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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

  1. A34 . 65   s
  2. B17 . 32   s
  3. C0 . 034   s
  4. 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.

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