A particle performs linear `SHM` of period `(2pi)/(omega)` about a point 'O' and is observed to have a velocity `b omega sqrt(3)` when at a distance `b` from 'O'. If the particle is moving towards the positive extermity at that instant, show that it will travel a further distance `b` in a time `(pi)/(3 omega)` before coming momentarily to rest.

Explanation

Given `T = (2pi)/(omega), v = b omega sqrt(3) at x = b`
Substituting in `v = omega sqrt(A^(2)-x^(2))`,
we get `b omega sqrt(3) = omega sqrt(A^(2) -b^(2))`
Squaring, `3b^(3) omega^(2) = omega^(2) = (A^(2)-b^(2))`
`3b^(2) = A^(2) - b^(2), 4b^(2) - A^(2) rArr A= +- 2b`
The time taken `(t)` to travel form the mean position to a distance `b` can be found form
`x = A Sin omegat` We have, `x = b, A = 2b`
`:. b = 2b sin omegat, sin omegat = (1)/(2), omegat = (pi)/(6) rArr t (pi)/(6 omega)`
`:.` Further time taken to reach exterme position
`=(T)/(4) -(pi)/(6omega) =(2pi)/(4omega) -(pi)/(6omega) = (pi)/(3omega)`
It will momentrily come to rest when it reaches the positive exterme position.
Further distance travelled `= A -b = 2b -b = b`