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

Options

A. `(t_(1))/(t_(2)) = 1`
B. `(t_(1))/(t_(2)) = (1)/(2)`
C. `(t_(1))/(t_(2)) = 2`
D. `(t_(1))/(t_(2)) = 3`

Answer

B

Explanation

`y = A sin omega t`
`(A)/(2) = A sin ((2pi)/(4) xx t_(1)) rArr (1)/(2) = sin ((pi)/(2) t_(1)) t_(1) = (1)/(3)`
`t_(2) = (T)/(4)-(1)/(3) = 1 - (1)/(3) = (2)/(3) " "rArr (t_(1))/(t_(2)) = (1)/(2)`