A block of mass $0.1\, kg$ is connected to an elastic spring of spring constant $640\, Nm^{-1}$ and oscillates in a damping medium of damping constant $10^{-2}\, kg\,s^{-1}$ . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to ..... $s$
JEE MAIN 2017, Diffcult
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Since system dissipates its energy gradually, and hence amplitude will also decreases with time according to
$a=a_{0} \mathrm{e}^{-\mathrm{bt/m}}$ $...(i)$
Energy of vibration drop to half of its in itial value $\left(\mathrm{E}_{0}\right),$ as $\mathrm{E} \propto \mathrm{a}^{2} \Rightarrow \mathrm{a} \propto \sqrt{\mathrm{E}}$
$a=\frac{a_{0}}{\sqrt{2}} \Rightarrow \frac{b t}{m}=\frac{10^{-2} t}{0.1}=\frac{t}{10}$
From $eq^n(i),$
$\frac{a_{0}}{\sqrt{2}}=a_{0} e^{-t / 10}$
$\frac{1}{\sqrt{2}}=e^{-t / 10}$ or $\sqrt{2}=e^{\frac{t}{10}}$
$\ln \sqrt{2}=\frac{t}{10} \quad \therefore t=3.5$ seconds
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