A block of mass $200\, g$ executing $SHM$ under the influence of a spring of spring constant $K=90\, N\,m^{-1}$ and a damping constant $b=40\, g\,s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $ln\,\frac{1}{2} = -0.693$)
Medium
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$\mathrm{X}=\mathrm{X}_{0} \mathrm{e}^{-\left(\frac{\mathrm{b}}{2 \mathrm{m}}\right)t}$
$\Rightarrow \frac{\mathrm{X}_{0}}{2}=\mathrm{X}_{0} \mathrm{e}^{-\left(\frac{40}{2\times 200}\right)t}$
$\Rightarrow \frac{1}{2}=\mathrm{e}^{\frac{1}{10} \mathrm{t}}$
$\Rightarrow \log \left(\frac{1}{2}\right)=-\frac{1}{10} \mathrm{t} \Rightarrow \mathrm{t}=10 \times 0.693=7 \mathrm{s}$
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