Two very small particles are executing `SHM` along very near parallel lines about the same mean position, same amplitude and the same frequency. At `t = 0` both the particle are moving with speed of `1m//sec` but on the opposite side of the mean position and having same direction of velocity at a distance of `5cm` from the mean position. At `t = 4sec` they are at the same position moving with speed of `1m//sec` but in opposite direction. Assuming that this is the minimum time form `t = 0` where they are meeting for the `1st` time. If the maximum speed of either particle is `(xpi)/(4sqrt(2))cm//sec` Find `x`

Answer

5

Explanation

`5cm = (A)/(sqrt(2)) rArr A = 5 sqrt(2) cm`
`T = 16 sec rArr omega = ((2pi)/(16))`

`V_(max) = (2pi)/(16) xx5 sqrt(2)`
`V_(max) = (10sqrt(2)pi)/(16) cm//sec`
`= (10sqrt(2))/(8xx2) pi = (5)/(4sqrt(2)) pi`