The mechanical energy of a damped oscillator becomes half of its initial energy in 4 seconds. In another $t$ seconds its mechanical energy becomes $12.5 \%$ of its initial mechanical energy. Then $t=$

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { at } \mathrm{t}=4, \mathrm{E}=\frac{\mathrm{E}_0}{2} \ & E=E_0 e^{-b t / m} \Rightarrow \frac{E_0}{2}=E_0 e^{-\frac{4 b}{m}} \ & \ln 2=\frac{4 b}{m} \ & \frac{\mathrm{b}}{\mathrm{m}}=\frac{\ln 2}{4} \ & 	ext { at } \mathrm{t}, \mathrm{E}=12.5 \% \mathrm{E}_0 \ & \frac{12.5}{100} \mathrm{E}_0=\mathrm{E}_0 \mathrm{e}^{-\frac{\mathrm{bt}}{\mathrm{m}}} \Rightarrow \frac{1000}{125}=\mathrm{e} \frac{\mathrm{bt}}{\mathrm{m}} \ & (2)^2=e \frac{b t}{m} \ & \frac{\mathrm{bt}}{\mathrm{m}}=2 \ln ^2 \ & 	ext { from equation (1), we have } \ & \frac{\ln 2}{4} \mathrm{t}=2 \ln 2 \ & 	herefore \mathrm{t}=8 \mathrm{~s} \ & \end{aligned}$

The mechanical energy of a damped oscillator becomes half of its initial energy in 4 seconds. In another $t$…

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