Question

If a partical moves in a potential energy held `U = U_(0) - ax + bx^(2)`, where are a and b partical constents obtian an expression for the force acting on if as a function of position. At what point does the force vanish? Is this a point of stable equilibriun ?
Calculate the force constant and friquency of the partical.

Answer 1

`F = - (dU)/(dx) = a - 2b x`
`F = 0, at x = a//2b`
If stable equilibrum is present at this position, `(d^(2(U)/(dx^(2))) gt 0)`
Hence, `(d^(2(U)/(dx^(2))) = 2b gt 0)`
i.e., `x = (a)/(2 b)` is a point of minimum potential energy
Hence, the equilibrium is stable. The partical will oscillate about `x = (a)/(2 b)`
The effective force constant of oscillation
`K_(eff) = ((d^(2)U)/(dx^(2))|_(x= (a)/(2b))) = 2b`
As `K_(eff) = m omeg^(2) = 2b = m omeg^(2)`
or `omega^(2) = (2 b)/(m) implies omega = 2 pi f = sqrt((2b)/(m)) implies f = (1)/(2pi) sqrt((k)/(m))`