A particle performs linear S.H.M. at a particular instant, velocity of the particle is ' u ' and acceleration is ' a₁ ' while at another instant velocity is ' V ' and acceleration is ' a ₂ ' ( o a ₁ a ₂ ) . The distance between the two position is

Step-by-step solution

When velocity is u and acceleration is a₁ , let the position of particle be x₁ . When velocity is v and acceleration is a ₂ , let the position of particle be x₂ . If is the angular frequency then, a ₁= ^2 x ₁ and a ₂= ^2 x ₂ a ₁+ a ₂= ^2 ( x ₁+ x ₂ )...(i) Also, velocity of particle at particular instant can be given as, u^2= ^2 A^2- ^2 x₁^2 and v^2= ^2 A^2- ^2 x₂^2 i.e., v^2-u^2= ^2 (x₁^2-x₂^2 )v^2-u^2= ^2 (x₁-x₂ ) (x₁+x₂ )...(ii) from equation (i) we get aligned & v ^2- u ^2 = & ( x ₁- x ₂ ) ( a ₁+ a ₂ ) & x ₁- x