Time period of a particle executing SHM is 8 s. At t = 0, it is at the mean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is

$\text{ Here }x=A\sin \omega t$

$x_1=A\sin \frac{2\pi }{8}\times 1=A\sin \frac{\pi }{4}=\frac{A}{\sqrt{2}}$

$\text{ and }{x}_2=A\sin \frac{2\pi }{8}\times 2=A$

Therefore, the distance traveled in 2nd second is

$\begin{array}{l}{x}_2^{\mathrm{\prime }}=x_2-x_1=A-\frac{A}{\sqrt{2}}=\frac{(\sqrt{2}-1)A}{\sqrt{2}}\\ \therefore \text{ Ratio }=\frac{x_1}{x_2^{\mathrm{\prime }}}=\frac{A/\sqrt{2}}{\frac{(\sqrt{2}-1)A}{\sqrt{2}}}=\frac{1}{(\sqrt{2}-1)}\\ =\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{(\sqrt{2}+1)}{2-1}=(\sqrt{2}+1)\end{array}$