The velocity #v# of a particle executing SHM is related with its displacement #x# from its equilibrium position as follows
#v=omegasqrt(a^2-x^2)......(1)#,
where #a and omega# respectively represent the ammplitude and the angular velocity of the imaginary particle moving in the reference circle associated with the SHM.
Its velocity is mximum when #x=0#, then #v_"max"=omegaa#
Given #v_"max"=100"cm/"s# and amplitude #a =10cm# we get
#omega=v_"max"/a=100/10=10"rad/"s#
We are to find out displacement #x# when velocity #v=50"cm/"s#,Insrting the values in equation (1)
#v=omegasqrt(a^2-x^2)......(1)#
#=>50=10sqrt(10^2-x^2)#
#=>5^2=10^2-x^2#
#=>x^2=100-25=75#
#=>x=sqrt75=pm5sqrt3cm#
So the velocity of the particle will be #50"cm/s"# at a distance #5sqrt3cm# from equilibrium position .#pm# sign represents both sides.