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. proportional to `a`
C. proportional to `sqrt(a)`
D. proportional to `a^(1//2)`

Answer

A

Explanation

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