Question:
A block of mass $1 \mathrm{~kg}$ attached to a spring is made to oscillate with an initial amplitude of $12 \mathrm{~cm}$. After 2 minutes the amplitude decreases to $6 \mathrm{~cm}$. Determine the value of the damping constant for this motion. (take In $2=0.693$ )
Correct Option:
Solution:
(Bonus)
$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\gamma \mathrm{t}}$
$\ln 2=\frac{b}{2 m} \times 120$
$\frac{0.693 \times 2 \times 1}{120}=b$
$1.16 \times 10^{-2} \mathrm{~kg} / \mathrm{sec}$
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