A block of mass 0 . 1 kg is connected to an elastic spring of spring constant 640 N m - 1 and oscillates in a damping medium of damping constant 10 - 2 kg s - 1 . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to-

Step-by-step solution

Given,   E = E o 2   ⇒ A = A o 2 A m p l i t u d e   o f   d a m p e d   o s c i l l a t i o n = A = A o e - b t 2 m ⇒ A o 2 = A o e - b t 2 m 2 = e b t 2 m   ln⁡ 2 = b t 2m = 10 - 2 2×0.1 t = 0.1 t 2 t 2 = 10 ln ⁡ 2 = 10 × 1 2 × 0.693 t = 10 × 0.693 = 6.93 t ≈ 7   s .