A particle performing `S.H.M.` undergoes displacement of `(A)/(2)` (where `A` = amplitude of `S.H.M.`) in one second. At `t = 0` the particle was located ast either extreme position or mean posdition. The time period of `S.H.M.` can be : (consider all possible cases)

Options

A. `12s`
B. `2.4`
C. `6s`
D. `1.2s`

Answer

A, B, C, D

Explanation

It `T` be the time period`,` time to fo from `O` to `Q` is `(T)/(12)` and from `M` to `P` is `(T)/(6)`

The displacement is `(A)?(2)` when particle goes from `O` to `Q`, from `O` to `N` to `Q`.
from `O` to `N` to `O` to `P`, then so on
`:. t=(T)/(12)`or `t=(T)/(4)+(T)/(6)=(5T)/(12)`
or `t=(T)/(2)+(T)/(12)=(7T)/(12)`
Hence possible time period `T` is
`T=12s ` or `T=(12xx1)/(5) 2.4 s ` or `T=(12xx1)/(7)s`
similarly displacements is `(A)/(2)` when particle goes from `M` to `P` or `M` to `N` to `P`
Hence the possible time period `T` is
`T=1xx6=6s` or `T=(6xx1)/(5)s=1.2s`
`Ans. T=1.2s,6s,2.4s,12s`