A partical in SHM ha a period of 4s .It takes time `t_(1)` to start from mean position and reach half the amplitude. In another case it taken a time `t_(2)` to start from mean position and reach half the amplitude. Find the ratio `t_(1)//t_(2)`

Explanation

Remember that where motion starts from origin or the mean position, then `x = a sin ( omega t)` and wherever it starts from exterme position then `x = a cos (omega t)`
Because at `t = 0, x = 0 for x = a sin (omega t)`
and at`t = 0, x = a for x = a cos (omega t)`
`(a)/(2) = a sin ( omega t_(1)), (a)/(2) = a cos ( omega t_(2))`
`sin (omega t_(1)) = (1)/(2),cos (omega t_(2)) = (1)/(2) implies (t_(1))/(t_(2)) = (1)/(2)`
`omega t_(1) = (pi)/(6),omega t_(2) = (pi)/(3)`