An object of mass m=80 kg moves in one dimension subject to the potential energy
U(x)=λ4(x2−a2)2+b2x2.(1)
Here we use λ= 1 kg/(m2s2), a= 7 m, and b= -11 kg/s2.
(a) How many equilibrium points (stable and unstable ones) does this potential have?
n=
0.1.2.3.4.
(b) Find a stable equilibrium point x0 such that x0 is positive. (in meters)
x0=
To determine the equilibrium points of the potential energy function U(x), we need to find the points where the force acting on the object is zero. Equilibrium points correspond to the points where the derivative of the potential energy with respect to x, dU/dx, is equal to zero.
(a) To find the equilibrium points, we differentiate the potential energy function U(x) with respect to x:
dU/dx = 8λ(x^2 - a^2)x + 2bx.
Setting dU/dx = 0, we can solve for x:
8λ(x^2 - a^2)x + 2bx = 0.
This equation can be factored as:
x(8λ(x^2 - a^2) + 2b) = 0.
For x to be non-zero, the expression in the parentheses must be equal to zero:
8λ(x^2 - a^2) + 2b = 0.
Simplifying this equation, we get:
8λx^2 - 8λa^2 + 2b = 0.
Now we can solve this quadratic equation for x. Plugging in the values for λ, a, and b, we have:
8x^2 - 56 + 22 = 0.
8x^2 - 34 = 0.
Dividing the equation by 2 gives:
4x^2 - 17 = 0.
Solving this quadratic equation, we find two possible values for x:
x = ±√(17/4).
Therefore, there are two equilibrium points.
Answer for (a): n = 2.
(b) To find a stable equilibrium point x0, we need to determine the sign of the second derivative of U(x) at the equilibrium points. If the second derivative is positive, it indicates a stable equilibrium point.
To find the second derivative, we differentiate dU/dx with respect to x:
d^2U/dx^2 = 8λ(2x^2 - a^2) + 2b.
Now we plug the equilibrium points x = ±√(17/4) into the second derivative expression:
For x = √(17/4):
d^2U/dx^2 = 8λ(2(√(17/4))^2 - a^2) + 2b
= 8(1)(2(17/4) - 49) - 22
= 8(8.5 - 49) - 22
= -300.
Since the second derivative is negative, this equilibrium point is unstable.
For x = -√(17/4):
d^2U/dx^2 = 8λ(2(-√(17/4))^2 - a^2) + 2b
= 8(1)(2(17/4) - 49) - 22
= 8(-8.5 - 49) - 22
= -300.
Again, the second derivative is negative, indicating an unstable equilibrium point.
To find a stable equilibrium point x0 that is positive, we need to find an interval where the potential energy function U(x) is minimized. In this case, since the potential energy function is a quartic function (power of 4), it has global minima at its equilibrium points. So the stable equilibrium point with a positive x0 will be the point where U(x) is minimized.
To find the value of x0, we need to calculate U(x) at the equilibrium points x = ±√(17/4):
For x = √(17/4):
U(√(17/4)) = λ/4(√(17/4)^2 - a^2)^2 + b^2(√(17/4))^2
= 1/4(289/4 - 49)^2 + 121(17/4)
= 1/4(65/4)^2 + 121(17/4)
= 4225/64 + 2057/16
= 65.9375 + 128.5625
= 194.5.
For x = -√(17/4):
U(-√(17/4)) = λ/4(-√(17/4)^2 - a^2)^2 + b^2(-√(17/4))^2
= 1/4(289/4 - 49)^2 + 121(17/4)
= 1/4(65/4)^2 + 121(17/4)
= 4225/64 + 2057/16
= 65.9375 + 128.5625
= 194.5.
Since both equilibrium points have the same potential energy value, we can choose either of them as the stable equilibrium point with a positive x0.
Answer for (b): x0 = √(17/4) = 2.59 meters.