The equations representing simple harmonic motion in terms of displacement is y = A sinωt and velocity is dy/dt = -A ω cos ωt = ω (A² – y²)^1/2 . The acceleration is given by Accelaration = - ω²y

We know that the velocity is maximum at the mean position and minimum at the extreme position.

 Given, Max velocity = v₀ and this occurs at y =0. dy/dt = ω (A² – y²)^1/2  becomes v₀ = ω (A² – 0²)^1/2  = ωA -------Eqn(1)

 Case 1: Finding y at v = v₀ / 2

            dy/dt = ω (A² – y²)^1/2  becomes v₀ / 2 = ω (A² – y²)^1/2 

            Put v₀ = ωA from Eqn (1) and simplify by squaring on both sides, we get A² = 4A² – 4y²

            Hence, y = A √3/4

Case 2:Finding dy/dt at y = A / 2

            dy/dt = ω (A² – y²)^1/2  becomes dy/dt = ω (A² – A²/4)^1/2 

            Hence, dy/dt = A ω √3/4