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 $a=a_{0} / 9$
$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$
