Time period of a particle executing $SHM$ is $8\, sec.$ At $t = 0$ it is at the mean position. The ratio of the distance covered by the particle in the $1^{st}$ second to the $2^{nd}$ second is :
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$x=A \sin \omega t$
$x(t=1 s)=A \sin \left(\frac{2 \pi}{T} t\right)=A \sin \left(\frac{2 \pi}{8}\right)=A \sin \left(\frac{\pi}{4}\right)=\frac{A}{\sqrt{2}}$
$x(t=2 s)=A \sin \left(\frac{2 \pi}{T} t\right)=A \sin \left(\frac{2 \pi}{8} \times 2\right)=A \sin \left(\frac{\pi}{2}\right)=A$
And the required ratio is$:$
$\frac{x(t=1 s)}{x(t=2 s)-x(t=1 s)}=\frac{\frac{A}{\sqrt{2}}}{A-\frac{A}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$
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