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JEE Mains · Physics · STD 11 - 13. oscillations
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 .... \(s\)
- A \(13\)
- B \(27\)
- C \(4\)
- D \(7\)
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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