify:
36x/(x^3) - 9/(x^2) = 0
Multiplying through by x^3 gives:
36x - 9x^3 = 0
Factor out x and divide both sides by 9:
x(4 - x^2) = 0
This equation implies two possibilities:
1. x = 0
2. 4 - x^2 = 0
The first possibility, x = 0, corresponds to the origin of the x-axis (since x > 0 in the given force field). However, at this point, the force is not defined, as it involves division by zero.
The second possibility, 4 - x^2 = 0, gives x^2 = 4. Solving for x, we have:
x = ±√4 = ±2
Therefore, there are two equilibrium positions for particle P: x = -2 and x = 2.
To determine the stability of these equilibrium positions, we need to analyze the behavior of the force field around each point.
Let's consider the point x = 2. To determine if it is stable, we can examine the force acting on a particle slightly displaced from this point. Let's denote the displacement as δx.
The net force acting on the particle can be approximated by expanding F(x) in a Taylor series around x = 2:
F ≈ F(x) + (dF/dx)|x=2 * δx
Evaluating the derivative of F with respect to x:
dF/dx = d(36/(x^3) - 9/(x^2))/dx = -108/(x^4) + 18/(x^3)
Substituting x = 2:
dF/dx| x=2 = -108/(2^4) + 18/(2^3) = -9 + 2.25 = -6.75
Substituting this back into the Taylor series expansion:
F ≈ 36/(x^3) - 9/(x^2) - 6.75δx
To analyze stability, we need to determine whether this force acts towards or away from the equilibrium position x = 2.
When δx is small, it is convenient to ignore the terms δx^2 and higher powers of δx. So, the expression becomes:
F ≈ 36/(x^3) - 9/(x^2) - 6.75δx
Let's plug in x = 2:
F ≈ 36/(2^3) - 9/(2^2) - 6.75δx
Simplifying:
F ≈ 4.5 - 2.25 - 6.75δx
F ≈ 2.25 - 6.75δx
Since 2.25 is positive, F will be positive for negative values of δx (displacement towards the equilibrium position) and negative for positive values of δx (displacement away from the equilibrium position).
Therefore, the equilibrium position at x = 2 is stable because the net force acts in the opposite direction of the displacement, tending to restore the particle to its equilibrium position.
The analysis for the point x = -2 would be similar, and you would find that it is also a stable equilibrium.
To find the period of small oscillations about this stable equilibrium point(x=2), we can use the formula for the period of a particle undergoing simple harmonic motion:
T = 2π√(m/k)
Where T is the period, m is the mass of the particle, and k is the spring constant.
In this case, we can approximate the force field for small oscillations around the equilibrium position, x = 2, as a simple harmonic oscillator. The force acting on the particle can be approximated as:
F ≈ -kδx
Comparing this to the approximation we derived earlier:
F ≈ 2.25 - 6.75δx
We can equate the two expressions to determine the spring constant:
-kδx = 2.25 - 6.75δx
Simplifying:
(6.75 - k)δx = 2.25
To achieve simple harmonic motion, the coefficient of δx must be equal to the spring constant:
6.75 - k = k
2k = 6.75
k = 6.75/2
k = 3.375
Now we can substitute the values of mass (m = 1) and the spring constant (k = 3.375) into the formula for the period:
T = 2π√(m/k)
T = 2π√(1/3.375)
T = 2π√(4/13.5)
T = 2π√(4/13.5) * √(13.5/13.5)
T = 2π√(54/175.5)
T = 2π√(6/19)
Therefore, the period of small oscillations about the equilibrium position x = 2 is given by 2π√(6/19).