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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2019 · 10 April · Morning Shift · Physics · Oscillations

The displacement of a damped harmonic oscillator is given by x(t) =...

The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos⁡(10\pit+ϕ).x\left(t\right) = {e}^{- 0.1t}\cos \left(10\pit + \phi\right) . Here tt is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

Options

  1. A27s27s
  2. B4s4s
  3. C13s13s
  4. D7s7s

Correct answer

D. 7s7s

Step-by-step solution

Amplitude A=A0e−ktA = {A}_{0}{e}^{- kt}
From given condition, A02=A0e−0.1×t\dfrac{{A}_{0}}{2} = {A}_{0}{e}^{- 0.1 \times t}
⇒ln⁡2=0.1×t⇒ \ln 2 = 0.1 \times t
t=ln⁡20.1=0.6930.10=6.935≈7st = \dfrac{\ln 2}{0.1} = \dfrac{0.693}{0.10} = 6.935 \approx 7s
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