# A block of mass

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 $2=0.693$ )

1. $0.69 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$

2. $3.3 \times 10^{2} \mathrm{~kg} \mathrm{~s} \mathrm{~s}^{-1}$

3. $1.16 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$

4. $5.7 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$

Correct Option:

Solution:

(Bonus)

$\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\gamma \mathrm{t}}$

$\ln 2=\frac{b}{2 m} \times 120$

$\frac{0.693 \times 2 \times 1}{120}=b$

$1.16 \times 10^{-2} \mathrm{~kg} / \mathrm{sec}$