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 $\ln \left(2\right)=0.693$).

• A. $3.3\times {10}^2{\mathrm{kgs}}^{-1}$
• B. $5.7\times {10}^{-3}\mathrm{Kg}{s}^{-1}$
• C. $1.16\times {10}^{-2}{\mathrm{Kgs}}^{-1}$
• D. $0.69\times {10}^2{\mathrm{Kgs}}^{-1}$

Answer 1

The correct option is C

$1.16\times {10}^{-2}{\mathrm{Kgs}}^{-1}$

Step 1: Given data

Mass of a given block, $\mathrm{m}=1\mathrm{Kg}$

Initial amplitude, ${\mathrm{A}}_0=12\mathrm{cm}$

Final amplitude, $\mathrm{A}=6\mathrm{cm}$

Time taken, $\mathrm{t}=2\mathrm{minutes}=120\mathrm{seconds}$

Step 2: Calculating the damping constant

The amplitude for the damped oscillation is given by:

$\mathrm{A}={\mathrm{A}}_0{\left(\mathrm{e}\right)}^{\frac{-\mathrm{b}}{2\mathrm{m}}\mathrm{t}}\left(1\right)$

Here we have, $\mathrm{A}=\mathrm{final}\mathrm{amplitude}$

$$\begin{gathered} {\mathrm{A}}_0=\mathrm{maximum}\mathrm{amplitude} \\ \mathrm{m}=\mathrm{mass}\mathrm{of}\mathrm{a}\mathrm{body} \\ \mathrm{t}=\mathrm{time}\mathrm{taken} \\ \mathrm{b}=\mathrm{damping}\mathrm{constant} \end{gathered}$$

Substitute the given values in equation $\left(1\right)$ we will get,

$$\begin{gathered} \mathrm{A}={\mathrm{A}}_0{\left(\mathrm{e}\right)}^{\frac{-\mathrm{b}}{2\mathrm{m}}\mathrm{t}} \\ \Rightarrow 6=12{\left(\mathrm{e}\right)}^{\frac{-\mathrm{b}}{2\times 1}\times 120} \\ \Rightarrow 6=12{\left(\mathrm{e}\right)}^{-60\mathrm{b}} \\ \Rightarrow \frac{1}{2}={\left(\mathrm{e}\right)}^{-60\mathrm{b}} \\ \Rightarrow \ln \left(\frac{1}{2}\right)=-60\mathrm{b} \\ \Rightarrow \ln \left(2\right)=60\mathrm{b} \\ \Rightarrow 0.693=60\mathrm{b} \\ \Rightarrow \mathrm{b}=\frac{0.693}{60} \\ \Rightarrow \mathrm{b}=0.01155 \\ \Rightarrow \mathrm{b}=1.16\times {10}^{-2}{\mathrm{Kgs}}^{-1} \end{gathered}$$

Hence, option(C) is correct answer.