A particle executes SHM with a time period of `4 s`. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

Explanation

`y = A sin (omega t + phi_(0))`
At, `t = 0 " "y = 0" "rArr A sin phi_(0) = 0 rArr phi_(0) = 0`
Hence `y = A sin omega t," ""at"" "y = A//2`
`(A)/(2) = A sin omega t" "rArr" "sin omega t = (1)/(2)`
`rArr" "omega t = (pi)/(6) rArr" "(2pi)/(T).t = (pi)/(6) rArr" "t = (pi)/(6).(T)/(2pi)=(1)/(3)s`