Answer
$0.102\;kg/s$
Work Step by Step
The amplitude of the damped oscillations is expressed as
$a=x_me^{-\frac{bt}{2m}}$
or, $e^{-\frac{bt}{2m}}=\frac{a}{x_m}$
or, $-\frac{bt}{2m}=\ln{\frac{a}{x_m}}$
or, $\frac{bt}{2m}=\ln{\frac{x_m}{a}}\;............(1)$
Now,
$\omega=\sqrt {\frac{k}{m}}$
$T=\frac{2\pi}{\omega}=2\pi\sqrt {\frac{m}{k}}$
Here, $t=4T=8\pi\sqrt {\frac{m}{k}}$
Now,
$\frac{bt}{2m}=\frac{b}{2m}\times 8\pi\sqrt {\frac{m}{k}}=4b\pi\sqrt {\frac{1}{mk}}$
Substituting in eq. $1$, we obtain
$4b\pi\sqrt {\frac{1}{mk}}=\ln{\frac{x_m}{a}}$
or, $b=\frac{\sqrt {mk}}{4\pi}\times\ln{\frac{x_m}{a}}$
Substituting the given values,
$b=\frac{\sqrt {2\times10}}{4\times\pi}\times\ln{\frac{4}{3}}$
or, $\boxed{b=0.102\;kg/s}$
Therefore, the value of $b$ is $0.102\;kg/s$
