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JEE Main 202116 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 16 March · Evening Shift · Physics · Oscillations

The amplitude of a mass-spring system, which is executing simple...

The amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass =500g= 500g, Decay constant =20gs−1= 20g{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)\left(\ln 2 = 0 . 693\right)

Options

  1. A34.65s34 . 65s
  2. B17.32s17 . 32s
  3. C0.034s0 . 034s
  4. D15.1s15 . 1s

Correct answer

A. 34.65s34 . 65s

Step-by-step solution

A=A0e−\gammat=A0e−bt2mA = {A}_{0}{e}^{- \gammat} = {A}_{0}{e}^{- \dfrac{bt}{2m}}

A02=A0e−bt2m\dfrac{{A}_{0}}{2} = {A}_{0}{e}^{- \dfrac{bt}{2m}}

bt2m=ln⁡2\dfrac{bt}{2m} = \ln 2

t=2mbln⁡2=2×500×0.69320t = \dfrac{2m}{b}\ln 2 = \dfrac{2 \times 500 \times 0 . 693}{20}

t=34.65t = 34 . 65 second.

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