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.
See April 26, Daphine
That doesnt answer the second part
To determine the equilibrium position(s) of the particle, we need to find the values of x for which the net force acting on the particle is zero. In other words, we need to solve the equation F = 0.
Given that F = 36/(x^3) - 9/(x^2), we can rewrite the equation as:
36/(x^3) - 9/(x^2) = 0
Multiplying both sides of the equation by x^3, we get:
36 - 9x = 0
Simplifying further, we have:
9x = 36
Dividing both sides by 9, we find:
x = 4
Therefore, the particle has a single equilibrium position at x = 4.
To determine the stability of the equilibrium position, we need to analyze the behavior of the force around the equilibrium point.
Taking the derivative of the force function F with respect to x, we have:
dF/dx = d(36/(x^3) - 9/(x^2))/dx
= d(36x^-3 - 9x^-2)/dx
= -108x^-4 + 18x^-3
Evaluating the derivative at x = 4, we get:
dF/dx (x=4) = -108(4^-4) + 18(4^-3)
= -108/256 + 18/64
= -27/64 + 9/64
= -18/64
= -9/32
Since the derivative of the force at the equilibrium point is negative (-9/32 < 0), the equilibrium position is stable.
To find the period of small oscillations about the equilibrium point, we can make a small displacement from x = 4 and consider the motion as a simple harmonic oscillator.
The force acting on the particle can be approximated by a linear function near the equilibrium point. Using Taylor expansion, we can write:
F ≈ F(x=4) + (dF/dx)(x-4)
Since F(x=4) = 0 (equilibrium position) and dF/dx (x=4) = -9/32 (as calculated above), we have:
F ≈ -9/32(x-4)
This approximation indicates that the force acting on the particle can be represented by Hooke's Law (F = -kx), where k = 9/32 is the spring constant.
The period of small oscillations for a simple harmonic oscillator can be determined using the formula:
T = 2π√(m/k)
Given that m = 1 and k = 9/32, we substitute these values into the formula:
T = 2π√(1/(9/32))
= 2π√(32/9)
= 2π(4/3)
= 8π/3
Therefore, the period of small oscillations around the equilibrium point x = 4 is 8π/3.
To determine the equilibrium position(s) for particle P, we need to find the point(s) where the net force acting on the particle is zero. In this case, the net force is given by the force field F = 36/(x^3) - 9/(x^2).
Setting F = 0, we have:
36/(x^3) - 9/(x^2) = 0
Multiplying through by x^3, we get:
36 - 9x = 0
Rearranging, we find:
36 = 9x
x = 4
So, the equilibrium position for particle P is when x = 4.
To determine if this equilibrium position is stable or not, we can analyze the behavior of the force field around x = 4.
Taking the derivative of F with respect to x, we have:
dF/dx = -108/(x^4) + 18/(x^3)
Evaluating this derivative at x = 4, we get:
dF/dx (x=4) = -108/(4^4) + 18/(4^3)
= -27/64 + 18/64
= -9/64
Since the derivative is negative at x = 4, it indicates that the force field is directing the particle towards x = 4 when it deviates from this point. Thus, the equilibrium position at x = 4 is stable.
To find the period of small oscillations about this equilibrium point, we can use the formula:
T = 2π√(m/k)
where T is the period, m is the mass of the particle P, and k is the spring constant. In this case, we can think of the force field as representing a spring force with a spring constant k.
To find k, we can calculate the second derivative of the force field at x = 4:
d²F/dx² = 432/(x^5) - 54/(x^4)
Evaluating this second derivative at x = 4, we have:
d²F/dx² (x=4) = 432/(4^5) - 54/(4^4)
= 432/1024 - 54/256
= 27/64 - 54/256
= 27/64 - 27/128
= 27/128
Since k is equal to the negative of the second derivative, we have k = -27/128.
Substituting the given mass m = 1 and the calculated k = -27/128 into the formula for the period, we get:
T = 2π√(1/(-27/128))
= 2π√(-128/27)
= 2π√(128/(-27))
= 2π√(-4.74...)
The period of small oscillations about the equilibrium point x = 4 is approximately 2π times the square root of -4.74 (which is an imaginary number), indicating that there are no real small oscillations about this point.