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

A particle performs linear SHM of period (2pi)/(omega) about a centre

1.	A particle performs linear SHM of period (2pi)/(omega) about a centre 'O' and is observed to have a velocity bomegasqrt(3) when at a distance b from O'. If the particle is moving towards the positive extremity at that instant, show that it will travel a further distance b in a time (pi)/(3omega) before coming momentarily to rest.
Answer» Solution :Given `T= (2PI)/(omega), v= bomegasqrt(3)` and x=b substituting in `v= omegasqrt(A^(2)-x^(2))` we get `bomegasqrt(3)= omegasqrt(A^(2)-b^(2))` Squaring , `3b^(2)omega^(2)= omega^(2)(A^(2)-b^(2))` `3b^(2) = A^(2)-b^(2) implies 4b^(2)= A^(2) implies A= +-2b` The time TAKEN (t) to travel from the mean position to a distance b can be found from `x= Asin omegat` we have, `x = b, A = 2b:. b=2b sin omegat` `sinomegat= (1)/(2), omegat= (pi)/(6) implies t= (pi)/(6omega)` `:.` Further time taken to reach the extreme position `= (T)/(4)- (pi)/(6omega)= (2pi)/(4omega)-(pi)/(6omega)= (pi)/(3omega)` It will momentarily come to REST when it reaches the positive extreme position. Further distance travelled = A-b=2b-b=b