Question #aa53f

1 Answer
P dilip_k
Mar 15, 2017

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.

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