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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

The displacement of a damped harmonic oscillator is given by \(x\left( t \right) = {e^{ - 0.1\,t}}\,\cos \left( {10\pi t + \varphi } \right)\)  The time taken for its amplitude of vibration to drop to half of its initial value is close to .... \(s\)

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Answer & Solution

Correct Answer

(D) \(7\)

Step-by-step Solution

Detailed explanation

\(A=A_{0} e^{-0.1 t}=\frac{A_{0}}{2}\)

\(\ln 2=0.1 \mathrm{t}\)

\(t=10\, \mathrm{ln} 2=6.93 \approx 7 \mathrm{sec}\)

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