A bead of mass `m` can slide on a frictionless wire as shown in figure Because of the given shape of the wire near `p` the bottom point , it can be approximated as near `p` the potential energy of the bead is given `U = cx^(2)` where `c` is a constant and `x` is measured from `p` The bead if displacement slightly from point `p` will oscillate about `p` The period of oscillation is

Options

A. `2 pi sqrt(c//m)`
B. `2 pi sqrt(m//2c)`
C. `2 pi sqrt(m//c)`
D. `2 pi sqrt(2c//m)`

Answer

B

Explanation

`U = Cx^(2)`
`F = -(dU)/(dx) rArr ma = -2cx rArr a = - (2c)/(m)x`
`rArr omega = sqrt((2c)/(m)) rArr T = (2pi)/(omega) = 2pi sqrt((m)/(2c))`