A particle execute SHM.If it has velocities V1&V2 at different displacements Y1&Y2then,Angular fequency of oscillation of the particle is??

use the relation v = w(a^2 - y^2)^0.5so that v1 = w(a^2 - y1^2)^0.5 andv2 = w(a^2 - y2^2)^0.5 .squaring both equations and subtracting we get:v1^2 - v2^2 = w^2(y2^2 - y1^2)so that w = {(v1^2 - v2^2)/(y2^2 - y1^2)}^0.5so this is the angular frequency of the shm.please approve.

we know the fact that velocity of particle executing simple harmonic motion is given by the expression v= w (A^2-x^2) ^ 1/2 where w is the angular frequency ,A is the amplitude and x is the displacement. Using this relationship we first calculate A according to the given information and then find w which is what we want and it comes out to be                                            ((V2 ^2 - V1^2)/Y2 ^2- Y1^2)) ^ 1/2.