A body is in simple harmonic motion with time period T = 0.5   s and amplitude A = 1   cm . Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Step-by-step solution

∴ x = A sin ⁡ ω t A 2 = A sin ⁡ ω t ω t = π 6 t = π 6 ω = π 6 × 2π t = T 12 ∴ T i m e   t a k e n   t o   r e a c h   f r o m   x = 0   t o   x = A 2   i s T 12 ∴ A v e r a g e   v e l o c i t y = D i s p l a c e m e n t T i m e = A / 2 T / 12 = A T × 6 = 6 × 1 0.5   = 12   c m / s