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 show that there is a single equilibrium position for particle P, we need to find where the net force acting on the particle is zero. In other words, we need to find the value of x for which F(x) = 0.
Let's start by setting F(x) equal to zero:
36/(x^3) - 9/(x^2) = 0
To simplify this equation, let's multiply both sides by x^3 to eliminate the denominators:
36 - 9x = 0
Now, let's solve for x:
36 = 9x
x = 36/9
x = 4
Therefore, the equilibrium position for particle P is at x = 4. This means that when the particle is at x = 4, the net force acting on it is zero.
To determine if this equilibrium position is stable, we need to analyze the behavior of the force as the particle deviates from x = 4. We can do this by examining the sign of the derivative of the force function.
The derivative of the force function is given by:
F'(x) = dF/dx = -108/(x^4) + 18/(x^3)
Now, let's evaluate F' at x = 4:
F'(4) = -108/(4^4) + 18/(4^3)
F'(4) = -108/256 + 18/64
F'(4) = -27/64 + 9/64
F'(4) = -18/64
F'(4) = -9/32
Since F'(4) is negative, this means that the force is increasing as the particle deviates from x = 4. This indicates that the equilibrium position is stable because the particle will experience a restoring force that brings it back towards x = 4 when it is displaced.
To find the period of small oscillations about the equilibrium position, we can use the formula:
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, k represents the stiffness of the force function in the vicinity of x = 4.
To find k, we can differentiate the force function twice:
F''(x) = d^2F/dx^2 = 432/(x^5) - 54/(x^4)
Now, let's evaluate F'' at x = 4:
F''(4) = 432/(4^5) - 54/(4^4)
F''(4) = 432/1024 - 54/256
F''(4) = 27/64 - 54/256
F''(4) = 27/64 - 27/128
F''(4) = 27/64 - 27/64
F''(4) = 0
Since F''(4) = 0, the stiffness (k) at the equilibrium position is zero. This indicates that the motion of the particle is not governed by a traditional spring force, but by a force field. Therefore, the period of small oscillations about the equilibrium position cannot be found using the formula mentioned earlier.
In this case, finding the period of small oscillations requires a more in-depth analysis of the behavior of the particle near x = 4 in the force field.