The displacement of a damped harmonic oscillator is given by $x(\mathrm{t})=\mathrm{e}^{-0.1 \mathrm{t}}…
The displacement of a damped harmonic oscillator is given by $x(\mathrm{t})=\mathrm{e}^{-0.1 \mathrm{t}} \cos (10 \pi \mathrm{t}+\varphi)$. Here $\mathrm{t}$ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
$27 \mathrm{~s}$
$4 \mathrm{~s}$
$13 \mathrm{~s}$
$7 \mathrm{~s}$
Solution
Amplitude $A=A_0 e^{-k t}$
From given condition, $\frac{A_0}{2}=A_0 e^{-0.1 \times t}$
$
\begin{aligned}
& \Rightarrow \ln 2=0.1 \times t \\
& t=\frac{\ln 2}{0.1}=\frac{0.693}{0.10}=6.935 \approx 7 \mathrm{~s}
\end{aligned}
$