Aparticle 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

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.

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length