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^(3//2)`

Answer

C

Explanation

As `U=k|x|^(3) therefore F=-(dU)/(dx)=-3k|x|^(2)" "...(i)`
The equation of SHM is given as `x=Asinomegat`
`therefore (d^(2)x)/(dt^(2))=-omega^(2)xorm(d^(2)x)/(dt^(2))=-momega^(2)x" "...(ii)`
Using (i) and (ii), we get
`3k|x|^(2)=momega^(2)x or omega=sqrt((3kx)/(m))`
`therefore T=(2pi)/(omega)=2pisqrt((m)/(3kx))=2pisqrt((m)/(3kAsinomegat))orTprop(1)/(sqrt(A))`