Initially $, \frac{a_{0}}{3}=a_{0} e^{-b \times 100 T}, \mathrm{T}=$ time of one oscillation or $\frac{1}{3}=e^{-100 b T} \ldots(i)$
Finally, $a=a_{0} e^{-b \times 200 T}$
or $a=a_{0}\left[e^{-100 b T}\right]^{2}$
or $a=a_{0} \times\left[\frac{1}{3}\right]^{2} \quad[\text { From Eq. (i) }]$
or $a=a_{0} / 9$
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{ m }{ k }}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{ m }{ k }}$.
$(D)$ the time at which the particle passes througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.
