Correct answer is C.
In $2 \mathrm{~s}$ (which is equal to $\frac{T}{4}$ ), one amplitude will be convered. In $1 \mathrm{st}$ second $x=a \sin \left(\frac{\pi}{4}\right)=\frac{a}{\sqrt{2}}$
Required ratio $=\frac{\frac{a}{\sqrt{2}}}{a-\frac{a}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}$