Question

A non-relativistic particle of mass m moves in one dimension x under the force

F (x) = ax3 − bx,
where a and b are positive constants.
(a) Does a potential energy function exist for this force? Why? Is the total energy conserved?
(b) Find an expression for the potential energy V (x), and sketch its graph. Choose the arbitrary constant so that the potential energy vanishes at x = 0.
(c) Find all of the turning points of the motion if the total energy is E.
(d) For what values of the total energy, E is the motion bounded?
(e) For positive values of E, less than that found in part (d), find the frequency for periodic oscillations. Do not assume that the oscillations are small.

Answer 1

U = - (x⁴ / 4 - b x² / 2) , c) The function is zero for x = 0 and √2

Step-by-step explanation:

a and b) Strength and potential energy are related

F = - dU / dx

Therefore to find the energy we must integrate

∫ dU = -∫ F dx

∫ dU = - ∫ (a x³ –b x) dx

Let's make the integration

U = - (x⁴ / 4 - b x² / 2)

We evaluate the integral between the value

U - U₀ = -x⁴ / 4 + x² / 2 - (-x₀⁴ / 4 + x₀² / 2)

The arbitrary constant is zero, so that U is zero in the zero position

U₀ = 0 for x₀ = 0

c) Mechanical energy is the sum of kinetic energy plus potential energy

Em = K + U

Em = ½ m v² + ½ (x² -x⁴ / 2)

E = ½ m v² + ½ x² (1 - x² / 2)

Energy is positive

2 (E –K) = x² (1-x² / 2)

At the return points K = 0

The zero points of this function are

x = 0

(1- x² / 2) = 0

x² = 2

x = √ 2

The function is zero

x = 0 and √2

d) the movement is bounded for energy values ​​less than or equal to

E <= ½ x² (1-x² / 2)

e) for this part we resolved Newton's second law

F = m a

ax³ - b x = m d²x / dt²

d²x / dt² = -b / m x + a / m x³3

The linear term gives a simple harmonic movement

w₀² = b / m

d²x / dt² = - w₀² x + a / m x³

The frequencies are the frequencies of the harmonic movement plus a small change due to the non-harmonic part of the movement