In the fig the pulled has a mass `M`, radius `r` and the string does not slip over it. The time period of oscillation is

Options

A. `2pi sqrt((2m+M)/(2k))`
B. `(2pi)/(R ) sqrt((2m+3MR^(2))/(2k))`
C. `2pi sqrt((m)/(k))`
D. `2pi sqrt((m+M)/(k))`

Answer

A

Explanation

from energy method
`E =(1)/(2)mv^(2) +(1)/(2)I omega^(2)+(1)/(2)k [(x+delta)^(2)-delta^(2)] -mgx`
=constant Where `I = (MR^(2))/(2), delta = (mg)/(k) =` extension in mean position. `(dE)/(dt) =0`. Use `(dv)/(dt) =a = omega^(2)x, T = (2pi)/(omega)`