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= 6 kg/(m2s2), a= 4 m, and b= 36 kg/s2.
------ then ,
(b) Find a stable equilibrium point x0 such that x0 is positive. (in meters)
x0=
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(c) Do a Taylor expansion of the force F(x) for x close to the equilibrium point, x≃x0, that is F(x)=F0−k(x−x0)+… What are the values for F0 (in Newton) and k (in kg/s2)?
hint: For help on using Taylor series to express the potential energy function near a stable minimum, we encourage you to look at the following page in the section on small oscillations:here
F0=
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k=
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(d) What is the period T of small oscillations (in seconds) of this mass around the equilibrium point x0? (Note that the parameter k found in the previous question acts like a spring constant that wants to pull small deviations back to the equilibrium point)
T=
To find the stable equilibrium point x0, we need to find the minimum of the potential energy function U(x).
To do this, we differentiate U(x) with respect to x and set it equal to zero:
dU/dx = 0
The potential energy function is given by:
U(x) = λ/4(x^2 - a^2)^2 + b^2x^2
Differentiating U(x) with respect to x gives us:
dU/dx = λ(x^2 - a^2)x + b^2x
Setting dU/dx to zero and solving for x gives us the equilibrium point x0:
λ(x^2 - a^2)x + b^2x = 0
x(x^2 - a^2)λ + b^2x = 0
x[(x^2 - a^2)λ + b^2] = 0
Since we are looking for a positive equilibrium point, x0 cannot be zero. Hence, we can solve for x0 as follows:
(x^2 - a^2)λ + b^2 = 0
(x0^2 - a^2)λ + b^2 = 0
(x0^2 - 4^2)6 + 36^2 = 0
(x0^2 - 16)6 + 1296 = 0
6x0^2 - 96 + 1296 = 0
6x0^2 + 1200 = 0
x0^2 = -1200/6
x0^2 = -200
Since x0 cannot be negative, there is no real solution for x0.
To find the stable equilibrium point x0, we need to find the minimum point of the potential energy function U(x).
To do this, we differentiate U(x) with respect to x and set it equal to zero to find the critical points:
dU(x)/dx = 0
Taking the derivative using the power rule and simplifying, we get:
dU(x)/dx = λ*4*(2x)*(x^2 - a^2) + 2b^2*x = 0
Now we can solve this equation to find the equilibrium point x0.
(b) To find the values for x0, we substitute the given values λ = 6 kg/(m^2s^2), a = 4 m, and b = 36 kg/s^2 into the equation and solve for x0:
0 = 6*4*(2x0)*(x0^2 - 4^2) + 2*36^2*x0
Simplifying the equation further, we have:
0 = 48x0*(x0^2 - 16) + 2592x0
0 = 48x0^3 - 48*16x0 + 2592x0
0 = 48x0^3 - 768x0 + 2592x0
0 = 48x0^3 - 768x0 + 2592x0
0 = 48x0^3 + 1824x0
0 = 48x0(x0^2 + 38)
From this equation, we can see that x0 = 0 is a solution, but we need to find a positive value for x0.
By solving the quadratic equation x0^2 + 38 = 0, we find that there are no real solutions. Therefore, the only stable equilibrium point for x0 such that x0 is positive is x0 = 0.
(c) To find the Taylor expansion of the force F(x) for x close to the equilibrium point x0, we first need to find the force F(x) from the potential energy function U(x).
The force is given by the negative derivative of the potential energy:
F(x) = -dU(x)/dx
Differentiating U(x) with respect to x, we have:
F(x) = -dU(x)/dx = -λ*4*(2x)*(x^2 - a^2) - 2b^2*x
Next, we need to find F0 and k by evaluating F(x) at the equilibrium point x0 and taking the derivative with respect to x:
F0 = F(x0) = -λ*4*(2x0)*(x0^2 - a^2) - 2b^2*x0
k = -dF(x)/dx = -d/dx(-λ*4*(2x)*(x^2 - a^2) - 2b^2*x)
We substitute the equilibrium point x0 = 0 into these equations to find the values of F0 and k.
(d) To find the period T of small oscillations around the equilibrium point x0, we can use the formula:
T = 2π * √(m / k)
where m is the mass of the object and k is the spring constant.
We were not given the mass of the object in the question, so we cannot calculate the period T without that information.