The period of a particle is $8s$. At $t = 0$ it is at the mean position. The ratio of distance covered by the particle in first second and 2nd second will be-

$x=A\sin \omega t$

If $t=1$ second then $x=A\sin \frac{2\pi }{T}\times 1$ or $x_1=A\sin \frac{2\pi }{8}=\frac{A}{\sqrt{2}}$

but in two second it will reach to extreme position, so distance travelled in next second

$x_2=A-\frac{A}{\sqrt{2}}=\frac{\left(\sqrt{2}-1\right)}{\sqrt{2}}$

ratio of distance $\frac{A/\sqrt{2}}{\left(\sqrt{2}-1\right)A/\sqrt{2}}=\frac{1}{\sqrt{2}-1}$