A partical starts SHM frommean position O executing SHM A and B are the two point at which its velocity is zero . It passes through a cartain point O at time `t_(1) = 0.5 and t_(2) = 1.5 s` with a speed of `3 m//s`.

i. The maximum speed .......
ii. ratio `AP//PB` .....

Explanation

`t_(OP) = 0.5 s, t_(OBP) = 1.5 s`
Time taken from p is R is
`(1.5 - 0.5)/(2) = 0.5 s`
Time taken from O to `B = 0.5 + 0.5 = 1 s`
Hence time period `= 4 s`
Let `x = A sin omega t`
`implies omega = (2 pi)/(T)`
`= (pi)/(2)`
`nu = (dx)/(dt) = A omega t`
` implies nu_(max) = A omega`
Given `nu = 3 m//s at t = 1//2 s`.
`3 = A xx (pi)/(2) xx cos ((pi)/(2) xx (1)/(2))`
`implies A = (6sqrt2)/(pi) cm`
`v_(max) = (6sqrt2)/(pi) xx (pi)/(2)3sqrt2 m//s`
`x_(OP) = A sin omega t = (6sqrt2)/(pi) sin ((pi)/(2)) xx (1)/(2) = (A)/(sqrt2)`
`PA = A - (A)/(sqrt2) = (A (sqrt2 -1))/(sqrt2)`
`PB = (A (sqrt2 + 1))/(sqrt2)`