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Question: Answered & Verified by Expert
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 ln2=0.693 )
PhysicsOscillationsJEE MainJEE Main 2021 (17 Mar Shift 2)
Options:
  • A 1.16×10-2 kg s-1
  • B 3.3×102 kg s-1
  • C 1.16×102 kg s-1
  • D 5.7×10-3 kg s-1
Solution:
2064 Upvotes Verified Answer
The correct answer is: 1.16×10-2 kg s-1

The amplitude of damped oscillation with time t and damping constant b is given by,

A=A0e-bt/2m

lnA0A=bt2m

ln2=b2m×120; here t=2 min=120 s

0.693×2×1120=b

b=1.16×10-2 kg s-1.

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