A block of mass $100 \mathrm{~g}$ is connected to an elastic spring of spring constant $450…
A block of mass $100 \mathrm{~g}$ is connected to an elastic spring of spring constant $450 \mathrm{Nm}^{-1}$ oscillates vertically. The block-spring system is in viscous surrounding medium with a damping constant $69.3 \mathrm{~g} \mathrm{~s}^{-1}$. The time in which the amplitude of oscillations drop to half of its initial value.
(take, $\ln 2=0.693)$
$6.93 \mathrm{~s}$
$2 \mathrm{~s}$
$20 \mathrm{~s}$
$69.3 \mathrm{~s}$
Solution
1 ) Amplitude of a damped oscillator varies with time as
$
A=A_{0 .} e^{-\alpha t}
$
Here, $\alpha=b / 2 m, b=$ damping constant and $m=$ mass of oscillator.
Here, given $A=A_0 / 2$
So, from eq. (i) we have
$
\begin{aligned}
& \frac{A_0}{2}=A_0 e^{-\alpha t} \\
& \Rightarrow \quad \frac{1}{2}=e^{-\alpha t} \\
& \Rightarrow \quad \log \left(\frac{1}{2}\right)=\log \left(e^{-\alpha t}\right) \Rightarrow-\log 2=-\alpha t \\
& \text { or } \\
& \log 2=\alpha t \\
& \alpha=b / 2 m \text { where， } \\
& b=69.3 \mathrm{gs}^{-1} \text { and } m=100 \mathrm{~g} \\
&
\end{aligned}
$
Now,
$
\alpha=b / 2 m \text { where, }
$
and $\log 2=0.693$ (given)
Substituting given values in eq. (ii), we have
$
\begin{aligned}
& 0.693=\frac{69.3}{2 \times 100} \times t \\
\Rightarrow \quad t & =2 \mathrm{~s}
\end{aligned}
$