JEE Main202116 Mar 2021Evening ShiftPhysicsOscillationsActual
The amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500 g , Decay constant = 20 g s - 1 then how much time is required for the amplitude of the system to drop to half of its initial value? ln 2 = 0 . 693
Options
- A34 . 65   s
- B17 . 32   s
- C0 . 034   s
- D15 . 1   s
Correct answer
A. 34 . 65   s
Step-by-step solution
A = A 0 e - γ t = A 0 e - b t 2 m A 0 2 = A 0 e - b t 2 m b t 2 m = ln 2 t = 2 m b ln 2 = 2 × 500 × 0 . 693 20 t = 34 . 65 second.