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

`V (x) = k|x|^(3)`
`:. [k] ([V])/([x]) = (ML^(2)T^(2))/(L^(3)) = ML^(-1)T^(-2)`
Now time period on `T` and `("mass")^(x) ("amplitude")^(y) (k)^(z)`
`[M^@L^@T]=[M]^(x)[L]^(y)[ML^(-1)T^(-2)]^(z)`
=`[M^(x+y)L^(y-x)T^(-2z)]`
Equating the powers, we get
`-2z =1` or `z = -1//2`
`y-z = 0 y = z = - 1//2`
Hence `T prop ("amplitude")^(-1//2)`
or `T prop (1)/(sqrt(a))`.