A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.

Options

A. proportional to `(1)/(sqrt(a))`
B. independent of a
C. proportional to `sqrt(a)`
D. proportional to `a^(3//2)`

Answer

A

Explanation

`U = k|x|^(3) rArr F = - (dU)/(dx)" "…(i)`
Also, for SHM `x = a sin omega t`
And, `(d^(2)x)/(dt^(2)) + omega^(2) x = 0`
`rArr` acceleration, `a = (d^(2)x)/(dt^(2)) =- omega^(2)x`
`rArr F = ma = m (d^(2)x)/(dt^(2)) = -m omega^(2) x" "...(ii)`
From Eqs. (i) and (ii), we get `omega = sqrt((3kx)/(m))`
`rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3kx))`
`= 2pi sqrt((m)/(3k (a sin omega t))) rArr T prop (1)/(sqrt(a))`