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

Answer

D

Explanation

Potential energy `U = k|x|^(2)`
Hence force `SHM F = (dU)/(dx) = -3k k|x|^(2)`….(i)
Also for `SHM x= a sin omega t`
and `d^(2)x)/(dt^(2)) + omega^(2) x= 0`
`rArr` Acceleration `omega = ((d^(2)x)/(dt^(2)) = - omega^(2)x`…(ii)
From (i) and (ii) we get `omega = sqrt((3ks)/(m))`
`rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3ks)) = 2pi sqrt((m)/(3k(a sin omega t)))`
`rArr T prop (1)/(sqrt(a))`