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JEE Main 202117 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 17 March · Evening Shift · Physics · Oscillations

A block of mass 1kg attached to a spring is made to oscillate with...

A block of mass 1kg1kg attached to a spring is made to oscillate with an initial amplitude of 12cm12cm. After 22 minutes the amplitude decreases to 6cm6cm. Determine the value of the damping constant for this motion. (take ln⁡2=0.693\ln 2 = 0 . 693 )

Options

  1. A1.16×10−2kgs−11 . 16 \times {10}^{- 2}kg{s}^{- 1}
  2. B3.3×102kgs−13 . 3 \times {10}^{2}kg{s}^{- 1}
  3. C1.16×102kgs−11 . 16 \times {10}^{2}kg{s}^{- 1}
  4. D5.7×10−3kgs−15 . 7 \times {10}^{- 3}kg{s}^{- 1}

Correct answer

A. 1.16×10−2kgs−11 . 16 \times {10}^{- 2}kg{s}^{- 1}

Step-by-step solution

The amplitude of damped oscillation with time tt and damping constant bb is given by,

A=A0e−bt/2mA = {A}_{0}{e}^{- bt / 2m}

⇒ln⁡(A0A)=bt2m⇒ \ln \left(\dfrac{{A}_{0}}{A}\right) = \dfrac{bt}{2m}

⇒ln⁡2=b2m×120⇒ \ln 2 = \dfrac{b}{2m} \times 120; here t=2min=120st = 2min = 120s

⇒0.693×2×1120=b⇒ \dfrac{0 . 693 \times 2 \times 1}{120} = b

⇒b=1.16×10−2kgs−1⇒ b = 1 . 16 \times {10}^{- 2}kg{s}^{- 1}.

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