Question

A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) Find the potential energy function U(x) (Take U(x=0) = 0). b) Find any equilibrium points and determine if they are stable or unstable. c) Sketch the potential as a function of position. d)What is the maximum total energy, Emax a particle may have if its motion is to remain bounded? e) For the case of bounded motion with E

Answer 1

a) The potential energy function U(x) is U(x) = -kx^2/2 + kx^4/(4α^2). b) Equilibrium points are x = 0, x = ±√(α^2/2), and the stability can be determined by examining the second derivative of the potential energy function. c) The sketch of the potential energy function will show a stable minimum at x = 0 and unstable maxima at x = ±√(α^2/2). d) The maximum total energy E_max occurs at the unstable equilibrium points, where E_max = kα^2/4. e) For bounded motion, the total energy E must be less than or equal to E_max, so E ≤ kα^2/4.

Step-by-step explanation:

a) To find the potential energy function U(x), we integrate the force function with respect to x. Since U(x=0) = 0, the constant of integration is also 0. The potential energy function is given by U(x) = -kx^2/2 + kx^4/(4α^2).

b) Equilibrium points are where the force is zero, so we set F(x) = 0 and solve for x. In this case, there are three equilibrium points: x = 0, x = ±√(α^2/2). To determine stability, we can examine the second derivative of the potential energy function at each equilibrium point. If the second derivative is positive, the equilibrium point is stable; if negative, it is unstable. For this potential energy function, the equilibrium point x = 0 is stable and the points x = ±√(α^2/2) are unstable.

c) The sketch of the potential energy function will show a symmetric curve with a stable minimum at x = 0 and unstable maxima at x = ±√(α^2/2).

d) The maximum total energy E_max occurs at the unstable equilibrium points. At these points, the kinetic energy is equal to the potential energy, so E_max = U(x = ±√(α^2/2)) = kα^4/(4α^2) = kα^2/4.

e) For bounded motion, the total energy E must be less than or equal to E_max, so E ≤ kα^2/4.

Answer 2

Step-by-step explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

= (-kx + kx³/a²) / m

If U be potential function

- dU / dx = (-kx + kx³/a²) / m

U(x) = ∫ (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx² = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U = k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x= ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max) = k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.