Question

A block of mass $1\text{kg}$ attached to a spring is made to oscillate with an initial amplitude of $12\text{cm}$. After $2$ minutes the amplitude decreases to $6\text{cm}$. Determine the value of the damping constant for this motion. (take $ln2=0.693$).

• A. $0.69\times {10}^2{\text{kg s}}^{-1}$
• B. $3.3\times {10}^2{\text{kg s}}^{-1}$
• C. $5.7\times {10}^{-3}{\text{kg s}}^{-1}$
• D. $1.16\times {10}^{-2}{\text{kg s}}^{-1}$

Answer 1

The correct option is D $1.16\times {10}^{-2}{\text{kg s}}^{-1}$

Amplitude of a damped oscillation is given by:$$A=A_0{e}^{\frac{-b}{2m}t}$$$\Rightarrow 6=12e^{\frac{-b}{2\times 1}\times 120}$

$\Rightarrow 6=12e^{-b\times 60}$

$\Rightarrow \frac{1}{2}=e^{-60b}$

$\Rightarrow ln\left(2\right)=60b$

$\Rightarrow b=\frac{ln\left(2\right)}{60}=1.16\times {10}^{-2}\text{kg/s}$