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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 .... \(s\)
Pick an option to check your answer.
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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