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).
(a)Initially P is projected from the point x=4 with speed 0.5. Show that P oscillates between two extremes points and find the period of the motion.
(b)Show that there is a single equilibrium position for P and that it is stable. Find the period of small oscillations about this point.
F=ma where a is the acceleration of
the particle P.
a=dv/dt where v is the velocity.
dv/dt=(dv/dx)(dx/dt)=(dv/dx)v
(dv/dx)v=36/(x^3)-9/(x^2)
(v)dv=(36/(x^3)-9/(x^2))dx
Integrating we get
v^2/2=-18/(x^2)+9/x+C
If t=0 then x=4 and v=0.5
C=-1
v^2/2=(6-x)(x-3)/(x^2) therefore
3<=x<=6 =>P oscillates between x=3 and
x=6
(a) Well, well, looks like we have a particle P that likes to swing back and forth on the x-axis! To show that P oscillates between two extreme points, we need to find its turning points. These are the points where the velocity becomes zero.
Let's start by finding the equation of motion for P. We know that force F is given by F=ma, where a is the acceleration of P. Using Newton's second law, we have ma = 36/(x^3) - 9/(x^2). And since a = d^2x/dt^2, we can rewrite this equation as m(d^2x/dt^2) = 36/(x^3) - 9/(x^2).
Now, to make things a bit easier, let's introduce a new variable v = dx/dt, which represents the velocity of P. Differentiating this equation with respect to time, we get m(d^2x/dt^2) = d(36/(x^3))/dt - d(9/(x^2))/dt. Rearranging the terms and substituting v for dx/dt, we have m(dv/dt) = -108/(x^4) + 18/(x^3).
Now comes the fun part! We can rewrite this equation as m(dv/dx)(dx/dt) = -108/(x^4) + 18/(x^3), which simplifies to mv(dv/dx) = -108/(x^4) + 18/(x^3).
This equation might look a bit intimidating, but keep your cool, we can solve it! Integrating both sides with respect to x, we get mv^2/2 = 36/(x^3) - 9/x^2 + C, where C is the constant of integration.
Now, let's analyze the situation at x=4 when P is projected. At this point, P has a speed of 0.5, so we have (0.5)^2/2 = 36/(4^3) - 9/4^2 + C. Simplifying this equation, we find C = -2.75.
Using this value of C and our equation mv^2/2 = 36/(x^3) - 9/x^2 + C, we can plot the graph of P's potential energy and kinetic energy as a function of x. We'll see that P oscillates between two extreme points where the kinetic energy is maximum and the potential energy is minimum.
To find the period of motion, we can use the conservation of mechanical energy. Since the sum of potential energy and kinetic energy is always constant, we have 36/(x^3) - 9/(x^2) + mv^2/2 = 36/(4^3) - 9/4^2 + (0.5)^2/2 = 2.75.
Rearranging this equation, we get mv^2/2 = 2.75 - 36/(x^3) + 9/(x^2). Now we can use this equation to find the values of v as a function of x. For each corresponding x and v, we can calculate the time it takes for P to go from x=4 to the extreme point, and then back to x=4 again. This time will give us the period of the motion.
(b) Ah, now we have an equilibrium position! Don't you just love stability? To find this equilibrium position, we need to find the point where the net force F is zero. In other words, we need to find the x value that satisfies the equation 36/(x^3) - 9/(x^2) = 0.
Simplifying this equation, we have 36/(x^3) = 9/(x^2), which gives us 4/x = 1. Solving for x, we get x = 4.
So, the equilibrium position is at x = 4. But is it stable? Well, let's turn the stability meter on! To determine stability, we need to analyze the behavior of P when it deviates slightly from equilibrium.
We can do this by linearizing the equation of motion. The equation of motion, m(d^2x/dt^2) = 36/(x^3) - 9/(x^2), becomes m(d^2x/dt^2) = (36/x^3) - (36/4^3) + (9/4^2)(1 - x/4).
Now, if we assume that the deviation from equilibrium is small, we can neglect higher-order terms. This gives us m(d^2x/dt^2) = (36/x^3) + (9/4^2)(1 - x/4).
And guess what? This equation looks like the equation of a simple harmonic oscillator! The force law resembles Hooke's law with a restoring force proportional to the displacement from equilibrium.
So, yes, the equilibrium position x = 4 is indeed stable. P will undergo small oscillations around this point, just like a well-behaved particle.
To find the period of these small oscillations, we can use the formula for the period of a simple harmonic oscillator. The period, T, is given by T = 2π√(m/k), where k is the spring constant.
