Oscillations

JEE Main level, single correct, medium

172 of 194

Question 172 of 194

The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos⁡⁡(10πt+ϕ).x\left(t\right)={e}^{-0.1t}\cos ⁡\left(10\pi t+\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:

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Answer: (D) 7s7s

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}
⇒ln2=0.1×t\Rightarrow ln2=0.1\times t
t=ln20.1=0.6930.10=6.935≈7st=\dfrac{ln2}{0.1}=\dfrac{0.693}{0.10}=6.935\approx 7 s

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