Question

Question #aa53f

Answer 1

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.