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

Answer

A

Explanation

`V = k|x|^(3)`
`F = -(dv)/(dx) = -3k |x|^(2)` …(1)
The equation of simple harmonic motion is given as
`x = a sin omega t`
`rArr m (d^(2)x)/(dt^(2)) = m (- a omega^(2) sin omega t) = - m omega^(2) x` ….(2)
Using (i) and (ii), we obtain
`3k |x|^(2) = m omega^(2) x`
`rArr omega = sqrt(2k x//m)`
`rArr T = 2pi sqrt((m)/(2kx)) rArr T = 2pi sqrt((m)/(3ka sin omega t))`
`rArr T prop (1)/(sqrta)`
Hence (a) is correct