A particle is executing SHM of amplitude `4cm` and time period `12s`. The time taken by the particle in going from its mean position to a position of displacement equal to `2cm` is `T_1`. The time taken from this displaced position to reach the extreme position on the same side is `T_2.T_1//T_2` is

Options

A. `2`
B. `1`
C. `1/2`
D. `1/3`

Answer

C

Explanation

`T_1=(T)/(12)`
`t^'=T/4`
`T_2=T/4-(T)/(12)=T/6`
`T_1//T_2=1//2`