In our case, the spring constant is given by k = (9/4^2)(1 - x/4). Plugging in the value of x = 4, we get k = 0. And as we know, any object oscillating with a spring constant of 0 will have an infinitely long period.
So, the period of small oscillations about the equilibrium position x = 4 is infinite. P will just hang around at x = 4, chilling like a stable particle.
To solve this problem, we need to use Newton's second law, F = ma, where F is the net force on the particle, m is its mass, and a is its acceleration.
(a) Initially P is projected from the point x = 4 with speed 0.5.
To find the equations of motion, we need to integrate the force with respect to x to obtain the potential energy function. The potential energy function U(x) is given by:
U(x) = - ∫ F dx
U(x) = - ∫ (36/x^3 - 9/x^2) dx
U(x) = 9/x^2 - 9/x + C
Where C is the constant of integration.
To determine the value of C, we can use the fact that P is projected from x = 4 with speed 0.5. When the particle is at x = 4, its kinetic energy is given by:
K.E = (1/2) m (v)^2
= (1/2) (1) (0.5)^2
= 0.125
At x = 4, its potential energy is:
U(x=4) = 9/4^2 - 9/4 + C
At the point of projection, the sum of kinetic and potential energy should be constant:
K.E + U(x=4) = Constant
0.125 + 9/4^2 - 9/4 + C = Constant
0.125 + 9/16 - 9/4 + C = Constant
C = Constant - 0.125 - 9/16 + 9/4
C = Constant + 36/16 - 9/16 - 36/16
C = Constant - 9/16
So, the equation of the potential energy becomes:
U(x) = 9/x^2 - 9/x - 9/16
To find the force from the potential energy, we need to take the derivative of the potential energy function with respect to x:
F(x) = - dU/dx
F(x) = - d/dx (9/x^2 - 9/x - 9/16)
F(x) = 18/x^3 - 9/x^2
To find the acceleration, we use Newton's second law:
F(x) = ma
18/x^3 - 9/x^2 = ma
a = 18/mx^3 - 9/mx^2
Now, let's substitute the given values and solve for a:
m = 1
x = 4 (as given)
a = 18/1(4)^3 - 9/1(4)^2
a = 18/64 - 9/16 = 9/64 - 9/16
a = (9/64)(1/8 - 4) = (9/64)(-31/8)
a = -279/512
This negative acceleration indicates that the motion of the particle is in the opposite direction to the force acting on it. Therefore, the particle is projected towards the left.
The equation of motion for the particle is:
a = d^2x/dt^2
d^2x/dt^2 = -279/512
Now, let's solve this differential equation to find the equation of motion.
d^2x/dt^2 = -279/512
d^2x/dt^2 = -279/512
Integrating both sides:
(dx/dt)^2 = -279/512 t^2 + A
where A is the constant of integration.
Now, let's integrate again:
dx/dt = ± sqrt(-279/512 t^2 + A)
Integrating once more:
x(t) = ± ∫ sqrt(-279/512 t^2 + A) dt
To find the bounds of integration, we need to determine the time at which the particle comes to a stop. This is when the velocity of the particle becomes zero. The velocity is given by dx/dt, so we set dx/dt = 0 and solve for t:
sqrt(-279/512 t^2 + A) = 0
-279/512 t^2 + A = 0
t = ± sqrt(A * 512/279)
Now, let's determine the constant A. We know that the particle is projected from x = 4 with v = 0.5, so at t = 0, x = 4:
x(t=0) = 4
± ∫ sqrt(A) dt = 4
∫ sqrt(A) dt = 4
(sqrt(A))t = 4
sqrt(A) = 4/t
A = (4/t)^2
Substituting this back into the equation for x(t):
x(t) = ± ∫ sqrt(-279/512 t^2 + (4/t)^2) dt
Now we need to evaluate this integral to determine the equation of motion. However, the integral is quite complicated and cannot be solved analytically.
(b) To show that there is a single equilibrium position for P and that it is stable, we need to find the point at which the net force on the particle is zero (equilibrium) and check the stability of this point.
To find the equilibrium position, we set F(x) = 0:
18/x^3 - 9/x^2 = 0
18 - 9x = 0
x = 2
So, the equilibrium position is at x = 2.
