# Question #aa53f

Mar 15, 2017

The velocity $v$ of a particle executing SHM is related with its displacement $x$ from its equilibrium position as follows

$v = \omega \sqrt{{a}^{2} - {x}^{2}} \ldots \ldots \left(1\right)$,

where $a \mathmr{and} \omega$ 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 ${v}_{\text{max}} = \omega a$

Given ${v}_{\text{max"=100"cm/}} s$ and amplitude $a = 10 c m$ we get

$\omega = {v}_{\text{max"/a=100/10=10"rad/}} s$

We are to find out displacement $x$ when velocity $v = 50 \text{cm/} s$,Insrting the values in equation (1)

$v = \omega \sqrt{{a}^{2} - {x}^{2}} \ldots \ldots \left(1\right)$

$\implies 50 = 10 \sqrt{{10}^{2} - {x}^{2}}$

$\implies {5}^{2} = {10}^{2} - {x}^{2}$

$\implies {x}^{2} = 100 - 25 = 75$

$\implies x = \sqrt{75} = \pm 5 \sqrt{3} c m$

So the velocity of the particle will be $50 \text{cm/s}$ at a distance $5 \sqrt{3} c m$ from equilibrium position .$\pm$ sign represents both sides.