For each of the following functions describing the net force on a particle, determine any and all positions of equilibrium and characterize each as stable or unstable. (a) F = 4x − 2x3, (b) F = x2 + x − 6, (c) F = 5x − 9.
Answer: a. x = 0, unstable; x = ±1.41, both stable
b. x = −3, stable; x = 2, unstable
c. x = −1.8, unstable
Please help me figure out how to get the answer. Thank you!
The second derivative. If f" is <0, it is unstable
a) f=4x-2x^3
f'=4-6x^2
f"=-12x unstable at zero,
now where is f=0?
0=x(4-2x^2) x=0 unstable
x=1.41 unstable
x=-1.41 stable
b. F = x2 + x − 6
f'=2x+1
f"=2 always stable.
https://en.wikipedia.org/wiki/Mechanical_equilibrium
wondering where you got these "answers", as they are wrong.
zeroes: 0=(x-3)(x+2)
x=3 stable, x=-2 stable
c. F = 5x − 9.
f'=5
f"=0 neutral
zeroes: x=1.8
The teacher gives us the answer, but still thank you!
To determine the positions of equilibrium and their stability for each given function, we need to find the points where the net force is equal to zero. Equilibrium occurs when the net force acting on a particle is zero, meaning there is no acceleration.
For function (a) F = 4x - 2x^3:
To find the positions of equilibrium, set the equation F = 4x - 2x^3 equal to zero and solve for x:
4x - 2x^3 = 0
Factor out an x:
x(4 - 2x^2) = 0
This equation is satisfied when either x = 0 or 4 - 2x^2 = 0.
For x = 0, the net force is zero. To determine stability, we need to examine the behavior of the force around this position. One way to do this is by taking the derivative of the force function with respect to x:
F' = 4 - 6x^2
Evaluate the derivative at x = 0:
F'(0) = 4 - 6(0)^2 = 4
Since the derivative is positive at x = 0, meaning the slope is positive, the position is unstable. This means any small disturbance away from x = 0 will cause the particle to move further away.
Now, let's find the other positions of equilibrium by solving 4 - 2x^2 = 0:
4 - 2x^2 = 0
2x^2 = 4
x^2 = 2
x = ±√2 ≈ ±1.41
For x = ±1.41, the net force is zero. To determine stability, we again examine the behavior of the force around these positions. Let's take the derivative again:
F' = 4 - 6x^2
Evaluate the derivative at x = ±1.41:
F'(1.41) ≈ 4 - 6(1.41)^2
F'(-1.41) ≈ 4 - 6(-1.41)^2
Both calculations give a negative value for F'. This indicates that the slope is negative, and therefore, the positions of equilibrium at x = ±1.41 are stable. Any small disturbance from these positions will cause the particle to return back towards them.
So, for function (a), the positions of equilibrium are x = 0, unstable, and x = ±1.41, both stable.
Now, let's go through functions (b) and (c) in a similar manner to find their positions of equilibrium and their stability.