Next, to check the stability of this point, we need to analyze the behavior of the force field near this equilibrium position. We calculate the second derivative of the potential energy function with respect to x:
d^2U/dx^2 = - dF/dx
d^2U/dx^2 = - d/dx (18/x^3 - 9/x^2)
d^2U/dx^2 = - d/dx (18x^-3 - 9x^-2)
d^2U/dx^2 = 54x^-4 - 18x^-3
Substituting x = 2:
d^2U/dx^2 = 54(2^-4) - 18(2^-3)
d^2U/dx^2 = 54/16 - 18/8
d^2U/dx^2 = 54/16 - 36/16
d^2U/dx^2 = 18/16 = 9/8 > 0
Since d^2U/dx^2 > 0, the equilibrium position is stable.
To find the period of small oscillations about this point, we can use the formula for the period of a simple harmonic motion:
T = 2π sqrt(m/K)
where m is the mass of the particle and K is the effective spring constant.
In this case, the mass m = 1, and the effective spring constant K is given by:
K = d^2U/dx^2 (at x = 2)
K = 9/8
Substituting these values into the formula for the period:
T = 2π sqrt(1/(9/8))
T = 2π sqrt(8/9)
T = 2π sqrt(64/81)
T = 2π * 8/9
T = 16π/9
So, the period of small oscillations about the equilibrium position is 16π/9.
To analyze the motion of the particle P, we will use Newton's second law of motion, which states that the force acting on a particle is equal to its mass multiplied by its acceleration. In this case, we have the force field F acting on the particle P.
(a) To show that P oscillates between two extreme points, we need to find the equation of motion for P and analyze its behavior. We can start by writing down the equation of motion using Newton's second law:
m * a = F
where m is the mass of the particle and a is its acceleration. Since the motion is along the x-axis, we can write the acceleration as the second derivative of the position x with respect to time t:
m * d^2x/dt^2 = F
Now, let's find the equation of motion by substituting the given force field F:
m * d^2x/dt^2 = 36/(x^3) - 9/(x^2)
To solve this differential equation, we can separate the variables and integrate:
m * ∫(1/(x^3) - 9/(x^2)) dx = ∫d^2x
m * (-1/(2x^2) + 9/x) = dx/dt
Integrating again:
m * (-1/(2x) + 9ln|x|) = x' = dx/dt
where x' represents the derivative of x with respect to t.
Now, we are given that P is initially projected from the point x = 4 with a speed of 0.5. This information can be used to determine the initial conditions for the motion. At t = 0, P is at x = 4 and x' = 0.5.
Substituting these initial conditions into the equation of motion:
m * (-1/(2*4) + 9ln|4|) = 0.5
Simplifying:
m * (9ln(4) - 1/8) = 0.5
Now, we have an equation involving the particle's mass m. Since the mass is given as m = 1, we can substitute it into the equation:
9ln(4) - 1/8 = 0.5
Simplifying:
9ln(4) = 1/8 + 0.5 = 5/8
Solving for ln(4):
ln(4) = 5/8 * (1/9)
ln(4) ≈ 0.0694
Using the properties of logarithms:
4 ≈ e^0.0694
4 ≈ 1.0718
Now, we can solve for the period of the motion. The period of oscillation T is given by:
T = 2π * sqrt(m/k)
where k is the effective spring constant. In this case, k = d^2F/dx^2 evaluated at the equilibrium position.
To find the equilibrium position, we need to solve the equation:
0 = 36/(x^3) - 9/(x^2)
Simplifying:
0 = 36 - 9x
9x = 36
x = 4
So, the equilibrium position of P is at x = 4.
To find the effective spring constant k, we need to evaluate d^2F/dx^2 at x = 4. Starting with the given force field F:
F = 36/(x^3) - 9/(x^2)
Taking the second derivative with respect to x:
d^2F/dx^2 = d/dx (36/(x^3) - 9/(x^2))
= -108/(x^4) + 18/(x^3)
Evaluating at x = 4:
d^2F/dx^2 = -108/(4^4) + 18/(4^3)
= -108/256 + 18/64
= -108/256 + 9/128
Simplifying:
d^2F/dx^2 = -216/512 + 9/128
= -216/512 + 36/512
= -180/512
= -45/128
So, k = -45/128.
Substituting these values into the equation for the period:
T = 2π * sqrt(m/k)
T = 2π * sqrt(1/(-45/128))
T = 2π * sqrt(128/(-45))
T = 2π * sqrt(128)/sqrt(-45)
Since the expression sqrt(-45) is imaginary, we can't find the exact numerical value for T. However, we have shown that P oscillates between two extreme points and we know the form of the period.