The velocity v and displacement x of a particle executing simple harmonic motion are related as
`v (dv)/(dx)= -omega^2 x`.
`At x=0, v=v_0.` Find the velocity v when the displacement becomes x.

Answer

B

Explanation

Given, `upsilon(dupsilon)/(dx)=-omega^(2)x`
Integrating it, withing the limits of motion,
we have `int_(upsilon_(0))^(upsilon)upsilondupsilon=int_(0)^(x)-omega^(2)xdx`
or `((upsilon^(2))/(2))_(upsilon_(0))^(upsilon)=-omega^(2)((x^(2))/(2))_(0)^(x)`
or `upsilon^(2)-upsilon_(0)^(2)=-omega^(2)x^(2) or upsilon=sqrt(upsilon_(0)^(2)-omega^(2)x^(2))`