Answer:
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.