A particle performs linear SHM of period (2pi)/(omega) about a centre 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 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 b omega sqrt(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=b omega sqrt(3)` and `x=b` SUBSTITUTING in `v=omega(A^(2)-x^(2))` we get `b omega sqrt(3)=omegasqrt(A^(2)-b^(2))` squaring `3b^(2)omega^(2)=omega^(2)(A^(2)-b^(2))` `3b^(2)=A^(2)-b^(2)implies4b^(2)=A^(2)=A=+-2b` The time taken (t) to travel from the mean position to a distance b can be found from `x=A sin omegat` We have `x=b,A=2b"":. b=2b sin omegat` `sin omegat=1/2,omegat=(pi)/6impliest=(pi)/(6 omega)` `:.` 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. Furtherdistance travelled `=A-b=2b-b=b`