JEE Main202117 Mar 2021Evening ShiftPhysicsOscillationsActual
A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm . After 2 minutes the amplitude decreases to 6 cm . Determine the value of the damping constant for this motion. (take ln 2 = 0 . 693 )
Options
- A1 . 16 × 10 - 2   kg   s - 1
- B3 . 3 × 10 2   kg   s - 1
- C1 . 16 × 10 2   kg   s - 1
- D5 . 7 × 10 - 3   kg   s - 1
Correct answer
A. 1 . 16 × 10 - 2   kg   s - 1
Step-by-step solution
The amplitude of damped oscillation with time t and damping constant b is given by, A = A 0 e - b t / 2 m ⇒ ln A 0 A = b t 2 m ⇒ ln 2 = b 2 m × 120 ; here t = 2   min = 120   s ⇒ 0 . 693 × 2 × 1 120 = b ⇒ b = 1 . 16 × 10 - 2   kg   s - 1 .