A particle executing linear S.H.M. has velocities V₁ and V₂ at distance x₁ and x₂ respectively, from the mean position, its angular velocity is

Step-by-step solution

The velocity of a particle executing linear S.H.M. at a distance x from the mean position is given by V = A^2 - x^2 , where A is the amplitude and is the angular velocity. Squaring both sides, we get V^2 = ^2 (A^2 - x^2) . For the two given positions, we can write: V₁^2 = ^2 (A^2 - x₁^2) V₂^2 = ^2 (A^2 - x₂^2) Subtracting the first equation from the second, we get: V₂^2 - V₁^2 = ^2 (A^2 - x₂^2) - ^2 (A^2 - x₁^2) V₂^2 - V₁^2 = ^2 (x₁^2 - x₂^2) ^2 = V₂^2 - V₁^2 x₁^2 - x₂^2 = V₂^2 - V₁^2 x₁^2 - x₂^2 Answer: V₂^2 - V₁^